Chapter 22: Electric Fields
Example Questions & Problems
E
1
4 ò0
dq
r
2
rˆ
Epo int 
q
4 ò0r
2
Edipole 
qd
4 ò0r
3
Eline 

2ò0r
Eplane 

2ò0
Example 22.1
a. Why do we use small test charges when measuring electric fields?
b. Why is it more difficult to charge an object by rubbing on a humid day than on a dry
day?
c. Explain why there can be a net force on an electric dipole placed in a nonuniform
electric field?
Example 22.2
A positive charge q = 7.81 pC is spread uniformly along a thin nonconducting
rod of length L = 14.5 cm. What are the (a) magnitude and (b) direction
(relative to the positive direction of the x axis) of the electric field produced at
point P, at distance R = 6.00 cm from the rod along its perpendicular bisector?
Example 22.3
Two curved plastic rods, one of charge +q and the other of charge –q, form a
circle of radius r = 8.50 cm in the xy plane. The x axis passes through both of
the connecting points, and the charge is distributed uniformly on both rods. If q
= 15.0 pC, what are the (a) magnitude and (b) direction (relative to positive
direction of the x axis) of the electric field E produced at P, the center of the
circle?
22-1
Example 22.4
Three very large square planes of charge are arranged as shown (on
edge). From left to right, the planes have charge densities of (σ1, σ2,
σ3) = (0.50 C/m2, +0.10 C/m2, 0.35 C/m2). Find the total electric
field (direction and magnitude) at points A, B, C, and D. Assume the
plates are much larger than the distance AD.
Example 22.5
A 10.0 g block with a charge of +8.00×105C is placed in an electric field E = (3000,
600) N/C. What are the (a) magnitude and direction (relative to the positive direction of
the x axis) of the electrostatic force on the block? If the block is released from rest at the
origin at time t = 0, what are its (c) x and y coordinates at t = 3.00 s?
22-2
Example A
The figure shows an uneven arrangement of electrons (e) and
protons (p) on a circular arc of radius r = 2.00 cm, with angles
1 = 30.0o, 2 = 50.0o, 3 = 30.0o, and 4 = 20.0o. What are the
(a) magnitude and (b) direction (relative to the positive
direction of the x axis) of the net electric field produced at the
center of the arc?
Solution
The field of each charge has magnitude
19
kq
e
C
9
2
2 1.60  10
E 2 k
 (8.99  10 N  m C )
 3.93  106 N C  E.
2
2
r
(0.020 m)
 0.020 m
The directions are indicated in standard format below. We use the magnitude-angle
notation (convenient if one is using a vector-capable calculator in polar mode) and write
(starting with the proton on the left and moving around counterclockwise) the
contributions to Enet as follows:
E0  E1  E2  E3  E4  E0   E210   E260   E130   E340 
In component form, it reads as
 E X  E0 X  E1X  E2X  E3 X  E4 X
 E  cos(0)  cos(210)  cos(260)  cos(130)  cos(340)   E  0.257 
E
Y
 E0Y  E1Y  E2Y  E3Y  E 4Y
 E  sin(0)  sin(210)  sin(260)  sin(130)  sin(340)   E  1.06 
a. The result above shows that the magnitude of the net electric field is
|Enet | E2X  E2Y 
E 0.257  E  1.06   3.92 10
2
2
2
2
6
N/C |Enet |
b. Similarly, the direction of Enet is –76.4 from the x axis.


 EY 
1 E  1.06 
  76.4  
  tan 
 EX 
 E  0.257  
  tan1 
Example B
A “semi-infinite” nonconducting rod (that is, infinite in one direction
only) has uniform linear charge density . Show that the electric field
Ep at point P makes an angle of 45o with the rod and that this result
is independent of the distance R. (Hint: Separately find the
component of Ep parallel to the rod and the component
perpendicular to the rod.)
Solution
Consider an infinitesimal section of the rod of length dx, a distance x
from the left end, as shown in the following diagram. It contains
charge dq =  dx and is a distance r from P. The magnitude of the
field it produces at P is given by
1 l dx
dE 
.
4p0 r 2
The x and the y components are
22-3
dEx  
1  dx
4 r
2
sin  and dEy  
1  dx
4 r 2
cos  ,
respectively. We use  as the variable of integration and substitute r = R/cos , x =
Rtan and dx = (R/cos2 ) d The limits of integration are 0 and /2 rad. Thus,
 
 



Ex  
sin

d


cos



40R 0
40R
40R
0
and
/2
 



Ey  
cos

d



sin


.

0
40R
40R
40R
0
We notice that Ex = Ey no matter what the value of R. Thus, E makes an angle of 45°
with the rod for all values of R.
Example C
An electron enters a region of uniform electric field with an initial velocity of 40 km/s in
the same direction as the electric field, which has magnitude E = 50 N/C. (a) What is the
speed of the electron 1.5 ns after entering this region? (b) How far does the electron
travel during the 1.5 ns interval?
Solution
a. Due to the fact that the electron is negatively charged, then (as a consequence of
Eq. 22-28 and Newton’s second law) the field E pointing in the same direction as the
velocity leads to deceleration. Thus, with t = 1.5  109 s, we find
eE
(1.6  10 19 C)(50 N/C)
v  v0  | a | t  v0 
t  4.0  10 4 m/s 
(1.5  10 9 s)
m
9.11 10 31 kg
 2.7  10 4 m/s  v
b. The displacement is equal to the distance since the electron does not change its
direction of motion. The field is uniform, which implies the acceleration is constant.
Thus,
v  v0
d
t  5.0  10 5 m
2
22-4