Two equal negative charges -q are fixed at points (0,a) and (0,-a) on the Y-axis. A positive charge Q is released from rest at the point (2a,0) on the X-axis. Will the charge Q execute simple harmonic motion?

Explanation

Charge Q will move to and fro along ROR' [Fig.2.72]. The point O is the equilibrium position of the charge Q. Now we shall have to see whether the force acting on the charge Q is proportional to the distance of Q from O.
Let the charge Q be at the point P(x,0) at any instant. At that time each of the two charges -q exerts force F' on Q.

Therefore, from Coulomb's law we get,
`F'prop-(1)/(a^(2)+x^(2))`
Now, `F=2F'costheta=2F'xx(x)/(sqrt(a^(2)+x^(2)))`
or, `F'=(sqrt(a^(2)+x^(2)))/(2x)F`
`:.(Fsqrt(a^(2)+x^(2)))/(2x)prop-(1)/(a^(2)+x^(2))`
or, `Fprop-(x)/((a^(2)+x^(2))^(3//2))`
The condition for a simple harmonic motion, `Fprop-x,` does not match with the above expression. So, the charge Q will not execute a simple harmonic motion.