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11 Electrostatics
E =
1
4πε 0
q −
q
2 2 +
q
4 2 −
q
8 2 +
q
16 2 · · ·
=
q
4πε 0
1 +
1
4 2 +
1
16 2 + · · ·
−
q
16πε 0
1 +
1
4 2 +
1
16 2 + · · ·
=
1
4πε 0
q −
q
4
1 +
1
4 2 +
1
16 2 + · · ·
=
3q
16πε 0
1
1 −
1
16
=
q
5πε 0
11.9 By prob. (11.3), E =
Qx
4πε 0
x 2 + R 2
3/2 , where we have replaced z by x and
substituted λ = Q/2π R. As x << R and F = qE,
F =
Qqx
4πε 0 R 3 = −kx, where q is negative and k =
Qq
4πε 0 R 3
=
10 −5 × 10 −6 × 9 × 10 9
1 3
= 0.09
Thus, the motion of the negatively charged particle is approximately simple
harmonic with angular frequency
ω =
k
m
=
0.09
0.9 × 10 −3 = 10
T =
2π
ω
=
2π
10
= 0.628 s
11.10 Let a be the side of the equilateral triangle. The forces on the charge q placed
at C due to the charges at A and B are repulsive and represented by CE and
CD, respectively, each given by q 2 /4πε 0 a 2 . The resultant of these two forces
is given by CP, the diagonal of the parallelogram CDPE, Fig. 11.20
CP = 2CD cos 30
◦
=
2q 2
4πε 0 a 2
√
3
2
=
√
3q 2
4πε 0 a 2
The force on q at C due to Q at the centre of the triangle is
Qq
4πε 0 (OC) 2 =
3 Qq
4πε 0 a 2
i. If Q = −q, this force will be attractive and will be directed along CO. As
the attractive force due to −q is greater than the combined repulsive force
due to charges +q at A and B, the charge at C will be attracted towards
O. Same is true for the charges placed at A and B.