Two equal negative charges `-q` are fixed at points `(0, -a)` and `(0,a)` on y-axis. A poistive charge Q is released from rest at point `(2a, 0)` on the x-axis. The charge Q will

Options

A. (a) execute simple harmonic motion about the origin
B. (b) move to the origin remain at rest
C. (c) move to infinity
D. (d) execute oscillatory but not simple harmonic motion

Answer

D

Explanation

Let us consider the positive charge Q at any instant of time t at a distance x from the origin. It is under the influence of two forces `vedF_1(=F)` and `vecF_2(=F)`. On resolving these two forces we find that `F sin theta` cancels out. The resultant force is

`F_R=2Fcos theta`
`=2xx(kQq)/((x^2+a^2))xx(x)/(sqrt(x^2+a^2)`
`=(2kQqx)/((x^2+a^2)^(3//2))`
Since `F_R` is not proportional to x, the motion is NOT simple harmonic. The charge Q will accelerate till the origin and gain velocity. At the origin the net force is zero but due to momentum it will cross the origin and more towards left. As it comes on negative x-axis, the force is again towards the origin.