Question

A parallel plate capacitor has a plate area $A$, separation $d$ and charge $Q$. A positive charge $q$ of mass $m$ is released from a point very near to the positive plate. The time taken by the charge to reach to the negative plate of capacitor is
(Neglect gravity)

• A. $\sqrt{\frac{3dm{ϵ}_{0}A}{qQ}}$
• B. $\sqrt{\frac{4dm{ϵ}_{0}A}{qQ}}$
• C. $\sqrt{\frac{2dm{ϵ}_{0}A}{qQ}}$
• D. None of these

Answer 1

The correct option is C $\sqrt{\frac{2dm{ϵ}_{0}A}{qQ}}$


We know the electric field inside the parallel plate capacitor is

$E=\frac{\sigma }{2{ϵ}_0}+\frac{\sigma }{2{ϵ}_0}=\frac{\sigma }{{ϵ}_0}$

Here, $\sigma =\frac{Q}{A}$

Force experienced by a small charge $q$ near the plate will be

$F=qE$

$\Rightarrow F=q\left(\frac{Q}{A{ϵ}_0}\right)=\frac{Qq}{A{ϵ}_0}$

Since force is constant i.e. acceleration of particle will be constant.

$\therefore a=\frac{F}{m}=\frac{Qq/A{ϵ}_0}{m}=\frac{Qq}{A{ϵ}_{0}m}$

Applying kinetic equation

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow d=(0\times t)+\frac{1}{2}\left(\frac{Qq}{mA{ϵ}_0}\right)t^2$

$\therefore t=\sqrt{\frac{2dmA{ϵ}_0}{Qq}}$

Hence, option $\left(c\right)$ is correct.