Question

A capacitor made of two parallel plates each of plate area $A$ and separations $d$ , being charged by an external ac source. How the displacement current inside the capacitor is the same as the current charging the capacitor.

Answer 1

In order to solve this question you should know the difference between the charging current and the displacement current inside the capacitor. While solving keep in mind that the source that is charging the capacitor is an ac source.

Complete step by step answer:
Here we are given with a parallel plate capacitor of plate area $A$ and separations $d$
This capacitor is charged by an ac source,
So the expression for the electric field or the voltage charging the capacitor is given by,
$E = {E_0}\sin \omega t$
Now the expression for the charge on the capacitor would be given by,
$Q = CE$
As Current is defined as the rate of change of charge that is
$\Rightarrow$ ${I_C} = \dfrac{{dQ}}{{dt}}$
This current ${I_C}$ is the current that is charging the capacitor.
Putting the expression for the charge in the equation of current we get,
$\Rightarrow$ ${I_C} = \dfrac{{d(C{E_o}\sin \omega t)}}{{dt}}$
On differentiating the above expression we get,
$\Rightarrow$ ${I_C} = C{E_o}\omega \cos \omega t$
This is the expression for current that is charging the capacitor.
Now we will find the expression for displacement current,
The displacement current is due to the electric flux between the plates of the capacitor which is given by the expression,
$\Rightarrow$ ${I_D} = {\varepsilon _o}\dfrac{{d{\phi _E}}}{{dt}}$
Here ${\phi _E}$ is the electric flux between the plates of the capacitor.
Now as ${\phi _E} = EA$
Putting this expression of Electric flux in the expression of displacement current we get,
$\Rightarrow$ ${I_D} = {\varepsilon _o}\dfrac{{d(EA)}}{{dt}}$
Electric Field between the plates of the capacitor is given by,
$\Rightarrow$ $E = \dfrac{Q}{{{\varepsilon _o}}}$
Putting this expression for electric field in the expression of displacement current we get,
$\Rightarrow$ ${I_D} = A{\varepsilon _o}\dfrac{{d(\dfrac{Q}{{{\varepsilon _o}A}})}}{{dt}}$
On simplifying this we get,
$\Rightarrow$ ${I_D} = A{\varepsilon _o} \times \dfrac{1}{{{\varepsilon _o}A}}\dfrac{{d(Q)}}{{dt}}$
As $I = \dfrac{{dQ}}{{dt}}$ so we have,
$\Rightarrow$ ${I_D} = C{E_o}\omega \cos \omega t$
As this expression is the same as the charging current.
Hence, we can say that the displacement current inside the capacitor is the same as the current charging the capacitor.

Note: In a capacitor, there is always a displacement current and never a conduction under the normal conditions. Conduction current is due to the flow of electrons in a circuit while displacement current is due to the time-varying electric field. In conduction current electrons actually move, while in displacement current no charge carriers are involved.