Question

A parallel plate capacitor has an electric field of ${10}^{5}V/m$ between the plates. If the charge on the capacitor plate is $1\mu C$, then the force on each capacitor plate is

• A. $0.1Nt$
• B. $0.05Nt$
• C. $0.02Nt$
• D. $0.01Nt$

Answer 1

Answer: (B)

$0.05Nt$

Force of attraction between capacitor plate is

$F=\frac{Q^2}{2{ϵ}_{o}A}=\frac{1}{2}QE$

putting values in above equation:

$F=0.5\times {10}^{-6}\times {10}^5$

$\Rightarrow 0.05N$