Chapter 25 – 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
W

F ds a
 a b
E

Work and Potential Energy for gravity dW
  dU

Work and Potential Energy dW

F ds
Electric Field Definition: E
 lim q

0
F q
F qE dW
 qE ds
Work Energy Theorem a b dW
  dU
 qE ds
 b a b dU
  q E ds a

E

a
Electric Potential Difference
Definition: b
E
 b a b dU
  q E ds a
 b
U b

U a
  q E ds a

U b

U a q
   b a
E ds
V ba

V b

V a

U b

U a q
   b a
E ds

What the heck is d s d s

What it means:
• Potential Difference, V b
-V a is the work per unit charge an external agent must perform to move a test charge from a
 b without a change in kinetic energy.
V ba

V b

V a

U b

U a q
 
W q ba a b
E

An example

Units of Potential Difference
V ba

V b

V a

U b
 q
U a
 
W ba q


Joules
  
Coulomb C

Volt

V
Because of this, potential difference is often referred to as “voltage”
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.
So what is an electron Volt (eV)?

Electron-Volts
• Another unit of energy that is commonly used in atomic and nuclear physics is the electronvolt
• 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

Example
Through 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 10 8 m/s)

Conventions for the potential “zero point”
V ba

V b

V a

U b

U a q
 
W ba q
Choice 1: V a
=0
Choice 2: V


0
“Potential”
0
V b

V a

U b

U a
0 q
V b

U b q
0
V b


V b

V


U b

U

0 q
   b

E ds
V b

U b
  q
 b

E ds

Potential difference for a uniform electric field a
+Q
-Q b d
V ba

V b

V a
   b a
E ds
E
 
E j o
ˆ ds
 dxi
ˆ  ˆ
V ba
V b
V a d
    
0
 o

ˆ  ˆ


0
 d
E dy o

E d o
U b

U a
 qE d o

Potential difference for a point charge
+Q
V ba

V b

V a
   b a
E
 kq ˆ
2 r r
E ds ds
 drr
ˆ    r sin d
ˆ
V ba

V b

V a
  r b  r a kq
ˆ r drr
ˆ r
2
  r b r a
 kq r
2 dr
  kq r b  r a dr r
2
V ba
V b
V a kq


1 r

 r b
      r a kq


1

1 r r b a



d s for a point charge

Recall the convention for the potential
“zero point”
V ba

V b

V a
 kq


1

1 r r b a


V


0 V b


V b

V

 kq

 r
1

1
 b


 kq 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:
Find:
Spherical conductor
Charge=Q
Radius=R
V(r)
R

The plots for a metal sphere

Determining the Electric Field from the Potential dV
 
E ds
 
E ds
E x
E y
E z
 

V
 x

V
 
 
 y

V
 z
E
  dV ds
E
 

V ˆ

V

V
 x i

 y
ˆ j

 z
E
 
V

Superposition of potentials
+Q
1
V
0

V
1

V
2

V
3

...
r
10 r
20
+Q
2 r
30
0
+Q
3
V
0
 kQ
1
 kQ r
10 r
20
2
 kQ
3
 r
30
...
V
0

N  kQ i i 1 r i0

+Q
1
+Q
2
+Q
3
Electric potential due to continuous charge distributions
 kQ r
Single charge dV
 kdq r
Single piece of a charge distribution r
20 r
10 r
30
0
V
0

N  kQ i i 1 r i0
Discrete charges
+
+ dq dV
0 +
+
V
 k
 all charge dq r
Continuous charge distribution

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 dV
 k e dq r

+Q
1
+Q
2
+Q
3
Electric field due to continuous charge distributions
E
0
 kQ ˆ
2 r r
Single charge dE
0
 kdq ˆ
2 r r
Single piece of a charge distribution r
20 r
10 r
30
E
0
 k
N  i 1 r i0
2
0 r i0
Discrete charges
E
03
E
02
E
01
+
+ dq dE
0
0 +
+
E
0
 k
 all charge dq ˆ
2 r r
Continuous charge distribution

Example: A ring of charge
+
+
 a
+
+
+ d

+
+ dq ds Rd x r
2  x
2 
R
2
 dV dV
 kdq r dV
 x
2  a
2
V
 x
2  a
2

2

0 d
  x
2  a
2
 kQ x
2  a
2

+
+
 a
+
+
+
Electric field from a ring of charge d
 dq ds Rd V
 kQ x
2  a
2
+
+ x r
2  x
2 
R
2
 dV
E dV ˆ i dx
E

 x
2 kQx ˆ
 a
2

3/ 2 i

+
+
Example: Electric field of a charged d

+
 a
+ ring directly dq ds Rd dE
 kdq ˆ
2 r r y-components cancel by symmetry r
2  x
2 
R
2
+ x
 dE x dE x
 kdq cos
 r
2
+ dE y dE dE
 k
 ad
 x x
2  a
2 x
2  a
2
+
E

 x
2  a
2
 3
2

2
0
 d
 
 x
2  a
2
 3
2 kQx
 x
2  a
2
 3
2

a r
Potential due to a charged disk x dV
V
 kQ x
2  r
2 dV
 kdq x
2  r dq
  dA
  rdrd 2 rdr
2
V
  a
0 x
2  r
2
 a
0 rdr x
2  r
2
V
   
 x
2  a
2  x



a r
Uniformly Charged Disk kQx
E

 x
2  r
2
 3
2 x dE kxdq dE

 x
2  r
2
 3
2 dq
  dA
  rdrd 2 rdr dE

 x
2  r
2
 3
2
E
  a
0
 x 2  r 2
 3
2
 kx
  a
0
2rdr
 x 2  r 2
 3
2
 kx
 x
2  a x

2
2 du u 2
3
 kx
 x
2  a
2

2 x
3
 u du
 kx
 u

1
2

1
2 x
2 x
2  a
2
 
2kx




1 x
2  a
2

1 x
2



 k 2 1
 x x
2  a
2




Electric Dipole p

Q2a
V
 kQ r
 k r
 
  r
 kQ


1

1 r r
  r


 kQ 
 r r r
  r
 r
  r r 2a cos

V
 kQ2a cos

 kp cos r
2 r
2


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
+Q
2
+Q
1
 kq r
V
 kq
1 r
12 r
21 U
 q V
 kq q
2 r
12
+Q
2 r
23 r
21
+Q
3
+Q
1 r
13
V
 kq
1
 kq r r
13 23
2
U
 kq q
1 2 r
12
 kq q
1 3
 kq q
2 3 r
13 r
23

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 5.41
k e
Q 2 / s.
U  U
1
 U
2
 U
3
 U
U U
12


U
13
 U
4
23
U s
2
 s
2 

U
14
 U
24
 U
1
2
 1


 s
2 

1 
34

1
2
 1

2
U  s
2 

4 
2
2


 5.41
s

Example P25.33
An 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? r
1
 r
2

1
2 m v
2 v 
2 k eQ


1

1 m  r
1 r
2

 v 
 

  2
C
2


9.11 10
 31
kg
  19
 


1

1
0.030 0 m 0.020 0 m

 v 

Example P25.37
The potential in a region between x = 0 and x = 6.00
At 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. x  0
V  10.0 V x  3.00 m V   11.0 V x  6.00 m V   32.0 V
E   dV dx b

7.00 V m

 7.00 N C in the  x direction

Example 25.43
A 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 = α x , where α is a positive constant.
(a) What are the units of α?
(b) Calculate the electric potential at A.
 
C m

1 m
 m
C
2
V  k e
 dq
 r k e

 dx r
 k e
 
L
0 xdx
 k e






L  d



Figure P25.43