PC1430 - 1-2 (Electric Charge_ Fields_ and Gauss's Law)_14012013111114982

Chapter 1: Electric Charge & Field

Chapter 1: ElectricCharge & Field

Electric Charge

Coulomb’s Law

Electric Fields andCalculations

Electric Dipole

Chap 2: Gauss’s Law

Application of the Law

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Electric Charge


Electric Charge❖Charge’s Conserved❖Conduction❖ Induction❖Uncharged Objects

Coulomb’s Law


Electric Dipole



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Electric charge is conserved



Coulomb’s Law


Electric Dipole



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Structure of a lithium atom

Principle of conservation ofcharge:The algebraic sum of all theelectric charges in any closedsystem is constant.

The magnitude of charge ofthe electron or proton is anatural unit of charge.

Charge is quantized.

Charging by conduction



Coulomb’s Law


Electric Dipole



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Copper is a good conductor ofelectricity; nylon is a goodinsulator.

(a) The copper wire conductscharge between the metal ball andthe charged plastic rod to chargethe ball negatively. Afterward, themetal ball is (b) repelled by anegatively charged plastic rod and(c) attracted to a positivelycharged glass rod.

Charging by induction

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(a) An uncharged metal ball is supported on an insulating stand.(b) A nearby negatively charged rod will cause the free electrons in the metal ball to

shift toward the right. The system reaches an equilibrium state in which the forcetoward the right on an electron, due to the charged rod, is just balanced by the forcetoward the left due to the induced charge.

(c) If you touch one end of a conducting wire to the right surface of the ball and theother end to earth, some of the negative charge flows to the earth.

(d),(e) Disconnecting the wire & removing rod leads to a net positive charge on ball.

Electric forces on uncharged objects

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A charged comb can pick up uncharged bits of paper or plastic with the comb. Thisinteraction is an induced-charge effect.The negatively charged plastic comb causes a slight shifting of charge within themolecules of the neutral insulator (polarization). The positive charges are closer to theplastic comb and so feel an attraction that is stronger than the repulsion felt by thenegative charges, giving a net attractive force.

Coulomb’s Law


Electric Charge

Coulomb’s Law❖Coulomb’s law❖Superposition❖Vector Addition❖Vector Addition 2❖Vector Addition 3


Electric Dipole



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Coulomb’s law


Electric Charge



Electric Dipole



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Coulomb’s law:Magnitude of electric forcebetween two point charges isdirectly / product of charges andinversely proportional to square ofdistance between them.

F D1

4!"0

jq1q2jr2

(1)

14!"0

D 8:988 ! 109 Nm2/C2.

Superposition of forces


Electric Charge



Electric Dipole



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The principle of superposition of forces states that when two ormore charges each exert a force on a charge, the total force onthat charge is the vector sum of the forces exerted by theindividual charges.

Vector addition of electric forces in a plane


Electric Charge



Electric Dipole



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ExampleTwo equal positive point charges q1 D q2 D 2:0 #C are located atx D 0, y D 0:30 m and x D 0, y D "0:30 m, respectively. Whatare the magnitude and direction of the total (net) electric force thatthese charges exert on a third point charge Q D 4:0 #C atx D 0:40 m,y D 0?

Vector addition of electric forces in a plane 2


Electric Charge



Electric Dipole



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SolutionFrom Coulomb’s law the magnitude F of this force is

F1 on Q D 9:0 ! 109 .4:0 ! 10!6C/.2:0 ! 10!6C/

.0:50m/2

D 0:29 N

The angle ˛ is below the x-axis, so the components of this forceare given by

.F1 on Q/x D .F1 on Q/ cos˛ D .0:29/0:40

0:50D 0:23 N

.F1 on Q/y D ".F1 on Q/ sin˛ D ".0:29/0:30

0:50D "0:17 N

Vector addition of electric forces in a plane 3


Electric Charge



Electric Dipole



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The lower charge q2 exerts a force with the same magnitude but atan angle a above the x-axis. From symmetry we see that itsx-component is the same as that due to the upper charge, but itsy-component has the opposite sign. So the components of thetotal force F on Q are

Fx D 0:23 C 0:23 D 0:46 NFy D "0:17 C 0:17 D 0

The total force on Q is in the +x-direction, with magnitude 0.46 N.

Electric Fields and Calculations


Electric Charge

Coulomb’s Law

Electric Fields andCalculations❖Electric field❖Electric field 2❖Superposition❖Strategy❖Field of a ring❖Field of a ring 2❖Field of a ring 3❖Field of a line❖Field of a line 2❖Field of a line 3❖Field of a line 4❖Two Sheets❖Two Sheets 2❖Electric Field Lines

Electric Dipole



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Electric field


Electric Charge

Coulomb’s Law


Electric Dipole



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A charged body creates E aroundit.

The repulsion between A and B isa two-stage process. Body Aproduces an electric field at pointP. Then body B, as a result of thecharge that it carries, experiencesthe force F0 exerted by the field.

Likewise, the point charge q0

produces an electric field in thespace around it and that thiselectric field exerts the force -F0

on body A.

Note: The electric force on acharged body is exerted by theelectric field created by othercharged bodies.

Electric field 2


Electric Charge

Coulomb’s Law


Electric Dipole



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Point charge q produces E at allpoints in space. Field strengthdecreases with increasing distance.

Define the electric field E at apoint as the electric force F0

experienced by a test charge q0 atthe point, divided by the chargeq0:

E DF0

q0(2)

Thus, the electric field of a pointcharge q is, from Coulomb’s lawEq. (1), given by

E D1

4!"0

q

r2Or (3)

Or : unit vector from source to fieldpoint.In electrostatics, E in a conductoris zero.

Superposition of Electric Fields


Electric Charge

Coulomb’s Law


Electric Dipole



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Illustrating the principle ofsuperposition of electric fields.

Assumed a charge distribution q1,q2, q3, # # # . At any given point P ,each point charge produces itsown electric field E1, E2, E3, # # # .

Thus, the electric field at point Pis given by

E D E1 C E2 C E3 C # # #

This is the principle ofsuperposition of electric fields.

E From Continuous Charge Distribution


Electric Charge

Coulomb’s Law


Electric Dipole



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Problem-Solving strategy for calculating the total electric field froma continuous charge distribution:

Draw a picture showing the location of the source chargedistribution and the field point P . Mark a generic chunk ofcharge dq on the source distribution. Draw at point P the dEvector produced by the charge dq.Write down an expression for the magnitude of dE usingdE D

1

4!"0

dqr2.

Resolve the dE vector into its components.Identify any special symmetry features of the problem and usethem to identify components that sum to zero.Perform the integration.

Field of a ring of charge


Electric Charge

Coulomb’s Law


Electric Dipole



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ExampleA ring-shaped conductor with radius a carries a total charge Q

uniformly distributed around it. Find the electric field at a point Pthat lies on the axis of the ring at a distance x from its center.

Field of a ring of charge 2


Electric Charge

Coulomb’s Law


Electric Dipole



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Solution

Assume the ring divided into infinitesimal segments of lengthds. Each segment has charge dQ and acts as a point-chargesource of electric field.Since P is on the symmetry axis of the ring, then, bysymmetry, the total field E will have only a component alongthe ring’s symmetry axis (the x-axis).The magnitude of this segment’s contribution to the electricfield at P is

dE D1

4!"0

dQx2 C a2

Field of a ring of charge 3


Electric Charge

Coulomb’s Law


Electric Dipole



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Using cos˛ D x=r D x=.x2 C a2/1=2, dEx is

dEx D dE cos˛ D1

4!"0

dQx2 C a2

xp

x2 C a2

D1

4!"0

x dQ.x2 C a2/3=2

To find the total x-component Ex of the field at P , we integrate thisexpression over all segments of the ring:

Ex DI

1

4!"0

x dQ.x2 C a2/3=2

Since x does not vary as we move from point to point around thering, all the factors on the right side except dQ are constant:

E D ExOi D

1

4!"0

Qx

.x2 C a2/3=2Oi

Field of a line of charge


Electric Charge

Coulomb’s Law


Electric Dipole



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ExamplePositive electric charge Q is distributed uniformly along a line withlength 2a, lying along the y-axis between y D "a and y D Ca.Find the electric field at point P on the x-axis at a distance x fromthe origin.

Field of a line of charge 2


Electric Charge

Coulomb’s Law


Electric Dipole



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SolutionDivide the line charge into infinitesimal segments. Let the length ofa typical segment at height y be dy. The linear charge density $ atany point on the line is equal to Q=2a. Hence the charge dQ in asegment of length dy is

dQ D $ dy DQ dy2a

The distance, from this segment to P is .x2 C y2/1=2, so themagnitude of field dE at P due to this segment is

dE D1

4!"0

dQr2

DQ

4!"0

dy2a.x2 C y2/

We represent this field in terms of its x- and y-components:dEx D dE cos˛; dEy D "dE sin˛.



Electric Charge

Coulomb’s Law


Electric Dipole



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Thus,

dEx DQ

4!"0

x dy2a.x2 C y2/3=2

dEy D "Q

4!"0

y dy2a.x2 C y2/3=2

Hence,

Ex D1

4!"0

Qx

2a

Z a

!a

dy.x2 C y2/3=2

DQ

4!"0

1

xp

x2 C a2

Ey D "1

4!"0

Q

2a

Z a

!a

y dy.x2 C y2/3=2

D 0

or, in vector form,

E D1

4!"0

Q

xp

x2 C a2Oi (4)



Electric Charge

Coulomb’s Law


Electric Dipole



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What is E at a distance x from a very long line of charge?

To find the answer we take the limit of Eq. (4) as a becomes verylarge:

E D$

2!"0xOi

The field magnitude depends only on the distance of point P fromthe line of charge. So at any point P at a perpendicular distance r ,from the line in any direction, E has magnitude

E D$

2!"0r(infinite line of charge)

Compare this to a point charge: 1=r2.

Field of 2 oppositely charged infinite sheets


Electric Charge

Coulomb’s Law


Electric Dipole



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ExampleTwo infinite plane sheets are placed parallel to each other,separated by a distance d . The lower sheet has a uniform positivesurface charge density % , and the upper sheet has a uniformnegative surface charge density "% with the same magnitude.Find the electric field between the two sheets, above the uppersheet, and below the lower sheet.

Field of 2 oppositely charged infinite sheets 2


Electric Charge

Coulomb’s Law


Electric Dipole



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SolutionLet sheet 1 be the lower sheet of positive charge, and sheet 2 bethe upper sheet of negative charge; the fields due to each sheetare E1, and E2, respectively. Both E1, and E2 have the samemagnitude at all points (will derive expression in Chapter 2):

E1 D E2 D%

2"0

At points between the sheets, E1, and E2 reinforce each other; atpoints above the upper sheet or below the lower sheet, E1, and E2

cancel each other. Thus the total field is

E D E1 C E2 D

8

<

:

0 above the upper sheet#"0

Oj between the sheets0 below the lower sheet

Electric Field Lines


Electric Charge

Coulomb’s Law


Electric Dipole



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Electric field lines for threedifferent charge distributions.

The direction of the electric field atany point is tangent to the fieldline through that point.

In general, the magnitude of E isdifferent at different points along agiven field line.

Electric Dipole


Electric Charge

Coulomb’s Law


Electric Dipole❖Force and Torque❖Force and Torque 2❖Potential Energy



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Force and Torque on an Electric Dipole


Electric Charge

Coulomb’s Law





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An electric dipole is a pair of pointcharges with equal magnitude andopposite sign (a positive charge qand a negative charge "q)separated by a distance d .

Electric dipole in a uniform electricfield.

The forces FC and F! on the twocharges both have magnitude qE ,but their directions are opposite.The net force on an electric dipolein a uniform external electric fieldis zero.

Since the two forces don’t actalong the same line, their torquesdon’t cancel. Let the anglebetween the electric field E andthe dipole axis be &; then themagnitude of the net torque aboutcenter of dipole is

' D .qE/.d sin&/

Force and Torque on an Electric Dipole 2


Electric Charge

Coulomb’s Law





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The product of the charge q and the separation d is the magnitudeof a quantity called the electric dipole moment, denoted by p:

p D qd Units: charge times distance (Cm)

and its direction is along the dipole axis from the negative chargeto the positive charge. Thus,

' D pE sin&

or, in vector form,

! D p ! E (5)

torque on an electric dipole

Potential Energy of an Electric Dipole


Electric Charge

Coulomb’s Law





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When a dipole changes direction in an electric field, theelectric-field torque does work on it. In a finite displacement from&1 to &2, the total work done on the dipole by the field is the workdone by an external force from &2 to &1. Thus

W DZ $1

$2

pE sin& d& D pE cos&2 " pE cos&1

The work is the negative of the change of potential energy i.e.W D U1 " U2. Thus, we define

U.&/ D "pE cos&

U D "p # E (6)

potential energy for a dipole in an electric field

Chapter 2: Gauss’s Law


Electric Charge

Coulomb’s Law


Electric Dipole

Chap 2: Gauss’s Law❖Electric Flux❖Electric Flux 2❖Flux of uniform E

❖Nonuniform Field❖Flux through Sphere❖NonsphericalSurface❖General Form


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Charge and Electric Flux


Electric Charge

Coulomb’s Law


Electric Dipole




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The electric field on the surface of boxes containing (a) a singlepositive point charge, (b) two positive point charges, (c) a singlenegative point charge, or (d) two negative point charges.

Electric Flux and Enclosed Charge


Electric Charge

Coulomb’s Law


Electric Dipole




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Three cases in which there is zero net charge inside a box and noelectric flux through the surface of the box. (a) An empty box withE D 0. (b) A box containing one positive and oneequal-magnitude negative point charge. (c) An empty boximmersed in a uniform electric field

Flux of a uniform Field

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Consider a flat area A perpendicular to a uniform electric field E . Define electricflux as the product of the field magnitude E and the area A: ˚E D EA. We canview ˚E in terms of field lines passing through A. Increasing A means morelines of E pass through area, and thus increasing ˚E . If area A is flat but notperpendicular to E , then

˚E D EA cos& D E # A (7)

Flux of a Nonuniform Electric Field


Electric Charge

Coulomb’s Law


Electric Dipole




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What happens if the electric field isn’t uniform but varies from pointto point over the area A? Or what if A is part of a curved surface?

We divide A into many small elements dA, each of which has aunit vector On perpendicular to it and a vector area dA D On dA.

Total flux is then

˚E DZ

E cos& dA DZ

E # dA (8)

general definition of electric flux

Electric flux through a sphere


Electric Charge

Coulomb’s Law


Electric Dipole




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A charge q is surrounded by a sphere with radius r .

Because the sphere is centered on the point charge, at any pointon the spherical surface, E is directed out of sphere perpendicularto surface. Using Eq. (8), the electric flux is thus given by

˚E DI

E # dA DI

E dA D EA Dq

"0

Point Charge Inside a Nonspherical Surface


Electric Charge

Coulomb’s Law


Electric Dipole




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Let us surround the sphere ofradius R by a surface of irregularshape. The electric flux throughthe spherical surface element isequal to the flux EdA cos&through the correspondingirregular surface element.

Total ˚E through irregularsurface, given by Eq. (8), must bethe same as total flux through asphere, which is equal to q="0.Thus, for irregular surface

˚E DI

E # dA Dq

"0(9)

General Form of Gauss’s Law


Electric Charge

Coulomb’s Law


Electric Dipole




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Suppose the surface encloses not just one point charge q butseveral charges q1; q2; q3; # # # . The total (resultant) electric field E

at any point is the vector sum of the E fields of the individualcharges. Let Qencl be the total charges enclosed by the surface:Qencl D q1 C q2 C q3 C # # # . Then we can write an equation likeEq. (9) for each charge and its corresponding field and add theresults. Thus, the general statement of Gauss’s law:

˚E DI

E # dA DQencl

"0(10)

Gauss’ law

The total electric flux through a closed surface is equal to the total(net) electric charge inside the surface, divided by "0.

Application of Gauss’ Law


Electric Charge

Coulomb’s Law


Electric Dipole


Application of the Law❖Enclosed Charge❖Strategy❖Field of a line❖Field of a line 2❖ Infinite Sheet❖Sphere❖Sphere 2❖Sphere 3❖Charges onConductors❖Field at the Surfaceof a Conductor

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Electric flux and enclosed charge


Electric Charge

Coulomb’s Law


Electric Dipole



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ExampleThe figure below shows the fieldproduced by two point chargesCq and "q of equal magnitudebut opposite sign (an electricdipole). Find the electric fluxthrough each of the closedsurfaces A, B , C , and D.

SolutionSurface A (shown in red)encloses the positive charge, soQencl D Cq; surface B (shown inblue) encloses the negativecharge, so Qencl D "q; surface C(shown in purple), which enclosesboth charges, hasQencl D Cq C ."q/ D 0; andsurface D (show in yellow), whichhas no charges enclosed within it,also has Qencl D 0. Hence, thetotal fluxes for the varioussurfaces are ˚E D Cq="0 forsurface A, ˚E D "q="0 forsurface B , and ˚E D 0 for bothsurface C and surface D.

Problem-Solving Strategy for Gauss’ Law


Electric Charge

Coulomb’s Law


Electric Dipole



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1. Calculate the flux:

Observe the symmetry of the electric field produced by thecharge distribution. If it’s spherical, select a sphericalGaussian surface. If it’s linear, cylindrical, or planar, selecta cylindrical Gaussian surface.Orient the Gaussian surface so that it surrounds all thecharge such that the electric field is constant over thesurface.Evaluate the flux through the Gaussian surface. Note theangle between E and the area vector pointing outwardfrom the Gaussian surface.

2. Calculate the charge and then apply the Gauss’ law.

Field of a line charge


Electric Charge

Coulomb’s Law


Electric Dipole



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ExampleElectric charge is distributed uniformly along an infinitely long, thinwire. The charge per unit length is $ (assumed positive). Find theelectric field.

Field of a line charge 2


Electric Charge

Coulomb’s Law


Electric Dipole



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SolutionThe system has cylindrical symmetry, which means that the fieldcan’t have any component parallel to the wire.

We break the surface integral for the flux ˚E into an integral overeach fiat end and one over the curved side walls. There is no fluxthrough the ends because E lies in the plane of the surface andE? D 0.

By symmetry, E has the same value everywhere on the walls. Thearea of the side walls is 2!rl . Hence, from gauss’ law,

˚E D .E/.2!rl/ D$l

"0and

E D1

2!"0

$

r(field of an infinite line of charge)

Field of an infinite plane insulating sheet ofcharge


Electric Charge

Coulomb’s Law


Electric Dipole



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ExampleFind the electric field caused by athin, flat, infinite insulating sheeton which there is a uniformpositive charge per unit area % .

SolutionPlanar symmetry ) charge distributiondoesn’t change if we slide it in anydirection parallel to sheet.

This implies, at each point, E isperpendicular to the sheet. Thesymmetry also tells us that thefield must have the samemagnitude E at any givendistance on either side of thesheet. Hence, Gauss’s law gives

2EA D%A

"0and

E D%

2"0(charged insulating sheet)

The field is uniform and directedperpendicular to the plane of thesheet, and its magnitude isindependent of the distance fromthe sheet.

Field of a uniformly charged sphere


Electric Charge

Coulomb’s Law


Electric Dipole



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ExamplePositive electric charge Q is distributed uniformly throughout thevolume of an insulating sphere with radius R. Find the magnitudeof the electric field at a point P a distance r from the center of thesphere.

Field of a uniformly charged sphere 2


Electric Charge

Coulomb’s Law


Electric Dipole



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Solution

From symmetry, E has the same value at every point on Gaussiansurface, and direction of E is radial at every point on surface.

The volume charge density is ( DQ

4!R3=3.

For r < R, amount of charge enclosed within Gaussian surface

Qencl D (Vencl D!

Q

4!R3=3

" !

4

3!r3

"

D Qr3

R3

From Gauss’ law, field inside a uniformly charged sphere is

4!r2E DQ

"0

r3

R3) E D

1

4!"0

Qr

R3

Field of a uniformly charged sphere 3


Electric Charge

Coulomb’s Law


Electric Dipole



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To find the field magnitude outside the charged sphere, we usea spherical Gaussian surface of radius r > R. This surfaceencloses the entire charged sphere, so Qencl D Q, andGauss’s law gives

4!r2E DQ

"0

E D1

4!"0

Q

r2

For any spherically symmetric charged body the electric fieldoutside the body is the same as though the entire charge wereconcentrated at the center.

Charges on Conductors

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(a) In an electrostatic situation, electric field at every point within a conductor is zero and that any excesscharge on a solid conductor is located entirely on its surface.

(b) For a cavity inside the conductor, if there is no charge within the cavity, a Gaussian surface A shows thatthe net charge on the surface of the cavity must be zero, * E D 0 everywhere on the Gaussian surface.

(c) The conductor originally has a charge qC , and the cavity contains an isolated charge q. Again E D 0

everywhere on surface A, so total charge inside this surface must be zero ) a charge !q distributed onthe surface of the cavity. As there can’t be any excess charge within the material of a conductor, chargeqC C q must appear on the outer surface.

Field at the Surface of a Conductor


Electric Charge

Coulomb’s Law


Electric Dipole



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While surface charge density % may vary from point to point on thesurface of a conductor, the direction of E is always perpendicularto the surface.Consider the small cylinder Gaussian surface A. The electric fieldis zero at all points within the conductor. According to Gauss’ law,the field at the surface of a conductor is

E?A D%A

"0and E? D

%

"0(11)

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PC1430 - 1-2 (Electric Charge_ Fields_ and Gauss's Law)_14012013111114982