Electrostatics

Objects can be charged by rubbing

Charge comes in two types, positive and negative; like charges repel and opposite charges attract

Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction.

Electric Charge in the Atom

Atom:

Nucleus (small, massive, positive charge)

Electron cloud (large, very low density, negative charge)

Atom is electrically neutral.

Rubbing charges objects by moving electrons from one to the other.

Polar molecule: neutral overall, but charge not evenly distributed

Insulators and Conductors

Conductor:

Charge flows freely

Metals

Insulator:

Almost no charge flows

Most other materials

Some materials are semiconductors.

Induced Charge; the Electroscope

Metal objects can be charged by conduction:

They can also be charged by induction:

The electroscope can be used for detecting charge:

The electroscope can be charged either by conduction or by induction.

The charged electroscope can then be used to determine the sign of an unknown charge.

Coulomb’s Law

Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them.

Coulomb’s law:

This equation gives the magnitude of the force.

The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same.

Unit of charge: coulomb, C

The proportionality constant in Coulomb’s law is then:

Charges produced by rubbing are typically around a microcoulomb:

k = 9.00x109N •m2 C2

The magnitude of the charge on the electron (the elementary charge) :

Electric charge is quantized in units of the electron charge.

The elementary charge was first measured by Millikan (1909) in his famous oil drop experiment.

The proportionality constant k can also be written in terms of , the permittivity of free space:

Coulomb’s law strictly applies only to point charges.

Superposition: for multiple point charges, the forces on each charge from every other charge can be calculated and then added as vectors.

The net electrostatic force on a charge is the vector sum of all the electrostatic forces acting on it.

When solving Coulomb’s law (2-d) problems, use the Coulomb’s law formula to calculate a magnitude of force (a + value), then use the principle of like and unlike charges to obtain the direction of the force vectors. At this point one can add the forces to find the net electrostatic force.

Example: a)  What is the electrostatic force between

two electrons 1.000 nm apart? ( 1 nm=1x10-9 m)

b)   What acceleration would this cause on each electron if they were free to move? (me= 9.11x10-31 kg)

Solution: a) We are really just interested in a magnitude. The force would be repulsive.

Fe = kq1q2r2

= 9x109 ⋅1.6x10−19 ⋅1.6x10−19

1x10−9( )2 = 2.30x10-10 N

b) FNet =ma , a = FNetm

= 2.53x1020 m/s2

Example: Determine the net electrostatic force exerted by q1, and q2 on charge q3. (q1= +3.00µC,q2= -4.00µC, q3= +5.00µC)

q3

q2

q110 cm

5 cm

900

Solution: Draw the FBD showing the forces acting on the +5µC charge. The direction of forces can be determined by knowing like charges repel, and unlike charges attract.

q3Fe1

Fe2

The magnitudes of the forces can be found by using coulomb’s law (drop signs).

Fe1 =(9x109 )(3x10−6 )(5x10−6 )

.12=13.5N

Fe2 =(9x109 )(5x10−6 )(4x10−6 )

.052= 72N

FNet =

Fe1 +

Fe2

You can now add forces to find the net electrostatic force acting on q3. The vector relation ship is:

The final result is: FNet= 73.3 N, 79.40 S of W

The Electric Field

The electric field is the electrostatic force on a small charge, divided by the charge:

Comparison to gravitational field:

g =Fgm

E =Feq

The units of electric Field: N/C , and also V/m The electric field is a vector

The electrostatic force on a point charge in an electric field:

Fe = q

E

A positive charge placed in an electric field will experience a force in the same direction, while a negative charge will experience a force in the opposite direction.

Field Lines-Point Charges The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.

The electric field is stronger where the field lines are closer together. The field can also be deduced by the behaviour of a positive “test charge”. Recall that a positive charge will experience a force in the direction of the field.

Below are diagrams showing the electric field line created by the presence of two charges equal in magnitude.

Two + charges Two oppositely charged charges Field lines always leave the positive charge

and enter a negative charge.

What happens if the charges are unequal??

Field Lines—Charged Parallel Plates

The electric field between two closely spaced, oppositely charged parallel plates is constant. This means that if you were to place a positive test charge anywhere between the plates it would experience the same force.

For the field to be a constant value the lines have to be evenly spaced. The field is not uniform at the edges (very top and bottom in this case).

Summary of field lines:

1.  Field lines indicate the direction of the field; the field is tangent to the line.

2.  The magnitude of the field is proportional to the density of the lines.

3.  Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge.

Electric Field due to Point Charges The force Fe between the two charges q

and q2 can be stated by Coulomb's Law:

Fe =kqq2r2

The electric field E, created by charge q at q2 's location is:

E = Feq2=kqq 2r2

÷ q2 ∴ E = k q

r2

Using this formula one can calculate the electric field a distance r from point charge q.

Example(1-D): What is the electric field 2.50 mm from an electron (qe= -1.6x10-19

C)?

qE

Solution: The electric field always points towards a single negative charge (think what would happen to a positive test charge ).

The intensity (magnitude of E field) can be calculated using the formula:

E = k q

r2=9x109( ) 1.6x10−19( )2.50x10−3( )

2 =

The sign is dropped because one is interested in a magnitude.

2.30x10-4 N/C

Example(2-D): Find the net electric field at location p, given that Q1= 4.50µC, Q2= -3.0 µC,d1= 8.00 cm, and d3 = 10.00 cm. You have to consider the electric field produced by each charge at location p.

Solution: The fields acting a location p can be drawn and all angles calculated:

E1

E2

The magnitudes of E1 and E2 can be calculate using the formula: E = k q

r2E1= 4.05x106 N/C, E2= 7.50x106 N/C

ENet =

E1 +E2The net electric field at p is given by :

At this point one could add the vectors “head to tail” or use the components. The final answer is:ENet= 6.02x106 N/C 57.40 N of E

Parallel plates--Revisited

Therefore,

Recall that the electric field is uniform between the plates.

E = ΔVd*This relationship is only true for a

uniform electric field.

Example: A negatively charged mass (m= 15 mg) was suspended between two charged plates with a voltage of 28 µV between them, and separated by a distance of 5.00 cm. It is the balance of the electrostatic force and gravity which keeps the mass suspended. How many excess electrons are on the mass? (Hint: find the total charge 1st)

As an aside note, this problem is similar to the principles of the famous oil drop experiment by Millikan.

m

negative plate

positive plate

Solution:

mFg

Fe

negative plate

positive plate

It is a static force problem.

The magnitude of the electrostatic force must equal the magnitude of the force of gravity.

Fg = Fe , mg = qE, also E = ΔVd

mg = q ΔVd

, q = mgdΔV

The total charge q= 0.2625 C

The number of electrons that make up this charge: n = q

1.6x10−19=1.64x1018electrons

Electric Potential and Electric Potential Energy

If one were to move a positive charge towards another positive charge it takes work to do so. One would be storing electric potential energy. V is the symbol for electric potential (J/C). It is the electric potential energy per unit charge, at a specific location. In the example below we are moving a charge q from position A to B

q+

fixed charge AB

q+

Let VA= electric potential at point A, and VB= the electric potential at point B(caused by fixed charge ).

The electric potential energy is related to electric potential by the following:

EP=qV , where EP= electric potential energy (J) q= charge (C) V= electric potential (J/C), or (V)

Let the electric potential energy at locations A and B are EPA, and EPB. The change in electric potential energy (ΔEp) in moving a charge q from A to B is: ΔEp= EPB- EPA = qVB-qVA =q(VB-VA)=qΔV

In summary, ΔEp=qΔV The change in electric potential energy is the product of the charge q, and the potential difference (or voltage) the charge moves through.

Conservation of Energy In the past we looked at conservation of energy taking into consideration the gravitational force. This principle applies to all forces, including the electrical forces. The general energy conservation (assuming no friction) is:

ΔEP +ΔEK = 0Where ΔEP is the change in electric potential energy, and ΔEK is the change in kinetic energy. We are assuming the energy changes caused by the gravitational force is negligible.

Example: An electron at rest is accelerated through a potential difference of 5.00mV. What was the final speed of the electron? Recall that me= 9.11x10-31 kg, and q=-1.6x10-19 C

Solution: ΔEp +ΔEk = 0 , qΔV+ 1

2mvf

2 −12mvi

2 = 0

Plugging in the numbers and solving for vf, one finds vf= 4.19x104 m/s.

This is a scalar relationship, so in general you should keep track of all signs such as the charge and potential difference. Sometimes in a problem, a voltage may be stated, but this is may be a magnitude of the potential difference and you may have to use common sense with signs.

The Electron-Volt (eV) The electron-volt (eV) is a unit of energy. It is the amount of energy gained by an electron moving through a potential difference of 1 volt.

1.6x10-19 J = 1 eV

This can be show with energy conservation: ΔEp+ΔEk=0, -qΔV=ΔEk= -(-1.6x10-19C)(1V)= 1.6x10-19 J

The Cathode Ray Tube (CRT)

The CRT consists of a vacuum tube, an electron gun, and a phosphor screen. Electrons are fired from the gun and the electrons strike the screen causing a phosphor pixel to fluoresce. Old TV’s used CRT’s.

The amount of deflection (d) of the electron beam turns out to be proportional to the deflecting voltage (Vd) and inversely proportional to the accelerating voltage (Va). Or one could state: d∝Vd

VaThis means that the deflection is equal to a constant (C) multiplied by the ratio Vd/Va.

d =C ⋅VdVa

∴ C = d ⋅VaVd

Example: A CRT has a deflection of 5.00 cm on the screen. If the accelerating voltage were reduce by two thirds, and the deflecting voltage were doubled, what would be the new deflection?

Solution: d1= initial deflection, d2= final deflection Vd1= initial deflecting voltage, Vd2=final deflecting voltage Va1= initial accelerating voltage, Va2= final accelerating voltage Va2= 1/3 Va1 and Vd2= 2Vd1

d1 ⋅Va1Vd1

= d2 ⋅Va2Vd2

d2 =Vd2Vd1

⋅Va1Va2

⋅d1 = (2)(3)d1 = (6)d1 = 30 cm

Electric Potential Energy of Point Charges Below are two charges separated by a distance r. There would be an electrostatic force exerted between the two charges (equal and opposite).

!

The energy stored between two point charges is given by the following relationship:

EP =kq1q2r

EP= electric potential energy (J) K= Coulomb's constant q= charge (C) r=distance between the two charge(m)

Energy is a scalar. It is important to keep track of the signs. One should notice that the relationship on the previous slide is very similar to the gravitational potential energy between two masses. The relationship can be found in a similar manner. Again, electric potential energy is defined to be zero (unbound) when the two charges are an infinite distance apart.

Example: What is the electric potential energy stored between and electron and proton 3.50 cm apart?

Solution:

EP =kq1q2r

=(9x109 )(1.6x10−19 )(-1.6x10−19 )

(0.035)

= -6.58x10-27 J

Energy Conservation and Point Charges Recall energy conservation can be stated:

ΔEP +ΔEK = 0

For two point charges it could also be stated by the following:

Epi +Eki = Epf +Ekf

kq1q2ri

+12mvi

2 =kq1q2rf

+12mvf

2

or

Example: An electron 15.0 cm away approaches another fixed electron with a speed of 4.5x103 m/s (v) . How close (D, distance of approach) will the two electrons get to each other?

ve- e-

15 cm

e- e-

D

Solution: (9x109 )(−1.6x10−19 )(−1.6x10−19 )

0.15+12(9.11x10−31)(4.5x103)2 = (9x10

9 )(−1.6x10−19 )(−1.6x10−19 )D

Solve for D: D= 2.50x10-5 m

Example: How much energy is stored between 3 protons 5.00 cm apart (equilateral triangle) from one another? It would be the same as the total energy required to assemble these charges when they are initially an infinite distance away.

p+

p+

p+

d d

d

Solution: It takes no energy to place the 1st charge as there is no electric force to work against.

p+

The energy required to bring the second proton from an infinite distance away to 5.00cm from the 1st charge would be:

p+

p+

5 cmE1 =

kq1q2r

=(9x109 )(1.6x10−19 )(1.6x10−19 )

(0.05)

Solution continued: When bringing the final 3rd charge from an infinite distance away one must work against the electrostatic forces of both the first two charges. The energy required to do this is:

p+

p+

5 cm

p+

5 cm

5 cm

E2 =(9x109 )(1.6x10−19 )(1.6x10−19 )

(0.05)+(9x109 )(1.6x10−19 )(1.6x10−19 )

(0.05)

The total energy required is E1+E2= 1.38x10-26 J *What if the distance between charges were not equal? How would this change things?

Electric Potential due to Point Charges

EP =kq1qr

Recall the electric potential energy (EP)of a charge q1 a distance r from a charge q would be given by:

The electric potential (V), due to charge q, would be given by:

V =Ep

q1

∴ V =kqr

Example: What is the electric potential 12 mm from an alpha particle (Helium nucleus)?

Solution: V =

kqr=(9x109 )(2)(1.6x10−19 )

0.012= 2.4x10-7 V

One may have to find the electric potential due to multiple charges. Remember that electric potential is a scalar so there is no direction associated with it. To find the electric potential due to multiple charges at a particular location one needs only to add numbers. These numbers could be negative or positive.

Example: Find the electric potential at location p (see below) due to the electron and proton.

p+ e-9006 cm

8 cm

p

Solution:

VP =(9x109 )(1.6x10−19 )

0.06+(9x109 )(−1.6x10−19 )

0.1= 9.6x10-9 V

Example: Two protons are positioned as shown on right. a)  Calculate the electric potential

at positions A and B (VA, and VB)

b)   Find the change in electric potential (Voltage) where ΔV =VB-VA

c)  If an electron were placed at position A, what speed would it acquire when it reached position B

A

B

10 cm

10 cm

7 cm 7 cm

p p

Solution: a) VA =

(9x109 )(1.6x10−19 )(2)0.1

=

VB =(9x109 )(1.6x10−19 )(2)

0.07=

b) V=VB-VA= 1.234x10-8V

2.88x10-8 V

4.114x10-8 V

c) ΔEk +ΔEp = 0

, in more detail: 12mvf

2 −12mvi

2 + qΔV = 0

vf =−2qΔVm

= 65.8 m/s

Equipotential Lines (Surfaces)

•  Equipotential lines are lines of constant electric potential

•  For a point charge the lines would be concentric circles about the charge.

•  For parallel plates the lines would be parallel to the plates.

Examples of Equipotential lines

Question to consider: How is the electric field related to the electric potential? Examining the electric field lines and equipotential lines on the previous slide might help to answer this question.