PHYS202 –SPRING 2009jdshaw/PHYS202/Lecture/Notes...18.28 –Four point charges have the same magnitude of 2.4 x 10‐12 C and are fixed to the corners of a square that is 4.0 cm

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PHYS202 ndash SPRING 2009Lecture notes ‐ Electrostatics

What is Electricity

bull Static effects known since ancient times

bull Static charges can be made by rubbing certain materials together

bull Described by Benjamin Franklin as a fluidndash Excess of the fluid was described as positive

ndash Deficit of the fluid was described as negative

bull Like charges repel unlike charges attract

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A Quantitative Approach

bull Charles Augustin de Coulomb first described relationship of force and charges

bull Coulomb invented the torsion balance to measure the force between charges

bull Basic unit of charge in SI system is called the Coulomb

bull One of the basic irreducible measurements

ndash Like mass length and time

Fundamental Charge

bull Due to atomic structure

bull Positively charged heavy nuclei and negative charged electrons

bull Charged objects have excess or deficit of electrons

bull Protons and electrons have equal and oppositecharge e = 1602 x 10‐19 C

bull Rubbing some objects causes electrons tomove from one object to another

bull Negatively charged objects have an excessof electrons

bull Positively charged objects have a deficit ofelectrons

bull Net change in charge in a closed system mustbe zero

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Fundamental Charge

Problem 186 ‐Water has a mass per molecule of 18 gmol and each molecule has 10 electrons How many electrons are in one liter of water

What is the net charge of the electrons

23 3

26

10electrons 6022times10 molecules 1mol 1g 10 mltimes times1molecule 1mol 18g 1ml 1liter

electrons33 10liter

N

N

= times times

= times

( )( )26 19

7

33 10 electrons 1602 10 C electron

54 10 C

q Ne

q

minus= = times minus times

= minus times

Conservation of Charge

bull Net change in charge in any closed system is zero

bull Or the change of charge in any volume equals the net charge entering and leaving the system

bull We denote the amount of charge by the letter Q or q

bull If Charge is created it must be equal and opposite

( ) ( ) ( ) ( )Charge Original Charge Charge in Charge out= + minus

0 in out= + -q q q q

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Two identical metal spheres of charge 50 C and ‐16 C are brought together and then separated What is the charge on each sphere

Since the charge is distributed equally each must have an equal charge

1 2 = 34Ctotalq q q= +

1 2 = 17C2totalqq q= =

Conductors and Insulators

bull Insulatorndash Charge stays where it is depositedndash Wood mica Teflon rubber

bull Conductorndash Charge moves ldquofreelyrdquo within the objectndash Metals exhibit high conductivity

bull Due to atomic structurendash Is the outermost electron(s) tightly bound

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Charging Objects

bull By frictionndash Rubbing different materialsndash Example fur and ebonite

bull By contactndash Bringing a charged object into contact with another (charged or uncharged) object

bull By inductionndash Bringing a charged object near a conductor temporally causes opposite charge to flow towards the charged object

ndash Grounding that conductor ldquodrainsrdquo the excess charge leaving the conductor oppositely charged

Force Between Charges

bull Coulomb found that the force between two charges is proportional to the product of their charges and inversely proportion to the square of the distance between them

Where q1 and q2 represent the charges and r12 the distance between them

1 22

12

q qFr

prop

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Coulombrsquos Law

bull By using the torsion balance Coulomb discovered the constant of proportionality k = 899 x 109 Nm2C2

bull If both q1 and q2 are both positive the force is positive

bull If both q1 and q2 are both negative the force is positive

bull If both q1 is positive and q2 is negative the force is negative

bull Hence the sign of the force is positive for repulsion and negative for attraction

1 22

12

q qF kr

=+ +

‐ ‐

‐ +

Coulombrsquos LawPrinciple of Superposition

bull Electric force obeys the law of linear superposition ndash the net force on a charge is the vector sum of the forces due to each charge

bull Add the vectors

The net force on charge4 is F14+F24+F34=Fnet

2î j

i iji j ij

q qk

rne

= sumF r

Red are positive chargesBlue are negative charges

1

2

3

4F14

F24

F34

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Coulombrsquos Law Principle of Superposition

bull The forces add for as many charges as you have

2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4

5

6

7

8



2î j

i iji j ij

q qk

rne

= sumF r

1

2

3

4

5

6

7

8

The net force on charge 5 is thevector sum of all the forces oncharge 5


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Coulombrsquos Law1814 ndash Two tiny conducting spheres are identical and carry charges of ‐200 μC and

+500 μC They are separated by a distance of 250 cm (a) What is the magnitude of the force that each sphere experiences and is the force attractive or repulsive

The net force is negative so that the force is attractive

(b) The spheres are brought into contact and then separated to a distance of 250 cm Determine the magnitude of the force that each sphere now experiences and state whether the force is attractive or repulsiveSince the spheres are identical the charge on each after being separated is one‐half the net charge so q1 + q2 = +15μC

The force is positive so it is repulsive

( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus

times sdot minus times times= =

times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus

times sdot times times= = = times

times

Coulombrsquos Law1821 ndash An electrically neutral model airplane is flying in a horizontal circle on a 30 m

guideline which is nearly parallel to the ground The line breaks when the kinetic energy of the plane is 500 J Reconsider the same situation except that now there is a point charge of +q on the plane and a point charge of minusq at the other end of the guideline In this case the line breaks when the kinetic energy of the plane is 518 J Find the magnitude of the charges

From the definition of kinetic energy we see that mv2 = 2(KE) so that Equation 53 for the centripetal force becomes

So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal

Centripetal forceforce

2 KE 2 KE1 2

k qT T

r rr+ = =

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Coulombrsquos Law1821 ndash (cont)

Subtracting Equation (2) from Equation (1) eliminates Tmax and gives

Solving for |q| gives

( ) ( )2charged uncharged

2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=

( ) ( ) ( ) ( )charged uncharged ndash59 2 2

2 KE ndash KE 2 30 m 518 J ndash 500 J35 10 C

899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

The Electric Fieldbull Electric charges create and exert and experience the electric force

bull We define the Electric Field as the force a test charge would feel if placed at that point

bull The electric field exists around all charged objects

2 2ˆ î j j

i ij i i iji j i jij ij

q q qk q k

r rne ne

rArr= =sum sumF r F r

2ˆj

i j

qk q

rne

= rArr =sumE r F E

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Electric Force Lines

bull Michael Faraday saw the force caused by one charge to be from lines emanating from the charges

bull Lines begin on Positive charges and end on Negative charges flowing smoothly through space never crossing


bull Faraday imagined these lines and the force was greater where the lines are denser weaker where they are spread out

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Electric Field Lines

bull Go out from positive charges and into negative charges

Positive Negative Neutral

The Inverse Square Law

bull Many laws based on the geometry of three‐dimensional space

1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

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The Electric Field

1828 ndash Four point charges have the same magnitude of 24 x 10‐12 C and are fixed to the corners of a square that is 40 cm on a side Three of the charges are positive and one is negative Determine the magnitude of the net electric field that exists at the center of the square

A

B C

D-

++

+

EA

EC

EB

ED

The drawing at the right shows each of the field contributions at the center of the square (see black dot) Each is directed along a diagonal of the square Note that ED and EB point in opposite directions and therefore cancel since they have the same magnitude In contrast EA and EC point in the same direction toward corner A and therefore combine to give a net field that is twice the magnitude of EA or EC In other words the net field at the center of the square is given by the following vector equation

2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A

E E E E EE E E EE EE

The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED

The magnitude of the net field is

In this result r is the distance from a corner to the center of the square which is one half of the diagonal distance d Using L for the length of a side of the square and taking advantage of the Pythagorean theorem we have With this substitution for r the magnitude of the net field becomes

A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =

times sdot timesΣ =

Σ =

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1834 ndash Review Conceptual Example 12 before attempting to work this problem The magnitude of each of the charges in Figure 18‐21 is 86 x 10‐12 C The lengths of the sides of the rectangles are 300 cm and 500 cm Find the magnitude of the electric field at the center of the rectangle in Figures 18‐21a and b

1 2

34

1

4

+ q

+ q + q

- qThe figure at the right shows the configuration given in text Figure 1821a The electric field at the center of the rectangle is the resultant of the electric fields at the center due to each of the four charges As discussed in Conceptual Example 11 the magnitudes of the electric field at the center due to each of the four charges are equal However the fields produced by the charges in corners 1 and 3 are in opposite directions Since they have the same magnitudes they combine to give zero resultant The fields produced by the charges in corners 2 and 4 point in the same direction (toward corner 2) Thus EC = EC2 + EC4 where EC is the magnitude of the electric field at the center of the rectangle and EC2 and EC4 are the magnitudes of the electric field at the center due to the charges in corners 2 and 4 respectively Since both EC2 and EC4 have the same magnitude we have EC = 2 EC2

Electric Field Lines1834 ndash (cont)

The distance r from any of the charges to the center of the rectangle can be found using the Pythagorean theorem

The figure at the right shows the configuration given in text Figure 1821b All four charges contribute a non‐zero component to the electric field at the center of the rectangle As discussed in Conceptual Example 11 the contribution from the charges in corners 2 and 4 point toward corner 2 and the contribution from the charges in corners 1 and 3 point toward corner 1

1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =

2Therefore 292 cm 292 10 m2dr minus= = = times

9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus

minustimes sdot minus times

= = = = minus timestimes

1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

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Notice also the magnitudes of E24 and E13 are equal and from the first part of this problem we know that

E24 = E13 = 181 times 102 NC The electric field at the center of the rectangle is the vector sum of E24 and E13 The x components of E24 and E13 are equal in magnitude and opposite in direction hence

(E13)x ndash (E24)x = 0

Therefore

From Figure 2 we have that

1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =

( ) ( )( )2 2C 132 sin 2 181 10 NC 0858 311 10 NCE E θ= = times = times

The Electric Field


bull This is shown byelectric fieldlines

bull Constructedidentically to theforce lines

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Drawing Electric Field Lines

bull Number of lines at a charge is proportional to the charge


bull The charge with ‐3qmust have three times as many lines as the charge with +q and 15 times the lines for +‐ 2q

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Parallel Plates of Charge

bull Adding fields from each point on surface causes field lines parallel to the plates to cancel

bull Away from the edges the field is parallel and perpendicular to the plate(s)

PositivelyCharged

NegativelyCharged

Electric Field amp Conductors

bull In conductors excess electrons will move when an electric field is present

bull For ldquosteady staterdquo (ie no moving charges) the net electric field must be zero

bull Therefore the charges move to the surface(s) of the conductor until they cancel the electric field inside the conductor

bull Field lines can start and end on a conductor at right angles to the surface of the conductor

0q rArr= = =F E F E

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1844 ndash Two particles are in a uniform electric field whose value is + 2500 NC The mass and charge of particle 1 are m1 = 14 x 10‐5 kg and q1 = ‐70 μC while the corresponding values for particle 2 are m2 = 26 x 10‐5 kg and q2 = +18 μC Initially the particles are at rest The particles are both located on the same electric field line but are separated from each other by a distance d When released they accelerate but always remain at this same distance from each other Find d

The drawing shows the two particles in the electric field Ex They are separated by a distance d If the particles are to stay at the same distance from each other after being released they must have the same acceleration so ax1 = ax2 According to Newtonrsquos second law the acceleration ax of each particle is equal to the net force ΣFx acting on it divided by its mass m or

Ex

d

+x

q1 = minus70 μC

m1 = 14 times 10minus5 kg

q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)

Note that particle 1 must be to the left of particle 2 If particle 1 were to the right of particle 2 the particles would accelerate in opposite directions and the distance between them would not remain constant

Note the signs indicated the direction of the vector pointing from charge 2 to charge 1 for the force on charge 1

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

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1844 ndash (cont)

The acceleration of each charge must be the same if the distance between the charges remains unchanged

Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus

Electric Fluxbull Flux (from Latin ndash meaning to flow)

bull Electric flux measures amount of electric field passing through a surface

E EAΦ =

E

Flux through surface is Area of surface times the field flowing through that surface

E

So to calculate find the normal to the surface

Then find the angle between the field and that normal vector φ

( )cosE E AφΦ =

φ

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Electric Flux

( )cosE E AφΦ =

E



This can be expressed as the vector dot product

EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux

1848 ndash A rectangular surface (016 m x 038m) is oriented in a uniform electric field of 580 NC What is the maximum possible electric flux through the surface

The maximum possible flux occurs when the electric field is parallel to the normal of the rectangular surface (that is when the angle between the direction of the field and the direction of the normal is zero) Then

( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

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Gaussrsquo Lawbull Electric Flux through a closed surface is proportional to the charge enclosed by that surface

bull The fundamental constant ε0 is called the permittivity of free space It is related to Coulombrsquos law by ε0 = 885 x 10‐12 C2(Nm2)

bull Gauss Law is most useful when the electric field on the entire surface is constant

bull Take a point charge Construct a sphere centered on the point charge (all points on the surface equidistant from the charge) ndash the electric field is constant

bull Since the field is radial it is perpendicular to the concentric spherical surface everywhere so φ = 0 cos φ = 1

bull The direction of the field is radially outward from the charge (or inward towards the charge if q is negative)

0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =

Gaussrsquo Law for a plate of chargebull In this case we take our closed surface to be a box oriented such that two faces are

parallel to the plate of charge and the others are perpendicular to the plate

bull Since the electric field from a plate of charge is perpendicular to the surface of the plate the sides that are perpendicular to the plate have no flux

E

bull The charge enclosed by the box is the surface charge density multiplied by the area the box passes through the plate

bull The area of the box with flux is 2A

Aσ

( )0 0 0

22E

q AE A Eσ σε ε ε

Φ = = =rArr rArr

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Gaussrsquo Law for a line of charge

1854 ndash A long thin straight wire of length L has a positive charge Q distributed uniformly along it Use Gaussrsquo law to find the electric field created by this wire at a radial distance r As the hint suggests we will use a cylinder as our Gaussian surface The electric field must point radially outward from the wire Because of this symmetry the field must be the same strength pointing radially outward on the surface of the cylinder regardless of the point on the cylinder thus we are justified in using Gaussrsquo Law

++++++ ++

Long straight wire Gaussian cylinder

12

3

L

r

End view

GausssiancylinderE


1854 ndash (cont) Let us define the flux through the surface Note that the field is parallel to the endcaps thus the angle between the normal from the surface (which is parallel to the wire) is perpendicular to the field thus φ = π2 so cos(φ)=0 for the ends For the side if the cylinder the flux is

Thus from Gaussrsquo Law

It is customary to refer to charge density λ = QL

End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

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Work

bull Let us take an object with charge q in a constant electric field E

bull If we release this charge the electric field causes a force to move the charge

bull Let us define the field as so there is only one dimension to consider Let us also set the initial velocity and position to be both zero

bull Let the object move through a distance d The equations of motion are

bull So after a time t the object moves

qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ

Conservation of Energy

bull The left hand side is the kinetic energy the particle has after moving a distance d

bull That means the right hand side must represent some form of potential energy the object had prior to being released

bull Types of Energy already covered

ndash Translational kinetic energy

ndash Rotational kinetic energy

ndash Gravitational potential energy

ndash Elastic (spring) potential energy

bull Electric Potential Energy

ndash Cutnell amp Johnson use (EPE) however it is often more convenient to use U

212

m qEd=υ

2 2212

1 )12

(2totalE kx Em h PI g Emω= + + + +υ

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Electric Potential Energy

bull Like gravitational potential energy EPE is defined with respect to the energy at some location

bull Importance lies in the difference between electric potential energy between two points

Example Problem

194 ndash A particle has a charge of +15 μC and moves from point A to point B a distance of 020 m The particle experiences a constant electric force and its motion is along the line of action of the force The difference between the particlersquos electric potential energy at A and B is UA ndash UB = +90 x 10‐4 J Find the magnitude and direction of the electric force that acts on the particle and the magnitude and direction of the electric field that the particle experiences The work done must be equal to the change in potential energy

(from A to B)

But we know that for a particle in an electric field F = qE

( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus

minus times= = = times

Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m

A B A BU U U UqE Ex q x

Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

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Electric Potential

bull The electric potential energy of an object depends on the charge of an object

bull Once again we can find a measurement independent of the objects charge

bull The SI unit for V is measured in Volts ( = JoulesCoulomb)

bull A common unit of energy is the electron Volt

bull Specifically How much potential energy does an electron get if moved through a potential of 1 Volt

( )EPEV

q=

191eV 1602 10 Jminus= times

Electric Potential

bull Like the Electric field the electric potential is independent of any test charge placed to measure a force

bull Unlike the electric field the potential is a scaler not a vector

bull By using calculus methods it can be shown that the electric potential due to a point charge is

bull Note that the value of the electric potential depends on the sign of the charge

kqVr

=

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Electric Potentialfor a point charge

bull Example What was the potential difference due to a charge qA = 100 nC at the distances r0 = 100 cm and r1 = 200 cm

0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus

⎛ ⎞Δ = minus⎜ ⎟

⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V

minus ⎛ ⎞Δ = times sdot minus⎜ ⎟⎝ ⎠

Δ =


bull Example A charge qA is held fixed A second charge qB is placed a distance r0 away and is released What is the velocity of qB when it is a distance r1 away

bull We use the conservation of energy

2 2 20 0 1 1

1 1 12 2 2

Energy m U m U m U= + rArr + = +υ υ υ

( )21 0 1 0 1

12 Bm U U q V V= minus = minusυ

( )1 0 12q V Vm

= minusυ

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bull Find the electric potential at r0 and r1

bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1

22 1 1A A A Bkq kq kq qqm r r m r r⎛ ⎞ ⎛ ⎞

= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ


bull For a single charge we set the potential at infinity to zero

bull So what is the potential energy contained in two charges

bull Bring in the charges from infinity

bull The first charge is free as it does not experience any forces

bull The second moves through the field created by the first

bull So the energy in the system must be

kqVr

=

2

1 1 1

12 12

rU q V V V Vkq kq kqVr r

infinΔ = Δ Δ = minus

Δ = minus =infin

1 2

12

kq qUr

=

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Potential Energyof a collection of charges

bull Letrsquos add a third charge to this group

bull Each successive charge experiences the field (ie potential) created by the previous charges Be careful not to double count

2 1 3 1 3 2

1 21 2

13 23

U q V q V q Vkq kqV Vr r

Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem

1913 ndash An electron and a proton are initially very far apart (effectively an infinite distance apart) They are then brought together to form a hydrogen atom in which the electron orbits the proton at an average distance of 529 x 10‐11 m What is the change in the electric potential energyThe first charge comes in for free the second charge experiences the field of the first so

( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin

times sdot timesΔ = minus times

timesΔ = minus times

352009

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Electric Field and Potential

bull Let a charge be moved by a constant electric field from points A to B (let Δd be that distance)

bull Now we know that the work done must be equal to the negative change in electric potential energy

bull But potential energy is simply

bull Thus

W F d W qE d= Δ rArr = Δ

ABU qE dminusΔ = Δ

ABq V qE dminus Δ = Δ

VEd

Δ= minus

Δ


bull This implies that the field lines are perpendicular to lines (surfaces) of equal potential

bull This equation only gives the component of the electric field along the displacement Δd

bull To truly find the electric vector field you must find the components in each direction

VEd

Δ= minus

Δ

x y zV V VE E Ex y z

Δ Δ Δ= minus = minus = minus

Δ Δ Δ

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bull If the potentials at the following points (x y) areV(11) = 30 V V(091)=15V and V(109)=40 V What is the (approximate) electric field at (11)



Δ Δ Δ

30V 15V 150 V m10m 09mxE minus

= minus = minusminus

30V 40V 100 V m10m 09myE minus

= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =

⎛ ⎞= = minus⎜ ⎟minus⎝ ⎠ x

E

(11)

y

So 34˚ clockwise of ndashx axis or 146˚ ccw of +x axis

Example

Problem 1934 ndash Determine the electric field between each of the points

VEx

Δ= minus

Δ

50V 50V 0 V m020m 00mABE minus

= minus =minus

30V 50V 100 V m040m 020mBCE minus

= minus =minus

10V 30V 50 V m080m 040mCDE minus

= minus =minus

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Equipotential Surfaces

bull Electric field lines are perpendicular to equipotential surfaces

bull Electric field lines are perpendicular to surfaces of conductors

bull Thus surfaces of conductors are equipotential surfaces

bull The interior of conductors have no electric field and are therefore equipotential

bull The Electric potential of a conductor is constant throughout the conductor

Example1968 ndash An electron is released at the negative plate of a

parallel plate capacitor and accelerates to the positive plate (see the drawing) (a) As the electron gains kinetic energy does its electric potential energy increase or decrease Why

(b) The difference in the electronrsquos electric potential energy between the positive and negative plates is Upos ndash Uneg How is this difference related to the charge on the electron (minuse) and to the difference in the electric potential between the platesThe change in the electronrsquos electric potential energy is equal to the charge on the electron (ndashe) times the potential difference between the plates

(c) How is the potential difference related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate

The electric potential energy decreases The electric force F is a conservative force so the total energy (kinetic energy plus electric potential energy) remains constant as the electron moves across the capacitor Thus as the electron accelerates and its kinetic energy increases its electric potential energy decreases

352009

31

Example1968 ndash (cont) (c) How is the potential difference related to

the electric field within the capacitor and the displacement of the positive plate relative to the negative plate

Problem The plates of a parallel plate capacitor are separated by a distance of 12 cm and the electric field within the capacitor has a magnitude of 21 x 106 Vm An electron starts from rest at the negative plate and accelerates to the positive plate What is the kinetic energy of the electron just as the electron reaches the positive plateUse conservation of energy

The electric field E is related to the potential difference between the plates and the displacement Δs by

Note that (Vpos ndash Vneg) and Δs are positive numbers so the electric field is a negative number denoting that it points to the left in the drawing

( )pos negV VE

s

minus= minus

Δ

Total energy at Total energy atpositive plate negative plate

pos pos neg n pos negeg pos negeV eVKE U KE U KE KErArr + = ++ = +

Example1968 ndash (cont)

But the electron starts from rest on the positive plate so the kinetic energy at the positive plate is zero

But so( )pos negV VE

s

minus= minus

Δ

pospos neg negKE KEeV eV+ = +

( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32

Lines of Equal Potential

bull Lines of equal elevation on amap show equal potentialenergy (mgh)

Example

1929 ndash Two equipotential surfaces surround a +15 x 10‐8 C point charge How far is the 190‐V surface from the 750‐V surfaceThe electric potential V at a distance r from a point charge q is V = kqr The potential is the same at all points on a spherical surface whose distance from the charge is r =kqV The radial distance r75 from the charge to the 750‐V equipotential surface is r75 = kqV75 and the distance to the 190‐V equipotentialsurface is r190 = kqV190 The distance between these two surfaces is

( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus

⎛ ⎞minus = minus = minus⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞sdot ⎛ ⎞minus = times + times minus =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

352009

33

Capacitors

bull A capacitor consists of two conductors at different potentials being used to store charge

bull Commonly these take the shape of parallel plates concentric spheres or concentric cylinders but any conductors will do

bull q = CV where C is the capacitance measured in SI units of Farads (Coulomb Volt or C2 J ) V is the voltage difference between the conductors and q is the charge on each conductor (one is positive the other negative)

bull Capacitance is depends only upon the geometry of the conductors

Parallel Plate Capacitor

bull Field in a parallel plate capacitor is

bull But the a constant electric field can be defined in terms of electric potential

bull Substituting into the equation for capacitance

VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors

bull Two concentric spherical shells

bull Cylindrical capacitor has a capacitance perunit length

02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b

πε=⎛ ⎞minus⎜ ⎟⎝ ⎠

Energy stored in a capacitor

bull A battery connected to two plates of a capacitor does work charging the capacitor

bull Each time a small amount of charge is added to the capacitor the charge increases so the voltage between the plates increases

qΔq

qc

q

V

V =qc

352009

35


bull Potential Energy is U = qV

bull We see that adding up each small increase in charge Δq x V is just the area of the triangle

Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

So the energy stored in ANY electric field can be calculated

Ad is the volume of the capacitor So the energy density is

20

12

Uu EAd

κε= =

Example

bull How much energy is stored in a parallel plate capacitor made of 100 cm diameter circles kept 20 mm apart when a potential difference of 120 V is applied

bull Find the capacitance

bull The energy stored is

20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd

πε πminussdot sdot

= = =

9209 10 JU minus= times

352009

36

Electric Polarization in Matter

bull Electric charge distribution in some molecules is not uniform

bull Example Water ndash both Hatoms are on one side

bull In the presence of anElectric Field water willalign opposite to the field


bull In the absence of a field Molecules will be oriented randomly

bull In an electric field they line up opposite to the field

352009

37


bull The resulting electric field is the sum of the external field and the fields generated by the molecules

bull The electric field strength through the matter is reduced

bull The reduction of the field depends on the energy in the molecules the strength of the field each one makes and the density of the molecules

bull This can be measured and the material is said to have a dielectric constant ‐ κ

Dielectrics

bull This dielectric constant changes the value of the permittivity in space (ε)

bull Coulombs law in a dielectric is

bull Where ε is related to ε0 by a constant κ

bull The dielectric constant for a vacuum is κ = 1

bull All other materials have a dielectric constant value greater than 1

bull Water at about 20˚C has κ = 804

bull Air at about 20˚C has κ = 100054

1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics

bull How does this affect the capacitance of parallel plates

0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=

Dielectric Breakdown

bull If the field in the dielectric becomes too strong (ie voltage is too great) the material begins to conduct current ndash this is called dielectric breakdown

bull When this happens the charge flows through the material to the other plate and neutralizes the capacitor

352009

39

Example1957 ndash Two hollow metal spheres are concentric with each other The inner sphere

has a radius of 01500 m and a potential of 850 V The radius of the outer sphere is 01520 m and its potential is 820 V If the region between the spheres is filled with Teflon find the electric energy contained in this spaceThe electric energy stored in the region between the metal spheres is

The volume between the spheres is the difference in the volumes of the spheres

Since E = ΔVΔs and Δs = r2 ndashr1 so E = ΔV(r2 ndashr1)

2 20 0

1 1 ( )2 2

u E E VolumU eκε κεrArr ==

( )3 3 3 32 2 2 2

4 4 4( )3 3 3

Volume r r r rπ π π= minus = minus

( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur

κε π minus⎛ ⎞Δ ⎡ ⎤minus = times⎜ ⎟ ⎢ ⎥minus ⎣ ⎦⎝ ⎠=

352009

2

A Quantitative Approach

bull Charles Augustin de Coulomb first described relationship of force and charges

bull Coulomb invented the torsion balance to measure the force between charges

bull Basic unit of charge in SI system is called the Coulomb

bull One of the basic irreducible measurements

ndash Like mass length and time

Fundamental Charge

bull Due to atomic structure

bull Positively charged heavy nuclei and negative charged electrons

bull Charged objects have excess or deficit of electrons

bull Protons and electrons have equal and oppositecharge e = 1602 x 10‐19 C

bull Rubbing some objects causes electrons tomove from one object to another

bull Negatively charged objects have an excessof electrons

bull Positively charged objects have a deficit ofelectrons

bull Net change in charge in a closed system mustbe zero

352009

3

Fundamental Charge



23 3

26


electrons33 10liter

N

N

= times times

= times

( )( )26 19

7


54 10 C

q Ne

q


= minus times








352009

4










352009

5

Charging Objects








1 22

12

q qFr

prop

352009

6

Coulombrsquos Law






1 22

12

q qF kr

=+ +

‐ ‐

‐ +





2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4F14

F24

F34

352009

7



2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4

5

6

7

8



2î j

i iji j ij

q qk

rne

= sumF r

1

2

3

4

5

6

7

8



352009

8






( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus


times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus


times




So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

3

Fundamental Charge



23 3

26


electrons33 10liter

N

N

= times times

= times

( )( )26 19

7


54 10 C

q Ne

q


= minus times








352009

4










352009

5

Charging Objects








1 22

12

q qFr

prop

352009

6

Coulombrsquos Law






1 22

12

q qF kr

=+ +

‐ ‐

‐ +





2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4F14

F24

F34

352009

7



2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4

5

6

7

8



2î j

i iji j ij

q qk

rne

= sumF r

1

2

3

4

5

6

7

8



352009

8






( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus


times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus


times




So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

4










352009

5

Charging Objects








1 22

12

q qFr

prop

352009

6

Coulombrsquos Law






1 22

12

q qF kr

=+ +

‐ ‐

‐ +





2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4F14

F24

F34

352009

7



2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4

5

6

7

8



2î j

i iji j ij

q qk

rne

= sumF r

1

2

3

4

5

6

7

8



352009

8






( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus


times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus


times




So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

5

Charging Objects








1 22

12

q qFr

prop

352009

6

Coulombrsquos Law






1 22

12

q qF kr

=+ +

‐ ‐

‐ +





2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4F14

F24

F34

352009

7



2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4

5

6

7

8



2î j

i iji j ij

q qk

rne

= sumF r

1

2

3

4

5

6

7

8



352009

8






( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus


times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus


times




So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

6

Coulombrsquos Law






1 22

12

q qF kr

=+ +

‐ ‐

‐ +





2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4F14

F24

F34

352009

7



2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4

5

6

7

8



2î j

i iji j ij

q qk

rne

= sumF r

1

2

3

4

5

6

7

8



352009

8






( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus


times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus


times




So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

7



2î j

i iji j ij

q qk

rne

= sumF r


1

2

3

4

5

6

7

8



2î j

i iji j ij

q qk

rne

= sumF r

1

2

3

4

5

6

7

8



352009

8






( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus


times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus


times




So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

8






( )( )( )( )

9 2 2 6 61 2

22 212

4

899 10 N m C 200 10 C 500 10 C

250 10 m

144 10 N

q qF kr

F

minus minus

minus


times

= minus times

( )( )( )( )

9 2 2 6 631 2

2 2212

899 10 N m C 150 10 C 150 10 C324 10 N

250 10 m

q qF k

r

minus minus

minus


times




So

( )2

c2 KEmF

r r= =υ

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =

352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

9





2

2 KE ndash KEk qrr

⎡ ⎤⎣ ⎦=



899 10 N m C

rq

k

⎡ ⎤⎣ ⎦= = = times

times sdot

( )( )

( )( )

2charged uncharged

max max2Centripetal


2 KE 2 KE1 2

k qT T

r rr+ = =




2 2ˆ î j j


q q qk q k

r rne ne


2ˆj

i j

qk q

rne

= rArr =sumE r F E

352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

10






352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

11






1 22

2 2

1 22

4 4

g

sound light

e

Gm mFr

P LI Ir r

q qF kr

π π

= minus

= =

=

352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

12

The Electric Field


A

B C

D-

++

+

EA

EC

EB

ED


2

Σ = + + +

= + + minus

= +

=

A B C D

A B C B

A C

A


The Electric Field

1828 ndash (cont)

A

B C

D-

++

+

EA

EC

EB

ED



A 22 2k q

E Er

Σ = =

2r L=

( )( )( )

( )

2 2

9 2 2 12

2

42

2

4 899 10 N m C 24 10 C

0040 m

54 NC

k q k qE

LL

E

E

minus

Σ = =


Σ =

352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

13



1 2

34

1

4

+ q

+ q + q





1 2

34

1

4

d

300 cm

500 cm

θ

2 2(300 cm) +(500 cm) 583 cmd = =


9 2 2 1222

2 2 2 22 2(899 10 N m C )( 860 10 C)2 181 10 NC

(292 10 m)C Ckq

E Er

minus



1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

14




(E13)x ndash (E24)x = 0

Therefore


1 2

34

1

4

- q

+ q + q

- q

C

E 13 E 24

C 13 24 13 13( ) ( ) 2( ) 2( )siny y yE E E E E θ= + = =

500 cm 500 cmsin 0858583 cmd

θ = = =


The Electric Field




352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

15





352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

16




PositivelyCharged

NegativelyCharged






0q rArr= = =F E F E

352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

17




Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


x xa F m= Σ


1844 ndash (cont)



Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2 1 1 2

1 21 2 1

11 1

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = minus

minusΣ= =

1 2 2 2 2

1 22 2 2

22 2

x x

xxx

q qF q E k

dq q

q E kF dam m

Σ = +

+Σ= =

352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

18


1844 ndash (cont)


Solving for d

Ex

d

+x

q1 = minus70 μC


q2 = +18 μC


1 2

1 2 1 21 22 2

1 2

x x

x x

a a

q q q qq E k q E k

d dm m

=

minus +=

( )( )1 2 1 2

1 2 2 165 m

x

kq q m md

E q m q m+

= =minus



E EAΦ =

E


E



( )cosE E AφΦ =

φ

352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

19

Electric Flux

( )cosE E AφΦ =

E




EΦ = sdotE A

cos

x x y y z z

E Aor

E A E A E A

φsdot =

sdot = + +

E A

E A

Electric Flux



( ) ( ) ( )2

( cos ) 580 N C (cos 0 ) 016 m 038 m

35 N m C E

E

E AφΦ = = deg

Φ = sdot

352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

20







0

encE

qε

Φ =

0

14

kπε

=

0 0

cosenc encE

q qEA φε ε

Φ =rArr=

( )22

0 0

44

encq qE r Er

πε πε

= rArr =




E



Aσ

( )0 0 0

22E


Φ = = =rArr rArr

352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

21



++++++ ++


12

3

L

r

End view

GausssiancylinderE





End view

GausssiancylinderE

( )2E sideEA E rLπΦ = =++++++ ++


12

3

L

r

( )0 0

0

2

2

encE

q QE rL

QErL

πε ε

πε

Φ = =

=

rArr

02E

rλπε

=

352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

22

Work






qqm

= rArr =F E a E

212

x at at= =υ

Ê=E x

x d=

2122

qE d m qEdm

= rArr =υ υ

21 2 2

d at t d a= rArr =

2 2

t a

d ada a

=

= rArr =

υ

υυ











212

m qEd=υ

2 2212

1 )12


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

23




Example Problem


(from A to B)


( )4

390 10 J 45 10 N020m

A B A B

A B

Work U U F x U U

U UFx

minusminus

= minus Δ = minus


Δ

rArr

( )( )4

36

90 10 J 30 10 N C15 10 C 020 m


Eminus

minus

minus minus= =

Δ Δ

= = timestimes

rArr

times

352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

24

Electric Potential






( )EPEV

q=


Electric Potential





kqVr

=

352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

25



0 10 1

0 1

1 1

A A

A

kq kqV V Vr r

V kqr r

Δ = minus = minus


⎝ ⎠

( )( )9 2 2 5 1 1899 10 N m C 10 C01m 02 m

450V

V

V


Δ =




2 2 20 0 1 1

1 1 12 2 2


( )21 0 1 0 1


( )1 0 12q V Vm

= minusυ

352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

26



bull Or

0 10 1

A Akq kqV Vr r

= =

10 1 0 1


= minus = minus⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

υ

( )1 0 12q V Vm

= minusυ

1 2 q V m= Δυ








kqVr

=

2

1 1 1

12 12



Δ = minus =infin

1 2

12

kq qUr

=

352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

27




2 1 3 1 3 2

1 21 2

13 23


Δ = Δ + Δ + Δ

Δ = Δ =

1 3 2 31 2

12 13 23

kq q kq qkq qUr r r

Δ = + +

Example Problem


( ) ( )( )

2

1 1 1

12 12

9 2 2 1919

11

18

899 10 N m C 1602 10 C1602 10 C

529 10 m435 10 J


U

U

infin

minusminus

minus

minus

Δ = Δ Δ = minus

Δ = minus =infin



352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

28





bull Thus




VEd

Δ= minus

Δ





VEd

Δ= minus

Δ



Δ Δ Δ

352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

29





Δ Δ Δ




= minus =minus

( ) ( )2 2

1

150V 100V 180V

100Vtan 34150VE

E

θ minus

= minus + =


E

(11)

y


Example


VEx

Δ= minus

Δ


= minus =minus


= minus =minus


= minus =minus

352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

30












352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

31






( )pos negV VE

s

minus= minus

Δ






s

minus= minus

Δ


( )

pos neg

pos

neg

neg neg

pos n gneg e

KE

KE

eV eV

eV eV

e V VKE

= +

minus

minus

=

=

( ) ( )( )( )19 6

15

1602 10 C 216 10 V m 0012m

40 10 J

neg

neg

KE

K

s

E

e E minus

minus

minus Δ =

times

= minus times

=

minus times

352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

32



Example


( )

75 19075 190 75 190

29 8

75 190 2

1 1

N m 1 1899 10 150 10 C 11 m750 V 190 VC

kq kqr r kqV V V V

r r minus


⎝ ⎠


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

33

Capacitors









VE V Edd

= minus rArr =

0

qEAε

=

0

0

q CV CEd

Aqq C d CA d

εε

= =

⎛ ⎞= rArr =⎜ ⎟

⎝ ⎠

352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

34

Other Capacitors



02

ln

CbLa

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

041 1

C

a b





qΔq

qc

q

V

V =qc

352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

35




Such that 22

12

12 2

U qV q CV

qU U CVV

= =

= =

qΔq

qc

q

V

V =qc

( )

( )

20

20

1212

AU Edd

U E Ad

κε

κε

⎛ ⎞= ⎜ ⎟⎝ ⎠

=



20

12

Uu EAd

κε= =

Example




20 0A rCd dε ε π

= =

( ) ( ) ( )( )

2 212 2 22 22 0

885 10 C (N m ) 005m 120V12 2 2 0002m

r VU CVd


= = =


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

36








352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

37






Dielectrics








1 22

12

ˆ4q q

rπε=F r

0ε κε=

352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

38

Dielectrics


0ε κε=

0

0

d

d

AACd d

C C

κεε

κ

= =

=




352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


352009

39





2 20 0

1 1 ( )2 2


( )3 3 3 32 2 2 2

4 4 4( )3 3 3


( )2

3 3 80 2 2

2 1

1 4 12 10 J2 3

V r rr

Ur


Documents

PHYS202 –SPRING 2009jdshaw/PHYS202/Lecture/Notes...18.28 –Four point charges have the same magnitude of 2.4 x 10‐12 C and are fixed to the corners of a square that is 4.0 cm