PHYS 222
Worksheet 5 – Electric Potential
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
1/24/12
Useful Equations
Wab  (Ub U a )
qq
U k 0
r
q
U  kq0  i
i ri
U
q
V  k
q0
r
q
V  k i
i ri
dq
V  k
r
Work done by a conservative force
Electric potential energy by two point charges q and q0
Potential energy of charge q0 due to the collection of charges qi
Potential due to a point charge. Units: [ 1 V = 1 volt = 1 J/C]
Potential due to a collection of point charges
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
Diagrams
Related Problems
1) A point charge q1 = +2.40 µC is held stationary at the origin. A second point charge
q2 = -4.30 µC moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m, y =
0.250 m. How much work is done by the electric force on q2?
W  (U b  U a )
Ua 
Ub 
kq1q2 9(109 )(2.4)(106 )( 4.3)(106 )

 0.6192 J
x1
0.150
kq1q2
x2 2  y2 2

9(109 )(2.4)(106 )( 4.3)(106 )
0.252  0.252
 0.2627 J
W  (U b  U a )  ((0.2627)  (0.6192))  0.3565 J
2) Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of
50.0 cm. An electron is released from rest at a point midway between the two charges
and moves along the line connecting the two charges. What is the speed of the electron
when it is 10.0 cm from the +3.00-nC charge?
KEi  U i  KE f  U f  KE f  U i  U f
q q 
 3(109 ) 2(109 ) 
17
U i  kqe  1  2   9(109 )(1.6)(1019 ) 

  2.88(10 ) J
r
r
0.25
0.25


 1i 2i 
q
q 
U f  kqe  1  2   5.04(1017 ) J
 r1 f r2 f 
KE  mv / 2  U i  U f  v 
2
2(U i  U f )
m
2(2.304(1017 )  (3.6(1017 )))

 6.89(106 ) m / s
31
9.11(10 )
3) A uniform electric field has magnitude E and is directed in the negative x direction.
The potential difference between point a (at x = 0.70 m) and point b (at x = 0.85 m) is
370 V.
(a) Which point is at a higher potential? a or b
E field points towards the direction of lower potential, thus potential at point a is lower
that at point b.
(b) Calculate the value of E
xb
Vb  Va    E dl   E ( xb  xa )  E 
xa
V
 2467 N / C pointing in the negative x
xb  xa
(c) A negative point charge q = -0.200 µC is moved from b to a. Calculate the work done
on the point charge by the electric field
W  q0 V  q(Va  Vb )  (0.2)(106 )(370)  7.4(105 ) J
4) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so
that the potential of its center, relative to infinity, is 1.50 kV? What is the potential of the
sphere's surface relative to infinity?
E
q
4 l 2 0
r
d /2


V    E  dl   
q
q
1
1
dl 
(
 )
2
4 l  0
4 0 d / 2 
 q  V 4 0 d / 2  1.5(103 )4 (8.85)(10 12 )(0.25 / 2)  20.8 nC
V on the surface equals 1.5 kV
5) A very long insulating cylinder of charge of radius 2.60 cm carries a uniform linear
density of 16.0 nC/m. If you put one probe of a voltmeter at the surface, how far from
the surface must the other probe be placed so that the voltmeter reads 180 V?
E

2 l 0
d r
r
V  

E  dl  
d r
 d  re

r
V 2 0



dl  
(ln(d  r )  ln(r ))
2 l 0
2 0
(180)2 8.85(1012 )
 r  (0.026)e
16(109 )
 0.026  2.26 cm