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

Options

A. execute simple harmonic motion about the origin
B. move to the origin and ramain at rest there .
C. move at infinity .
D. execute oscillatory but not simple harmonic motion .

Answer

D

Explanation

`:. ` The Net force on Q (F) ` = F_(1) cos theta +F_(2) cos theta ` …(1)
` :. F_(1) = F_(2) = (qQ)/(4piin_(0)r^(2))`

Let AP = BP = r & OP = x
From eqn , (1) , `F = (2pQ)/(4pi in_(0)r^(2)) cos theta = (2qQ)/(4piin_(0)r^(2))(x/r)`
` = (2qQx)/(4piin_(0)(a^(2)+x^(2))^(3//2)) " " ....(2)`
Eqn . (2) is not form F = kx
Because ` k = (2qQ)/(4pi in_(0) (a^(2)+x^(2))^(3//2)) ` is not constant here .
So , it execites oscillatory motion about pt. O but not S.H.M