Figure shows a large conducting ceiling having uniform charge density $\sigma$ below which a charge particle of charge $q_0$ and mass m is hung from point O, through a small string of length $L$. Calculate the minimum horizontal velocity v required for the string to become horizontal.

The correct option is A $v=\sqrt{\frac{2mgl+2q_0\times \frac{\sigma }{2{ϵ}_0}\times L}{m}}$
Electric field due to a long conducting ceiling = $\frac{\sigma }{2{ϵ}_0}$ (where $\sigma$ is the uniform charge density).
Force due to this electric field on charge $q_0$(in downward direction) = $q_0\times \frac{\sigma }{2{ϵ}_0}$----(i)
The coordinates of $q_0$ when the string becomes horizontal is given by
$mg$ also acts on charge $q_0$ in downward direction while the charge moves from $(0,0$ to ($L,L$).
At point ($L,L$), velocity of $q_0$= 0 as we gave minimum velocity to $q_0$ so that the string becomes horizontal.
By applying work-energy theoream
$\sum W=\Delta K.E$
$mg.(-L)+\frac{\sigma }{2{ϵ}_0}.(-L)=0-\frac{1}{2}m{v}^2$
$⟹v^2=\frac{2mgl+2q_0\times \frac{\sigma }{2{ϵ}_0}\times L}{m}$
$\therefore v=\sqrt{\frac{2mgl+2q_0\times \frac{\sigma }{2{ϵ}_0}\times L}{m}}$