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. $3QE_{0}$
• B. $\frac{2 Q E_{0}}{3}$
• C. $\frac{3}{2}QE_{0}$
• D. $\frac{Q E_{0}}{2}$

Answer 1

Answer: (C)

The net electric field between the plates due to both the plates is
$E_{0}=\frac{\sigma }{\epsilon _{0}}=\frac{Q}{A \epsilon _{0}}$
So the electric field due to each plate is
$E=\frac{E_{0}}{2}=\frac{Q}{2 A \epsilon _{0}}$
So the force on one plate due to the other one is
$F=QE=3Q\times \left[\frac{Q}{2 A \epsilon _{0}}\right]=\frac{3 Q^{2}}{2 A \epsilon _{0}}$
$=\frac{3 Q E_{0}}{2}$