Question

Choose the correct answer from the alternatives given.
A parallel plate capacitor with plate area $A$ and separation between the plates $d$ is charged by a constant current $I$. Consider a plane surface of area $\frac{A}{2}$ parallel to the plate and drawn between the plates. The displacement current through the area is :

• A. I
• B. $\frac{I}{2}$
• C. $\frac{I}{4}$
• D. $\frac{I}{8}$

Answer 1

The correct option is B $\frac{I}{2}$

Charge on capacitor plates at time $t$ is , $q=It$.

Electric field between the plates at this instant is

$E=\frac{Q}{A{\epsilon }_0}=\frac{It}{A{\epsilon }_0}$

The Electric flux through the given area is:

$\varphi =A.E$

The Electric flux through the area $\frac{A}{2}$ is:

${\varphi }_E=\left(\frac{A}{2}\right)E$

${\varphi }_E=\frac{A}{2}\frac{It}{A{\epsilon }_0}=\frac{It}{2{\epsilon }_0}$

The displacement current through the area is given as:

$I_D={\epsilon }_0\frac{d{\varphi }_E}{dt}$

$⟹{\epsilon }_0\frac{d}{dt}\left(\frac{It}{2{\epsilon }_0}\right)=\frac{I}{2}$

Option $\left(B\right)$ is correct.