Two infinite non-conducting sheets with uniform surface charge densities $+\sigma$ and $-\sigma$ are placed parallel to each other at a distance $d$. A particle of mass $m$ and charge $q$ is released from rest from the positive plate. The time taken to reach the negative plate is

Solution

1. The electric field $E$ between two infinite parallel sheets with surface charge densities $+\sigma$ and $-\sigma$ is given by $E=\frac{\sigma }{{ϵ}_0}$.
2. Using Newton's second law, the acceleration $a$ of the particle is $$\begin{array}{rl}a& =\frac{F}{m}\\ & =\frac{qE}{m}\\ & =\frac{q\sigma }{m{ϵ}_0}\end{array}$$.
3. Applying the kinematic equation $d=\frac{1}{2}a{t}^2$ for a particle starting from rest, we solve for time: $$\begin{array}{rl}t& =\sqrt{\frac{2d}{a}}\\ & =\sqrt{\frac{2d}{q\sigma /m{ϵ}_0}}\end{array}$$.
4. Simplifying the expression yields $t=\sqrt{\frac{2md{ϵ}_0}{q\sigma }}$.
5. Hence the answer is (A).