875
Chapter 21
1. The magnitude of the force of either of the charges on the other is given by
F
q Q
q
r
1
4
0
2
b
g
where
r
is the distance between the charges. We want the value of
q
that maximizes the
function
f
(
q
) =
q
(
Q
–
q
). Setting the derivative
/
dF
dq
equal to zero leads to
Q
– 2
q
= 0,
or
q
=
Q
/2. Thus,
q
/
Q
= 0.500.
2. The fact that the spheres are identical allows us to conclude that when two spheres are
in contact, they share equal charge. Therefore, when a charged sphere (
q
) touches an
uncharged one, they will (fairly quickly) each attain half that charge (
q
/2). We start with
spheres 1 and 2, each having charge
q
and experiencing a mutual repulsive force
2
2
/
F
kq
r
. When the neutral sphere 3 touches sphere 1, sphere 1’s charge decreases to
q
/2. Then sphere 3 (now carrying charge
q
/2) is brought into contact with sphere 2; a total
amount of
q
/2 +
q
becomes shared equally between them. Therefore, the charge of sphere
3 is 3
q
/4 in the final situation. The repulsive force between spheres 1 and 2 is finally
2
2
2
(
/ 2)(3
/ 4)
3
3
3
0.375.
8
8
8
q
q
q
F
F
k
k
F
r
r
F
3. Equation 211 gives Coulomb’s law,
F
k
q
q
r
1
2
2
, which we solve for the distance:
9
2
2
6
6
1
2
8.99 10 N m
C
26.0 10
C
47.0 10
C



1.39 m.
5.70N
k q
q
r
F
4. The unit ampere is discussed in Section 214. Using
i
for current, the charge
transferred is
4
6
2.8
10 A
20
10
s
0.56 C.
q
it
5. The magnitude of the mutual force of attraction at
r
= 0.120 m is
6
6
1
2
9
2
2
2
2
3.00
10
C
1.50
10
C
8.99
10 N m
C
2.81N.
(0.120 m)
q
q
F
k
r
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CHAPTER 21
876
6. (a) With
a
understood to mean the magnitude of acceleration, Newton’s second and
third laws lead to
7
2
7
2
2
1
1
2
2
6.3 10
kg
6.0m s
4.2
10
kg.
9.0m s
m a
m a
m
(b) The magnitude of the (only) force on particle 1 is
2
1
2
9
2
2
1
1
2
2
8.99
10 N m
C
.
(0.0032 m)
q
q
q
F
m a
k
r
Inserting the values for
m
1
and
a
1
(see part (a)) we obtain 
q
 = 6.6
10
–11
C.
7. With rightward positive, the net force on
q
3
is
1
3
2
3
3
13
23
2
2
23
12
23
.
q q
q q
F
F
F
k
k
L
L
L
We note that each term exhibits the proper sign (positive for rightward, negative for
leftward) for all possible signs of the charges. For example, the first term (the force
exerted on
q
3
by
q
1
) is negative if they are unlike charges, indicating that
q
3
is being
pulled toward
q
1
, and it is positive if they are like charges (so
q
3
would be repelled from
q
1
). Setting the net force equal to zero
L
23
=
L
12
and canceling
k
,
q
3
, and
L
12
leads to
1
1
2
2
0
4.00.
4.00
q
q
q
q
8. In experiment 1, sphere
C
first touches sphere
A
, and they divided
up their total charge
(
Q
/2 plus
Q
) equally between them. Thus, sphere
A
and sphere
C
each acquired charge
3
Q
/4. Then, sphere
C
touches
B
and those spheres split up their total charge (3
Q
/4 plus –
Q
/4) so that
B
ends up with charge equal to
Q
/4. The force of repulsion between
A
and
B
is therefore
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 Spring '11
 Charge, Force, Electric charge, +X

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