PHYS 222
Worksheet 6 – Electric Potential
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
1/30/12
Useful Equations
V  k
dq
r
Potential due to a continuous distribution of charge
b
Vb  Va    E dl
Potential difference as an integral of E
E  V
E in terms of V. E is the gradient of V.
a
Related Problems
1) Two metal spheres of different sizes are charged such that the electric potential is the
same at the surface of each. Sphere A has a radius 6 times that of sphere B. Let Qa
and Qb be the charges on the two spheres, and let Ea and Eb be the electric-field
magnitudes at the surfaces of the two spheres. Find Qb/Qa and Eb/Ea. (Book 23.81)
V  k 
dq
r
QA
rA
Q
VB  k B
rB
V  0  VA  VB
Q
Q
Q r
k A k B  B  B
rA
rB
QA rA
Q 1
rA  6rB  B 
QA 6
VA  k
QA
4 0rA2
QB
EB 
4 0rB 2
QB
EB 4 0rB 2 rA 6rB

 
6
QA
EA
rB rB
4 0rA2
EA 
2) A thin insulating rod is bent into a semicircular arc of radius a, and a total electric
charge Q is distributed uniformly along the rod. Calculate the potential at the center of
curvature of the arc if the potential is assumed to be zero at infinity. (Book 23.70)
dq 
Q

d
dq  Q
V  k   k  d
r
a
0
kQ
kQ
V 
(  0) 
a
a
3) A ring of charge has a radius of 1 m and a linear charge density of 3 mC/m. If
V(∞) = 0, what is the electric potential 2 m from the center of the ring along the axis of
symmetry (along the line perpendicular to the ring, going through its center)?
A) 1.2 x 107 V
B) 3.3 x 107 V
C) 7.5 x 107 V
D) 8.4 x 107 V
E) 17 x 107 V
dq   ad
2
dq
 a d
V  k  k 
r
d
0
V k
2 (3)(1)
2 a
2 a
k
 (9)(109 )
 7.5(107 )V
2
2
2
2
d
a P
1 2
4) Electric charge is distributed uniformly along a thin rod of length a, with total charge
Q. Take the potential to be zero at infinity. Find the potential at the following points:
Q
dx
a
a x
dq
Q
VP  k   k  dx
r
a
x
dq 
 VP 
dq 
kQ  a  x 
ln
a  x 
Q
dx
a x2  y 2
0
dq
Q
VR  k   k 
dx
2
2
r
a
x

y
a
 VR 
0
kQ  2 2
ln  x  y  x 
a 
 a
 VR 

kQ 
y

ln


2
2
a
 y a a 