Chapter 23 – Electric Potential

What is the value of employing the concept of energy when solving physics problems?

Potential energy can be defined for conservative forces. Is the electrostatic force conservative?

For conservative forces, the work done in moving an object between two points is independent of the path taken.

b

a

W F d

a

b

E

Work and Potential Energy for gravity

dW dU

Work and Potential EnergydW F d

q 0

FE lim F qE

q

Electric Field Definition:

dW qE d

dW dU qE d

Work Energy Theorem

b b

a a

dU q E d

a

b

E

Electric Potential Differenceb b

a a

dU q E d

b

b a

a

U U q E d

bb a

a

U UE d

q

Definition:

bb a

ba b a

a

U UV V V E d

q

a

b

E

What the heck is dl

dl

What it means:

• Potential Difference, Vb-Va is the work per unit charge an external agent must perform to move a test charge from ab without a change in kinetic energy.

b a baba b a

U U WV V V

q q

a

b

E

An example

Units of Potential Difference

Joules JVolt V

Coulomb C

Because of this, potential difference is often referred to as “voltage”

b a baba b a

U U WV V V

q q

So what is an electron Volt (eV)?

In addition, 1 N/C = 1 V/m - we can interpret the electric field as a measure of the rate of change with position of the electric potential.

Electron-Volts

• Another unit of energy that is commonly used in atomic and nuclear physics is the electron-volt

• One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt– 1 eV = 1.60 x 10-19 J

ExampleThrough what potential difference would one need to accelerate an electron in order for it to achieve a velocity of 10% of the velocity of light, starting from rest? (c = 3 x 108 m/s)

Conventions for the potential “zero point”

b a baba b a

U U WV V V

q q

Choice 1: Va=0b a

b a

U UV V

q

0 0

bb

UV

q

Choice 2:

“Potential”

V 0 b

bb b

U UV V V E d

q

bb

b

UV E d

q

00

Potential difference for a uniform electric field

+Q

-Q

ab d

b

ba b a

a

V V V E d

oˆE E j

ˆ ˆd dxi dyj

d d

ba b a o o o

0 0

ˆ ˆ ˆV V V E j dxi dyj E dy E d

b a oU U qE d

Potential difference for a point charge

+Q

b

ba b a

a

V V V E d

2

kqˆE r

r

ˆ ˆˆd drr rd r sin d

b b b

a a a

r r r

ba b a 2 2 2r r r

kq kq drˆ ˆV V V r drr dr kq

r r r

b

a

r

ba b ar b a

1 1 1V V V kq kq

r r r

dl for a point charge

Recall the convention for the potential “zero point”

V 0

ba b ab a

1 1V V V kq

r r

b bb

1 1V V V kq

r

kqV r

r

Equipotential surfaces are concentric spheres

Electric Potential of a Point Charge

• The electric potential in the plane around a single point charge is shown

• The red line shows the 1/r nature of the potential

E and V for a Point Charge

• The equipotential lines are the dashed blue lines

• The electric field lines are the brown lines

• The equipotential lines are everywhere perpendicular to the field lines

Potential of a charged conductor

Given: Spherical conductorCharge=QRadius=R

Find: V(r)

R

The plots for a metal sphere

Determining the Electric Field from the Potential

dV E ds E ds

dVE

ds

x

VE

x

y

VE

y

z

VE

z

V V Vˆ ˆ ˆE i j kx y z

E V

Superposition of potentials

0 1 2 3V V V V ...

+Q3

+Q2

+Q110r

20r

30r0

31 20

10 20 30

kQkQ kQV ...

r r r

Ni

0i 1 i0

kQV

r

Electric potential due to continuous charge distributions

kQV r

r

all charge

dqV k

r

Discrete charges Continuous charge distribution

Single charge

kdqdV

r

Single piece of a charge distribution

+Q3

+Q2

+Q110r

20r

30r 0 0

++

++

dVdq

Ni

0i 1 i0

kQV

r

Electric Potential for a Continuous Charge Distribution• Consider a small

charge element dq– Treat it as a point

charge

• The potential at some point due to this charge element is

e

dqdV k

r

Electric field due to continuous charge distributions

Ni

0 i02i 1 i0

QˆE k r

r

0 2all charge

dqˆE k r

r

Discrete charges Continuous charge distribution

0 2

kQˆE r

r

Single charge

0 2

kdqˆdE r

r

Single piece of a charge distribution

+Q3

+Q2

+Q1

01E

03E

02E

10r

20r

30r 0 0

++

++

0dE

dq

Example: A ring of charge

dVx

dq ds Rd

2 2 2r x R

a

d

+

+

+

+ +

+

+

kdqdV

r

2 2

k RddV

x a

2

2 2 2 2 2 20

k a k 2 a kQV d

x a x a x a

Electric field from a ring of charge

2 2

kQV

x a

dVx

dq ds Rd

2 2 2r x R

a

d

+

+

+

+ +

+

+

dV ˆE V idx

3/ 22 2

kQx ˆE ix a

Example: Electric field of a charged ring directly

dE

x

dq ds Rd

2 2 2r x R

xdE

ydE

a

d

+

+

+

+ +

+

+

2

kdqˆdE r

r

y-components cancel by symmetry

x 2

kdqdE cos

r

2 2 2 2

k ad xdE

x a x a

2

3 3 32 2 2 2 2 202 2 2

k xa k xa kQxE d 2

x a x a x a

Potential due to a charged disk

dVx

ra

dq dA rdrd 2 rdr

2 2

kQV

x r

2 2

kdqdV

x r

a a

2 2 2 20 0

k 2 rdr rdrV k 2

x r x r

2 2V k 2 x a x

Uniformly Charged Disk

dE

x

3

2 2 2

kQxE

x r

r

3

2 2 2

kxdqdE

x r

dq dA rdrd 2 rdr

3

2 2 2

kx 2 rdrdE

x r

a

2 2

2

a a x a

3 3 32 2 2 20 02 2 2x

kx 2 rdr 2rdr duE kx kx

x r x r u

2 2

2 2

2

2

x a1

x a 3 22

2 2 2 2 2x

x

u 1 1 xkx u du kx 2kx k 2 1

1 x a x x a2

Electric Dipole

k QkQ 1 1 rV kQ kQ

r r r r r r r r r

r r r 2a cos

2 2

kQ2a cos kpcosV

r r

p Q2a

Electric Potential of a Dipole

• The graph shows the potential (y-axis) of an electric dipole

• The steep slope between the charges represents the strong electric field in this region

E and V for a Dipole

• The equipotential lines are the dashed blue lines

• The electric field lines are the brown lines

• The equipotential lines are everywhere perpendicular to the field lines

Potential energy due to multiple point charges

+Q1

21r

kqV r

r 1

12

kqV

r+Q2

1 22

12

kq qU q V

r

+Q3

+Q1+Q2

1 2

13 23

kq kqV

r r

21r

13r23r1 3 2 31 2

12 13 23

kq q kq qkq qU

r r r

Irregularly Shaped Objects

• The charge density is high where the radius of curvature is small– And low where the radius of

curvature is large

• The electric field is large near the convex points having small radii of curvature and reaches very high values at sharp points

Problem P25.23

Show that the amount of work required to assemble four identical point charges of magnitude Q at the corners of a square of side s is 5.41keQ2/s.

1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 5.41

2

e e e

e e

U U U U U

U U U U U U U

kQ kQ kQU

s s s

kQ kQU

s s

Example P25.33An electron starts from rest 3.00 cm from the center of a uniformly charged insulating sphere of radius 2.00 cm and total charge 1.00 nC. What is the speed of the electron when it reaches the surface of the sphere?

2

1 2

12

ee k qQk eQmv

r r

1 2

2 1 1ek eQvm r r

9 2 2 19 9

31

2 8.99 10 N m C 1.60 10 C 10 C 1 10.0300 m 0.0200 m9.11 10 kg

v

67.26 10 m sv

Example P25.37

The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 10.0 V and b = –7.00 V/m. Determine

(a) the potential at x = 0, 3.00 m, and 6.00 m, and

(b) the magnitude and direction of the electric field at x = 0, 3.00 m, and 6.00 m.

0x 10.0 VV

3.00 mx 11.0 VV

6.00 mx 32.0 VV

At

7.00 V m 7.00 N C in the directiondV

E b xdx

Example 25.43A rod of length L (Fig. P25.43) lies along the x axis with its

left end at the origin. It has a nonuniform charge density λ = αx, where α is a positive constant.

(a) What are the units of α? (b) Calculate the electric potential at A.

Figure P25.43

2

C 1 Cm m mx

0

ln 1L

e e e edq dx xdx L

V k k k k L dr r d x d