# Two positive point charges Q are held on the x-axis at x = a and at x = -a

S.P.P
I've been doing this question, but cant seem to get past a particular bit. heres the question:

'Two positive point charges Q are held on the x-axis at x = a and at x = -a. A third positive point charge q, of mass m, is placed on the x-axis away from the origin at coordinate x such that |x| << a. The charge q, which is free to move along the x-axis, is then released.

Find the frequency of oscillation of the charge q (hint, use the binomial expansion (1+z)^n = 1 + nz + (n(n+1)Z^2)/2, valid for the case |z| < 1.)

Right, ive no idea where the binomial expansion comes into play, but heres what i got so far:

in SHM F = kx where k is the spring constant. And in electric fields, F = KqQ/r^2 where K = 9*10^9 N.m^2/C^2

if q is placed slightly to the right, of the origin, it will have two forces acting on it. A force from Q nearest to it, and a force from the other Q on the left side of the origin. They will both be repealing it.

The net force at any given position on q would be:

F = KqQ( 1/(a-d)^2 - 1/(a+d)^2 ) where d is the distance of q from the origin. (Call this eqn 1)

Since F = kx where x = d ie F = kd, i should be able to divide eqn 1 by d and have k. But im pretty sure ive done something wrong up to here, as k is constant for a given system, but varying the length d, gives a different answer of k.

Any help? or any other ways to do this would be very helpful.

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## Answers and Replies

Staff Emeritus
Gold Member

Originally posted by S.P.P
The net force at any given position on q would be:

F = KqQ( 1/(a-d)^2 - 1/(a+d)^2 ) where d is the distance of q from the origin. (Call this eqn 1)

You're almost there. All you have to do is set F equal to m(d2x/dt2). Of course, you won't be able to solve that exactly, but this is where the binomial expansion comes in handy.

For |d|<<a, we have:

(a-d)-2~1+2d
(a+d)-2~1-2d

Can you try to solve the equation of motion from there?