Two equal positive charges , `Q` , are fixed at points `(0 , d)` and `(0 , -d)` on the `y`-axis. A charged particle having charge `-q` and mass `m` is released from rest at point `(d,0)` on the `x`-axis . Discuss the motion of charged particle.

Explanation

In the given figure, both charges Q are equidistant from -q. Hence, both will apply the force of same magnitude as shown in the figure.
`F=1/(4piepsilon_0)(Qq)/([sqrt(d^2+y^2)]^2)=1/(4piepsilon_0)(Qq)/([d^2+y^2])`
Net force acting on the particle is `2Fcos theta` towards the centre of line AB. Let a be the acceleration of particle then,
ma=2F cos `theta`
`rArr a=2/m 1/(4piepsilon_0)(Qq)/([d^2+y^2])y/sqrt(d^2+y^2)`
`rArr a=1/(4piepsilon_0)(2Qqy)/(m[d^2+y^2]^(3//2))`
If y < < d , then `y^2` can be neglected in comparison to `d^2`. We can rewrite the above equation as :
`a=1/(4piepsilon_0)(2Qqy)/(md^3)`
We can write above magnitude of acceleration as follows :
`a=omega^2y`, where `omega=sqrt((2Qq)/(4piepsilon_0md^3))`
Above acceleration is restoring by nature , so we can write `a=-omega^2y`, hence this is S.H.M.
To calculate the time period , we have:
`T=(2pi)/omega rArr T =2pi sqrt((4piepsilon_0md^3)/(2Qq))`
`rArr T=sqrt((8pi^3 epsilon_0md^3)/(Qq))`