Question

Let Q represent the charge on a parallel plate capacitor and E the electric field between the plates, then each plate of the capacitor experiences a force of magnitude

• A. zero
• B. $ QE $
• C. $ \frac{1}{2}QE $
• D. $ 2QE $

Answer 1

Answer: (C)

: Intensity of field $ E=\frac{q}{{{\varepsilon }_{0}}A}=\frac{Q}{{{\varepsilon }_{0}}A} $ Force between plates $ =\frac{{{q}^{2}}}{2{{\varepsilon }_{0}}A}=\frac{{{Q}^{2}}}{2{{\varepsilon }_{0}}A} $ $ =\frac{q}{2}\times \frac{q}{{{\varepsilon }_{0}}A}=\frac{qE}{2}=\frac{QE}{2} $