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881
HOMEWORK SOLUTIONS
PHY108 Fall 08 (Due on 93)
4. 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
22
/
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)(3 / 4)
3
3
'
3
0.375.
8
8
8
q
q
q
F
F
k
k
F
r
r
F
5. 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.
7. We assume the spheres are far apart. Then the charge distribution on each of them is
spherically symmetric and Coulomb’s law can be used. Let
q
1
and
q
2
be the original
charges. We choose the coordinate system so the force on
q
2
is positive if it is repelled by
q
1
. Then, the force on
q
2
is
F
q q
r
k
q q
r
a
1
4
0
1
2
2
1
2
2
where
r
= 0.500 m. The negative sign indicates that the spheres attract each other. After
the wire is connected, the spheres, being identical, acquire the same charge. Since charge
is conserved, the total charge is the same as it was originally. This means the charge on
each sphere is (
q
1
+
q
2
)/2. The force is now one of repulsion and is given by
F
r
k
q
q
r
b
q
q
q
q
1
4
4
0
2
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 Spring '08
 IASHVILI
 Physics, Charge, Work

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