Homework 1
Chapter 19
Problem 7. Two identical conducting small spheres are placed with their centers r = 0.300 m apart. One is given a charge
of q1 = 12.0 nC and the other a charge of q2 = −18.0 nC. (a) Find the electric force exerted by one sphere on the other.
(b) Next, the spheres are connected by a conducting wire. Find the electric force between the two after they have come to
equilibrium.
(a)
F
q1
q2
r
q1 q2
12.0 · 10−9 C · (−18.0) · 10−9 C
F = ke 2 = 8.99 · 109 N·m2 /C2
= −2.16 · 10−5 N
r
(0.300 m)2
And the force is towards the other sphere for each sphere because opposites attract.
(b)
(1)
F
Q/2
Q/2
r
The total charge on the both spheres is Q = q1 + q2 = −6.0 nC. The spheres are identical, so at equilibrium, there will be
Q/2 = −3.0 nC on each sphere. The repulsive (since now they have the same charge sign) force between them is given by
2
−3.0 · 10−9 C
(Q/2)2
9
2
2
F = ke
=
8.99
·
10
N·m
/C
= 8.99 · 10−7 N
(2)
r2
0.300 m
Problem 13. Three point charges are arranged as shown in Figure P19.13.
(a) Find the vector electric field E that q2 and q3 together create at the origin. (b) Find the vector force F on q1 .
0.300 m
E21
0.100 m
ĵ
î
q1 = 5 nC
q2 = 6 nC
E31
q3 = −3 nC
(a)
X qi
q3
q2
9
2
2
E = ke
2 r̂i = ke x2 (−î) + y 2 ĵ = 8.99·10 N·m /C
r
2
3
i
i
3.00ĵ
−6.00î
−
2
0.300
0.1002
!
·10−9 C/m2 = −0.599î − 2.70ĵ kN/C (3)
(b)
F = q1 E = −3.00î − 13.5ĵ µN
(4)
Problem 16. Consider the electric dipole shown in Figure P19.16. Show that the electric field at a distant point on the +x
axis is Ex ≈ 4ke qa/x3 .
-q
q
0
E−q
Eq
x
a
a
Let us assume the point in question has a positive x value (just reverse the sign if x < 0).
X qi
q
−q
E = ke
r̂i = ke
î +
î
ri2
(x − a)2
(x + a)2
i
(5)
For |x| |c|,
n
(x + c) = x
n
c n
c
n(n − 1) c 2
c
n
1+
=x 1+n +
·
+ . . . ≈ xn (1 + n ) ,
x
x
2
x
x
(6)
because (c/x)2 is very, very small. (We are Taylor expanding (x + c)n as a function of c/x aout c/x = 0, and keeping only
the first two terms.) In our case, n = −2 and c = ∓a
q
−a
−q a
q a
a
4ke qa
E = ke 2 1 − 2
+ 2 1−2
î = ke 2 1 + 2 − 1 + 2
î =
î
(7)
x
x
x
x
x
x
x
x3