P22_007.PDF

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Chapter 22 - 22.7
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 a =
−
1
4 πε 0
q 1 q 2
r 2
=
−
k q 1 q 2
r 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 b =
1
4 πε 0
q 1 + q 2
2
q 1 + q 2
2
= k ( q 1 + q 2 ) 2
4 r 2
.
r 2
We solve the two force equations simultaneously for q 1 and q 2 . The first gives the product
q 1 q 2 =
−
r 2 F a
k
=
−
(0 . 500 m) 2 (0 . 108 N)
8 . 99
×
10 9 N
·
m 2 / C 2 =
−
3 . 00
×
10 − 12 C 2
,
and the second gives the sum
q 1 + q 2 =2 r F b
k
=2(0 . 500 m) 0 . 0360 N
8 . 99
×
10 9 N
·
m 2 / C 2 =2 . 00
×
10 − 6 C
wherewehavetakenthepositiveroot(whichamountstoassuming q 1 + q 2 ≥
0). Thus, the product
result provides the relation
q 2 = −
(3 . 00
×
10 − 12 C 2 )
q 1
which we substitute into the sum result, producing
q 1 −
3 . 00
×
10 − 12 C 2
q 1
=2 . 00
×
10 − 6 C .
Multiplying by q 1 and rearranging, we obtain a quadratic equation
q 1 −
(2 . 00
×
10 − 6 C) q 1 −
3 . 00
×
10 − 12 C 2 =0 .
The solutions are
q 1 = 2 . 00
×
10 − 6 C
±
(
−
2 . 00
×
10 − 6 C) 2
−
4(
−
3 . 00
×
10 − 12 C 2 )
.
2
If the positive sign is used, q 1 =3 . 00
×
10 − 6 C, and if the negative sign is used, q 1 =
1 . 00
×
10 − 6 C.
Using q 2 =(
−
3 . 00
×
10 − 12 ) /q 1 with q 1 =3 . 00
×
10 − 6 C, we get q 2 =
−
1 . 00
×
10 − 6 C. If we instead work
10 − 6 C. Since the spheres are identical,
the solutions are essentially the same: one sphere originally had charge
−
1 . 00
×
10 − 6 C root, then we find q 2 =3 . 00
×
−
1 . 00
×
10 − 6 C and the other
0? If
the signs of the charges were reversed (so q 1 + q 2 < 0), then the forces remain the same, so a charge of
+ 1 . 00
×
10 − 6 C. What if we had not made the assumption, above, that q 1 + q 2 ≥
×
10 − 6 C on one sphere and a charge of
−
3 . 00
×
10 − 6 C on the other also satisfies the conditions
of the problem.
−
with the q 1 =
had charge +3 . 00
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