Question

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 plates and drawn symmetrically between the plates. Find the displacement current through this area.

• A. $\frac{i}{3}$
• B. $\frac{i}{2}$
• C. $i$
• D. $\frac{i}{5}$

Answer 1

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

We know that displacement current is given by
$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}\text{Now we know that}{\varphi }_E=E.\frac{A}{2}$
Since
$E=\frac{Q}{{\epsilon }_{0}A}$
Replacing this to get ${\varphi }_E$
$\Rightarrow {\varphi }_E=\frac{Q}{{\epsilon }_{0}A}\times \frac{A}{2}=\frac{Q}{2{\epsilon }_0}$
Now
$\frac{d{\varphi }_E}{dt}=\frac{1}{2{\epsilon }_0}\frac{dQ}{dt}$
$\Rightarrow \frac{d{\varphi }_E}{dt}=\frac{1}{2{\epsilon }_0}i$
Now
$i_d={\epsilon }_0\frac{1}{2{\epsilon }_0}i$
$\Rightarrow i_d=\frac{i}{2}$