# Electrostatics problems

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Solutions for Chapter 2 Problems

1. Vectors in the Cartesian Coordinate System
P2.1: Given P(4,2,1) and APQ=2ax +4ay +6az, find the point Q.

2. Coulomb’s Law, Electric Field Intensity, and Field Lines
P2.4: Suppose Q1(0.0, -3.0m, 0.0) = 4.0nC, Q2(0.0, 3.0m, 0.0) = 4.0nC, and Q3(4.0m,
0.0, 0.0) = 1.0nC. (a) Find the total force acting on the charge Q3. (b) Repeat the problem
after changing the charge of Q2 to –4.0nC. (c) Find the electric field intensity for parts (a)
and (b).

Q1Q2
(a) F13              a13 , where R13 = 4 ax + 3 ay =, R13 = 5m, a13 = 0.8 ax + 0.6 ay.
4 o R13
2

so

 4 x10 C 1x10 C  
9         9   4a  3a 
x       y
5     FV NM
F13   
4 10 F 36 m   5m 
9                2
C VC
 1.15 x109 ax  0.86 x109 a y N .
Similarly, F23  1.15x109 ax  0.86 x109 ay N , so FTOT  2.3ax nN

(b) with Q2 = -4 nC, F13 is unchanged but F23  1.15x109 ax  0.86x109 ay N , so
FTOT  1.7ay nN .

FTOT  2.3x10 ax N  VC
9
V
(c) Ea                          2.3ax .
Q3     1x10 C  Nm
-9
m
V
Likewise, Eb  1.7a y       .
m

Fig. P2.4

4. Line Charges and the Cylindrical Coordinate System
P2.12: Convert the following points from Cartesian to cylindrical coordinates:
a. P(0.0, 4.0, 3.0)
b. P(-2.0, 3.0, 2.0)
c. P(4.0, -3.0, -4.0)
4
(a)   02  42  4,   tan 1    90o , z  3, so P(4.0,90o ,3.0)
0
 3 
(b)   22  32  3.6,   tan 1    124o , z  2, so P(3.6,124o , 2.0)
 2 
2-2

 3 
(c)   42  32  5,   tan 1    37 o , z  4, so P(5.0, 37 o , 4.0)
 4 

P2.13:   Convert the following points from cylindrical to Cartesian coordinates:
a.    P(2.83, 45.0, 2.00)
b.    P(6.00, 120., -3.00)
c.    P(10.0, -90.0, 6.00)

(a)
x   cos   2.83cos 45o  2.00
y   sin   2.83sin 45o  2.00
z  z  2.00
so P(2.00, 2.00, 2.00).
(b)
x   cos   6.00 cos120o  3.00
y   sin   6.00sin120o  5.20
z  z  3.00
so P(3.00,5.20, 3.00).
(c)
x   cos   10.0cos(90.0o )  0
y   sin   10.0sin(90.0o )  10.0
z  z  6.00
so P(0, 10.0, 6.00).

P2.15: A line charge with charge density 2.00 nC/m exists at y = -2.00 m, x = 0.00. (a) A
charge Q = 8.00 nC exists somewhere along the y-axis. Where must you locate Q so that
the total electric field is zero at the origin? (b) Suppose instead of the 8.00 nC charge of
part (a) that you locate a charge Q at (0.00, 6.00m, 0.00). What value of Q will result in a
total electric field intensity of zero at the origin?

(a) The contributions to E from the line and point charge must cancel, or E  E L  EQ .

For the line: EL 
L
a 
 2nC / m             V
a y  18 a y
2 o 
9
2 10 F
36 m  
2m         m

and for the point charge, where the point is located a distance y along the y-axis, we
 8nC   a 
have: EQ 
Q
 a y   109 F y 2  72  a y 
4 o y 2

4             
36 m
y y2

Therefore:
2-3

72               72
2
 18, or y      2m.
y                18
So Q  0,2.0m,0 

(b)
Q
 18,
4 o  6 
2

Q
18  36   72nC.
9
Fig. P2.15

P2.16: You are given two z-directed line charges of charge density +1 nC/m at x = 0, y =
-1.0 m, and charge density –1.0 nC/m at x = 0, y = 1.0 m. Find E at P(1.0m,0,0).

The situation is represented by Figure P2.16a. A better 2-dimensional view in Figure
P2.16b is useful for solving the problem.
 L1                       a  ay 
E1          a  , and  a   2  x      .
2 o                          2 

E1 
1x10 C 
9
a   x    a y  FV
 9  a x  a y  , and E2  9  -a x  a y  .
V                           V
9
2 10 F
36 m             2m              2      C                   m                           m

So ETOT = 18 ay V/m.

P2.18: A segment of line charge L =10 nC/m exists on the y-axis from the origin to y =
+3.0 m. Determine E at the point (3.0, 0, 0)m.

It is clear from a sketch of the problem in Figure P2.18a that the resultant field will be
directed in the x-y plane. The situation is redrawn in a temporary coordinate system in

Fig. P2.16b
Fig. P2.16a
2-4

Figure P2.18b.

Fig. P2.18a                                                         Fig. P2.18b

 L dz a   za z
We have from Eqn (2.34) E                                             E a   Ez a z .
4 o  2  z 2 3 2
        
For E we have:
3
                    dz                          z        
E  L
4 o                                  L                   
4 o   2 z 2   2 
        z2 
3
               0
2                2

With  = 3, we then have E = 21.2 V/m.
For Ez:
3
L                  zdz                        1                 V
Ez 
4 o                                 L 
4 o   2  z 2
  8.79

        z2 
3
2                2
              z 0    m

Thus we have ETOT = 21 a – 8.8 az V/m.
Converting back to the original coordinates, we have ETOT = 21 ax – 8.8 ay V/m.

5. Surface and Volume Charge
P2.19: In free space, there is a point charge Q = 8.0 nC at (-2.0,0,0)m, a line charge L =
10 nC/m at y = -9.0m, x = 0m, and a sheet charge s = 12. nC/m2 at z = -2.0m.
Determine E at the origin.

The situation is represented by Figure P2.19, and
the total field is ETOT = EQ + EL + ES.

EQ 
Q
aR 
8x109 C  a x
4 o R 2           9
4 10 F      
36 m  
2m
2

V
 18a x
m

Fig. P2.19
2-5

EL 
L
a 
10 x10 C m  a y           9

2 o             9
2 10 F         
36 m  
9m                        
V
 20a y
m

Es 
s
aN 
12 x10 C m  a        9         2

2 o

9

z
2 10 F
36 m
V
 679a z
m
So: Etot = 18 ax + 20 ay + 680 az V/m.

P2.20: An infinitely long line charge (L = 21 nC/m) lies along the z-axis. An infinite
area sheet charge (s = 3 nC/m2) lies in the x-z plane at y = 10 m. Find a point on the y-
axis where the electric field intensity is zero.

We have ETOT = EL + ES.
a      a
EL  L   L y
2 o  2 o y


 21 x10   9
C / m
ay 
378

                       y
9
ay
2 10 F                                         y
36 m
a
Es  s N 
3x10 C m           9             2
  a  y

2 o        9
2 10 F                         36 m     
 54 a y
Fig. P2.20
so
378
 54  0, or y  7.
y
Therefore, P(0, 7m, 0).

6. Electric Flux Density
P2.29: Given D = 2a + sin  az C/m2, find the electric flux passing through the surface
defined by 2.0 ≤ ≤ m, 90. ≤ ≤ 180, and z = 4.0 m.

   E dS, dS   d  da z
4    
    2 a   sin a z   d  d a z    d                        sin  d  6C
2    2
2-6

7. Gauss’s Law and Applications
P2.31: Given a 3.00 mm radius solid wire centered on the z-axis with an evenly
distributed 2.00 coulombs of charge per meter length of wire, plot the electric flux
density D versus radial distance from the z-axis over the range 0 ≤  ≤ 9 mm.

2 C m            C
For a 1 m length, v             70.7 x103 3 ,  a  3mm  .003m 
a 2
m
Qenc     D dS   D a  d dza  2 LD , where L is the length of the Gaussian
surface. Note that this expression for Qenc is valid for both Gaussian surfaces.
GS1 ( < a):
      2   L
Qenc   v dv  v  d   d  dz  v  2 L
0      0    0

v   L v
2
so D                   for   a.
2 L      2
GS2 ( > a):
 a2 1
Qenc   v a 2 L, D  v      for   a.
2 

P2.32: Given a 2.00 cm radius solid wire centered on the z-axis with a charge density v
= 6 C/cm3 (when  is in cm), plot the electric flux density D versus radial distance
from the z-axis over the range 0 ≤  ≤ 8 cm.

Choose Gaussian surface length L, and as usual we have
Qenc   D dS =  D a   d dza   2 L D , valid for both Gaussian surfaces.
In GS1 ( < a): Qenc   v dv   6 2 d  d dz  4 L 3 ,
4 L  3
so D              2  2 for   a.
2 L 
2a 3
For GS2 (> a): Qenc  4 La 3 , D              for   a.

2-7

Fig. P2.32

9. Electric Potential
P2.42: A sheet of charge density s = 100 nC/m2 occupies the x-z plane at y = 0. (a)
Find the work required to move a 2.0 nC charge from P(-5.0m, 10.m, 2.0m) to M(2.0m,
3.0m, 0.0). (b)Find VMP.

M
(a) W  Q  E dL; so we need E for the sheet charge.
P


E  s aN 
100 x10 C  FV a  5.65x103 V a
9

2 o    2  8.854 x1012 F m  C
y            y
m
Notice that we are only concerned with movement in the y-direction. We then have:
y 3
      V            J 
W  2 x109 C   5.65 x103 a y  dya y       79 J
y 10       m            CV 

(b) VMP 
W

 79 J  CV  39.5kV ; so V  40kV .
Q  2 x109 C  J
MP

P2.45: A 100 nC point charge is located at the origin. (a) Determine the potential
difference VBA between the point A(0.0,0.0,-6.0)m and point B(0.0,2.0,0.0)m. (b) How
much work would be done to move a 1.0 nC charge from point A to point B against the
electric field generated by the 100 nC point charge?

A
(a) VBA   E dL.
A
2-8

The potential difference is only a function of radial distance from the origin. Letting ra =
6m and rb = 2m, we then have
Q 1 1
rb
Q
VBA              a drar            300V .
ra
4 o r 2 r
4 o  rb ra 

(b) W  Q2VBA  109 C   300V 
J
 300nJ
CV

P2.47: For an infinite length line of charge density L = 20 nC/m on the z-axis, find the
potential difference VBA between point B(0, 2m, 0) and point A(0, 1m, 0).

B
L
VBA    E dL; E               a  , dL  d  a  ,
A
2 o 
B
L                 L
so VBA             a  d a         ln  2   250V
A
2 o               2 o

P2.48: Find the electric field at point P(0.0,0.0,8.0m) resulting from a surface charge
density s = 5.0 nC/m2 existing on the z = 0 plane from = 2.0 m to  = 6.0 m. Assume
V = 0 at a point an infinite distance from the origin.

(Method 1)
For a ring of charge it was previously found that
 L aha z
E                        .
2 o  a  h 
3
2      2   2

We can then break up our disk into differential rings (see Figure P2.48), each
contributing dE as:
 h     d 
dE  S                   a z , where we've used  L   S d  .
2 o  2  h 2 3 2
        
So we then have
 ha        d 
E S z                              .
2 o
        
3
 2  h2            2

This is easy to integrate if we let u = 2 + h2, then du = 2 d, and we have
b
 ha    3      ha 2   S ha z                      1
E  S z  u 2 du  S z    
4 o           4 o u   2 o                         2  h2   a
Solving, we arrive at
 h      1           1     
E S                           .
2 o  b2  h2     a 2  h2 
2-9

Upon inserting the appropriate values we find E = 48 V/m az.

Fig. P2.48

(method 2)
Find an expression for potential and then evaluate the gradient at the point.
dQ                                                   S  d  d
V           , R   2  h2 , dQ  S  d  d , so V= 
4 o R                                             4 o  2  h2
S     b
d            S           b

V=                                 2  h2  S  b2  h2  a 2  h2  .
2 o   a    2  h2       2 o         a 2 o                   
Now we let h = z and E  V ;
                                  
E   S   b2  z 2  2   a 2  z 2  2  a z
1                 1

2 o  z            z               
S 1 2 2  12         1 2 2  12              S             
 2  b  z  2 z  2  a  z  2 z  a z   2  2 2  2 2  a z
z    z

2 o
                                              o  b z a z 
Plugging in the values we find E = 48 V/m az.

12. Boundary Conditions
P2.62: For y < 0, r1 = 4.0 and E1 = 3ax + 6ay + 4az V/m. At y = 0, s = 0.25 nC/m2. If
r2 = 5.0 for y > 0, find E2.
E1 = 3ax + 6ay + 4az V/m                    (g) E2 = 3ax + 20.7ay + 4az V/m
(a) EN1 = 6ay                               (f) EN2 = DN2/5o = 20.7ay
(b) ET1 = 3ax + 4az                          (c) ET2 = ET1 = 3ax + 4az
(d) DN1 = r1oEN1 = 24o ay                (e) DN2 = 0.92 ay

(e) a21  D1  D2   s , -a y  DN1  DN 2  a y  s , DN 2  DN1  s
nC        109 F  nC        nC
DN 2   s  DN 1  0.25            24           2  0.92 2
 36 m  m
2
m                             m
2-10

P2.63: For z ≤ 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes a 30 angle with a normal
to the surface, what angle does E2 make with a normal to the surface?

Refer to Figure P2.63.
ET 1  E1 sin 1 , ET 2  E2 sin  2 , and ET 1  ET 2
also
DN1   r1 o E1 cos1, DN 2   r 2 o E2 cos2 , and DN1  DN 2 since s  0
Therefore
ET 1 ET 2                                                             
       , and after routine math we find  2  tan 1  r 2 tan 1 
DN 1 DN 2                                                    r1       
Using this formula we obtain for this problem 2 = 14°.

Fig. P2.63

P2.64: A plane defined by 3x + 2y + z = 6 separates two dielectrics. The first dielectric,
on the side of the plane containing the origin, has r1 = 3.0 and E1 = 4.0az V/m. The
other dielectric has r2 = 6.0. Find E2.

We first use gradient to find a normal to the planar surface.
Let F = 3x + 2y + z – 6 = 0.
F  3a x  2a y  a z , and F  14,
F
so a N         0.802a x  0.534a y  0.267a z .
F
Now we can work the boundary condition problem.
E1  4az , EN1   E1 aN  aN  0.857ax  0.570a y  0.285az .
ET 1  E1  E N 1  0.857a x  0.570a y  3.715a z , ET 2  ET 1
DN1   r1 o EN1   o 2.571a x  1.710a y  0.855a z  , and DN 2  DN1
                               
DN 2           DN 2
EN 2                        0.429a x  0.285a y  0.143a z
 r 2 o       6 o
V
Finally we have E2  ET 2  E N 2  0.43a x  0.29a y  3.8a z       .
m

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