Two small equal point charges of magnitude $$q$$ are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle $$\theta$$ from the vertical. If the mass of each charge is $$m$$, then the electrostatic potential at the centre of line joining them will be $$\left(\frac{1}{4\pi\epsilon_0} = k\right)$$.

Let distance between charges be

2r

So midpoint is at distance

r

from each charge.

Potential at midpoint due to one charge:

$$V_1=\frac{kq}{r}$$

From both charges,

$$V=\frac{2kq}{r}$$

Now find r.

For one charge in equilibrium:

Horizontal balance:

$$T\sin\theta=F_e$$

Vertical balance:

$$T\cos\theta=mg$$

Dividing,

$$\tan\theta=\frac{F_e}{mg}$$

Electrostatic repulsion between charges:

$$F_e=\frac{kq^2}{(2r)^2}=\frac{kq^2}{4r^2}$$

So

$$mg\tan\theta=\frac{kq^2}{4r^2}$$

Thus

$$r^2=\frac{kq^2}{4mg\tan\theta}$$

$$r=\frac{q}{2}\sqrt{\frac{k}{mg\tan\theta}}$$

Substitute into

$$V=\frac{2kq}{r}$$

$$V=\frac{2kq}{\frac{q}{2}\sqrt{\frac{k}{mg\tan\theta}}}$$

$$=4k\sqrt{\frac{mg\tan\theta}{k}}$$

$$=4\sqrt{kmg\tan\theta}$$