Question

Two large conducting plates of a parallel plate capacitor are given charges $Q$ and $3Q$ respectively. If the electric field in the region between the plates is $E_o$ , then the force of interaction between the plates is :

• A. $\frac{3Q{E}_o}{2}$
• B. $3Q{E}_o$
• C. $\frac{2Q{E}_o}{3}$
• D. $\frac{Q{E}_o}{2}$

Answer 1

The correct option is A $\frac{3Q{E}_o}{2}$


Electric field between the plates due to plate having charge $Q$ is :
$E_1=\frac{{\sigma }_1}{{ϵ}_o}=\frac{Q}{{ϵ}_{o}A}$
Here, $A$ is the area of the plates.

Electric field between the plates due to plate having charge $3Q$ is :
$E_2=\frac{{\sigma }_2}{{ϵ}_o}=\frac{3Q}{{ϵ}_{o}A}$

As shown in the figure, $E_1$ and $E_2$ are opposite to each other.
The net electric field is given by :
$E_{net}=E_o=E_2-E_1$
$E_o=\frac{3Q}{{ϵ}_{o}A}-\frac{Q}{{ϵ}_{o}A}=\frac{2Q}{{ϵ}_{o}A}$
$\Rightarrow \frac{Q}{{ϵ}_{o}A}=\frac{E_o}{2}....\left(1\right)$

Force on plate having $Q$ charge due to the electric field of plate having $3Q$ charge is :
$F=Q{E}_2=Q\times \frac{3Q}{{ϵ}_{o}A}$
$F=\frac{3Q^2}{{ϵ}_{o}A}$

From equation $\left(1\right)$ :
$F=\frac{3Q{E}_o}{2}$

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