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 $E_0$ , then the force of interaction between the plates is

• A. $3Q{E}_0$
• B. $\frac{2Q{E}_0}{3}$
• C. $\frac{3}{2}Q{E}_0$
• D. $\frac{Q{E}_0}{2}$

Answer 1

Answer: (C)

The net electric field between the plates due to both the plates is
$E_0=\frac{\sigma }{{ϵ}_0}=\frac{Q}{A{ϵ}_0}$
So the electric field due to each plate is
$E=\frac{E_0}{2}=\frac{Q}{2A{ϵ}_0}$
So the force on one plate due to the other one is
$F=QE=3Q\times \left[\frac{Q}{2A{ϵ}_0}\right]=\frac{3Q^2}{2A{ϵ}_0}$
$=\frac{3Q{E}_0}{2}$