Download Chapter 21 #8

Review for exam #1
Chapter 21 #8
In Fig. 21-23, four particles form a square. The charges are
and
.
(a) What is
if the net electrostatic force on particles 1 and 4 is zero? (b) Is there any
value of q that makes the net electrostatic force on each of the four particles zero? Explain.
8. For ease of presentation (of the computations below) we assume Q > 0 and q < 0
(although the final result does not depend on this particular choice).
(a) The x-component of the force experienced by q1 = Q is


Q  Q 
| q |  Q   Q | q |  Q / | q | 



F1x 

cos 45 


 1
2
 4 a 2  2 2
4 0 
a2

0

2a


1


which (upon requiring F1x = 0) leads to Q / | q | 2 2 , or Q / q  2 2  2.83.
(b) The y-component of the net force on q2 = q is


| q |  Q  

1  | q |2
| q |2  1
Q
F2 y 
sin
45



 
2
2
2 


4 0  2a
a
 4 0 a  2 2 | q | 




which (if we demand F2y = 0) leads to Q / q  1/ 2 2 . The result is inconsistent with
that obtained in part (a). Thus, we are unable to construct an equilibrium configuration
with this geometry, where the only forces present are given by Eq. 21-1.
Chapter 21 #28
Two tiny, spherical water drops, with identical charges of
, have a centerto-center separation of 1.00 cm. (a) What is the magnitude of the electrostatic force acting
between them? (b) How many excess electrons are on each drop, giving it its charge
imbalance?
28. (a) Eq. 21-1 gives
F
2
8.99  10 9 N  m
(1.00  10 16 ) 2 C 2
C2
 8.99  10 19 N .
2 2
2
(1.00  10 ) m
(b) If n is the number of excess electrons (of charge –e each) on each drop then
q
100
.  1016 C
n 
 625.
e
160
.  1019 C
Chapter 22 #7
Two particles are fixed to an x axis: particle 1 of charge
C at
cm and
particle 2 of charge
at
cm. At what coordinate on the axis is the net
electric field produced by the particles equal to zero?
7. At points between the charges, the individual electric fields are in the same direction
and do not cancel. Since charge q2= 4.00 q1 located at x2 = 70 cm has a greater
magnitude than q1 = 2.1 108 C located at x1 = 20 cm, a point of zero field must be closer
to q1 than to q2. It must be to the left of q1.
Let x be the coordinate of P, the point where the field vanishes. Then, the total electric
field at P is given by
| q1 | 
1  | q2 |
E


.
4 0  ( x  x2 ) 2  x  x1 2 
If the field is to vanish, then
| q2 |
| q1 |
| q2 | ( x  x2 ) 2



.
( x  x2 )2  x  x1 2
| q1 |  x  x1 2
Taking the square root of both sides, noting that |q2|/|q1| = 4, we obtain
x  70 cm
 2.0 .
x  20 cm
Choosing –2.0 for consistency, the value of x is found to be x = 30 cm.
Chapter 22 #9
In Fig. 22-32 , the four particles form a square of edge length
cm and have charges
. In unit-vector
notation, what net electric field do the particles produce at the square's center?
9. The x component of the electric field at the center of the square is given by
 | q1 |
| q3 |
| q2 |
| q4 | 



cos 45

2
2
2
2
(
a
/
2)
(
a
/
2)
(
a
/
2)
(
a
/
2)


1
1
1

| q1 |  | q2 |  | q3 |  | q4 |
2
4p 0 a / 2
2
Ex 
1
4p 0
 0.
Similarly, the y component of the electric field is
| q3 |
| q1 |
| q2 |
| q4 | 
1 




 cos 45
2
2
2
4 0  (a / 2) (a / 2) ( a / 2) ( a / 2) 2 
1
1
1

  | q1 |  | q2 |  | q3 |  | q4 |
2
4 0 a / 2
2
Ey 

8.99 10
9
N  m 2 / C2  (2.0 10 8 C) 1
 1.02 105 N/C.
2
(0.050 m) / 2
2
ˆ
Thus, the electric field at the center of the square is E  Ey ˆj  (1.02 105 N/C)j.
Chapter 22 #11
Figure 22-34 shows two charged particles on an x axis:
at
and
at
. What are the (a) magnitude and (b)
direction (relative to the positive direction of the x axis) of the net electric field produced at
point P at
?
11. (a) The vertical components of the individual fields (due to the two charges) cancel,
by symmetry. Using d = 3.00 m and y = 4.00 m, the horizontal components (both
pointing to the –x direction) add to give a magnitude of
Ex ,net
2|q|d
2(8.99 109 N  m2 C2 )(3.20 1019 C)(3.00 m)


4 0 (d 2  y 2 )3/ 2
[(3.00 m) 2  (4.00 m) 2 ]3/ 2
.
 1.38 1010 N/C .
(b) The net electric field points in the –x direction, or 180 counterclockwise from the +x
axis.
Chapter 23 #2
An electric field given by
pierces a Gaussian cube of edge length
and positioned as shown in Fig. 23-5. (The magnitude
is in Newton per coulomb and
the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face,
(c) left face, and (d) back face? (e) What is the net electric flux through the cube?

2. We use   E dA and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m.
(a) On the top face of the cube y = 2.0 m and dA   dA ĵ . Therefore, we have


2
E  4iˆ  3  2.0   2 ˆj  4iˆ  18jˆ . Thus the flux is

top
E  dA  
top
 4iˆ 18jˆ   dA ˆj  18
top
dA   18 2.0  N  m2 C  72 N  m2 C.
2


(b) On the bottom face of the cube y = 0 and dA  dA( j ) . Therefore, we have
. Thus, the flux is

bottom
E  dA  
bottom
 4iˆ  6jˆ   dA  ˆj  6
bottom
dA  6  2.0  N  m2 C  24 N  m 2 C.
2
 
(c) On the left face of the cube dA   dA î . So
   Eˆ  dA  
left
left
 4iˆ  E ˆj   dA  ˆi   4
y
bottom
dA  4  2.0  N  m2 C  16 N  m2 C.
2
 
(d) On the back face of the cube dA   dA k̂ . But since E has no z component
E  dA  0 . Thus,  = 0.
(e) We now have to add the flux through all six faces. One can easily verify that the flux
through the front face is zero, while that through the right face is the opposite of that
through the left one, or +16 N·m2/C. Thus the net flux through the cube is
 = (–72 + 24 – 16 + 0 + 0 + 16) N·m2/C = – 48 N·m2/C.