Question

The charge on a parallel plate capacitor is varying as $q=q_0\sin \left(2\pi nt\right)$. The plates are very large and close together. neglecting the edge effects, the displacement current through the capacitor is:

• A. $\frac{q}{{\epsilon }_{0}A}$
• B. $\frac{q_0}{{\epsilon }_0}\sin 2\pi nt$
• C. $2\pi n{q}_0\cos 2\pi nt$
• D. $\frac{2\pi n{q}_0}{{\epsilon }_0}\cos 2\pi nt$

Answer 1

Answer: (C)

$2\pi n{q}_0\cos 2\pi nt$

Displacement current through a surface is ${ϵ}_0$ times the rate of change of flux through the surface.

Since $E=\frac{q}{A{ϵ}_0}$.

Therefore electric flux through the surface is $\varphi =EA=\frac{q}{{ϵ}_0}$

So real current though the capacitor $i=\frac{d\varphi }{dt}={ϵ}_0\frac{d}{dt}\left(\frac{q}{{ϵ}_0}\right)=\frac{dq}{dt}=i$

Thus displacement current is $\frac{dq}{dt}=2\pi n{q}_0\cos 2\pi ft$