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PHYS 222
Worksheet 7 – Equipotential Surfaces
Supplemental Instruction
Iowa State University
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
1/31/12
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) Draw the electric potential lines for the following diagrams: (all are conductive
materials)
a)
c)
b)
d)
2) A proton is released from rest at point a. It then travels past point b. What is its speed
at point b?
mp = 1.67(10-27) kg
qp =| qe |= 1.6(10-19) C
W  qV
KE  W
mv 2
KE 
2
2qV
mv 2
 qV  v 
2
m
2(1.6)(1019 )(20  60)
v
 1.24(105 ) m / s
27
1.67(10 )
3) A very long cylinder of radius 2.60 cm carries a uniform charge density of 2.10 nC/m.
Describe the shape of the equipotential surfaces for this cylinder. (Book 23.51)
The potential difference depends only on r, and not the direction. Therefore all points at
the same value of r will be at the same potential. Thus the equipotential surfaces are
cylinders coaxial with the given cylinder.
(a) Taking the reference level for the zero of potential to be the surface of the cylinder,
find the radius of equipotential surfaces having potentials of 13.0 V.
E

2 l 0
rA
V   E dl   
r
 rA  re
 V 2 

0





 dl    ln  rA 
 
2 l 0
2 0  r 
 0.026e

12 ) 
  ( 13)2 (8.85)(10




9


2.1(10 )


 3.67 cm
(Remember, E-field points towards lower potential, that is why ∆V is negative)
(b) Find the radius of equipotential surfaces having potentials of 26.0 V.
Repeat equations from part a and substitute rB for rA:
rB  5.17 cm
(c) Find the radius of equipotential surfaces having potentials of 39.0 V.
rC  7.30 cm
(d) Are the equipotential surfaces equally spaced?
No. The change between the radii are not constant even though the change in voltage
is constant.
4) An infinite conducting slab with 8.85 nC/m2 on each of the two surfaces is perpendicular to
the x axis and located between x = 0 and x = +1.0 cm (i.e., the thickness of the slab is 1.0 cm).
The V = 20 V equipotential is at x = −1.0 cm. Find the potential at x = + 2.0 cm.
a) –20 V
b) –10 V
c) 20 V
d) 30 V
e) 50 V
The system is symmetric with basically 2 infinite plates, so the electric field at the two
points is the same. The voltage is them the same.
5) An insulating sphere with a uniformly distributed charge of 5.00 C has a radius of 1.00 m. A
concentric, thin walled metallic shell has a charge of 10.0 C and a radius of 3.00 m. What is the
electric potential of a point located 2.00 m from their shared center? Take V(∞) = 0
a) (10 C) k / (2 m)
b) (15 C) k / (3 m)
c) (20 C) k / (3 m)
d) (20 C) k / (6 m)
e) (35 C) k / (6 m)
Outside the shell:
15C
15C
;
V

k
r
r2
15C
15C
V (r3 )  k
k
r3
3m
Ek
Between shell and sphere:
Ek
5C
r2
V (r2 ) V (r3 )  
r2 2 m
5C
k 2

r 3m r
k
5C
5C
k
2m
3m
2
Total V is the summation:
 5 5 15 
35C
V (r2 )  kC      k
6m
2 3 3 