Question

Two parallel conducting plates, each of area $A$, are separated by a distance $d$. Now, the left plate is given a positive charge $Q$. A positive charge $q$ of mass $m$ is released from a point near the left plate. Find the time taken by the charge to reach the right plate.

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

Answer 1

The correct option is B $\sqrt{\frac{4dm{\in }_{0}A}{qQ}}$

Let us consider a capacitor with charge $Q$ on the left plate and a charge $q$ of mass $m$ is placed near its left plate. The separation between two plates is $d$.
Now, as there is no charge on the right plate so the electric field will only be due to the left plate .
The electric field due to a charge sheet at any point is given by
$E=\frac{\sigma }{2{ϵ}_0}$
$⟹E=\frac{Q}{2A{ϵ}_0}$------($\because \sigma =\frac{Q}{A}$)
The acceleration of the charge particle due to this electric field can be given by
$a=\frac{qE}{m}$
$⟹a=\frac{qQ}{2A{ϵ}_{0}m}$
Using second equation of motion we can write
$s=ut+\frac{1}{2}a{t}^2$
$s=d$ (i.e separation between the plates)
Putting values of $s$ and $a$ we get
$t=\sqrt{\frac{4Adm{ϵ}_0}{Qq}}$