Question

Two charges each of magnitude Q are fixed at 2a distance apart and a third charge q of mass m is placed at the
mid point of the line joining the two charges; all charges being of the same sign. If a charge is slightly displaced
from its position along the line and released then determine its time period.​

Answer 1

Answer: you displace q then

Explanation:

The answer is in the picture attached.

Answer 2

Given that,

Two charges each of magnitude Q are fixed.

Distance = 2a

Third charge = q

Mass = m

When q is displaced x towards A

Then force experienced by q due to charge at A

$F_{1}=\dfrac{kQq}{(a-x)^2}$

The force experienced by q due to charge at B

$F_{2}=\dfrac{kQq}{(a+x)^2}$

Net force on q

$F=F_{1}-F_{2}$

$F=\dfrac{kQq}{(a-x)^2}-\dfrac{kQq}{(a+x)^2}$

$F=\dfrac{kQq(a+x)^2-(a-x)^2}{(a-x)^2(a+x)^2}$

$F=\dfrac{KQq(a^2+x^2+2ax)-(a^2+x^2-2ax)}{(a^2-x^2)^2}$

$F=\dfrac{KQq(4ax)}{(a^2)^2}$       (A>>>>x)

$F=\dfrac{4KQqx}{a^3}$.......(I)

The restoring force is

$F=kx$

$F=m\omega^2 x$......(II)

From equation (I) and (II)

$m\omega^2 x=\dfrac{4KQqx}{a^3}$

$\omega=\sqrt{\dfrac{4KQq}{ma^3}}$

We need to calculate the time period

Using formula of time period

$T=\dfrac{2\pi}{\omega}$

$T=2\pi\sqrt{\dfrac{ma^3}{4KQq}}$

Hence, The time period is $2\pi\sqrt{\dfrac{ma^3}{4KQq}}$