Question

A parallel-plate capacitor is being charged. If $I_D$ represents the displacement current between the plates of the capacitor and $I_C$ represents the conduction current flowing in the connecting wire, then

• A. $I_D=2I_C$
• B. $I_D=\frac{I_C}{2}$
• C. $I_D=I_C$
• D. $I_D\text{and}{I}_C$ are not related.

Answer 1

The correct option is C $I_D=I_C$

Let the area of the plates of the capacitor is $A$ and charge on the capacitor at any instant is $Q$.

The electric field between the plates is

$E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{{\epsilon }_{0}A}$

The flux of this field between the plates of the capacitor is

${\varphi }_E=EA=\frac{Q}{{\epsilon }_0}$

The displacement current is given by

$I_D={\epsilon }_0\frac{d{\varphi }_E}{dt}={\epsilon }_0\frac{d}{dt}\left(\frac{Q}{{\epsilon }_0}\right)$

$\Rightarrow I_D=\frac{dQ}{dt}$

But, $\frac{dQ}{dt}=I_C,$ is the conduction current flowing in the connecting wire.

$\therefore I_D=I_C$

Hence, option $\left(C\right)$ is the correct answer.