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 $A/2$ parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Answer 1

Hint A capacitor is a device that stores electric energy drawn from the electric field. It usually consists of 2 conductor plates separated by a region filled with vacuum or a dielectric material. The capacitance or ability to store energy of such a plate is defined by $C = \dfrac{Q}{V} = \dfrac{{\varepsilon A}}{d}$ .
Thus, the capacitance of a capacitor is dependent directly on the area of the plates.
Formulas used: We will be using the formula to find the electric field between the plates of the capacitor, $E = \dfrac{q}{{{\varepsilon _0}A}}$ where $E$ is the electric field between the capacitor plates, $q$ is the charge flowing in between the plates, ${\varepsilon _0}$ is a constant called permittivity, and $A$ is the area between the plates.
We will also be using the formula to find the displacement current, ${I_d} = {\varepsilon _0}\dfrac{{\partial {\Phi _E}}}{{\partial t}}$ where ${I_d}$ is the displacement current, ${\Phi _E}$ is the electric flux and $t$ is the time taken.

Complete Step by Step Answer:
We know that the capacitor is a device that is capable of storing electrical energy. The capacitor looks like 2 parallel plates arranged with a dielectric substance in between them. It stores the electrical energy from the electric field formed when charges are passed through the plates.
From the problem we can infer that the capacitor is charged by a constant current $i$ and a plane of surface area $A/2$ is introduced in between them.
Now let us calculate the electric field due to the charge $i$ , which is given by $E = \dfrac{q}{{{\varepsilon _0}A}}$ . We know that the electric flux is the rate of flow of electric field in a given area. In the area between the newly introduced plate, the electric flux will be,
${\Phi _E} = E.A$
${\Phi _E} = \dfrac{q}{{{\varepsilon _0}A}} \times \dfrac{A}{2}$
Solving the equation, we get, electric flux in between the plates of the capacitor to be, ${\Phi _E} = \dfrac{q}{{2{\varepsilon _0}}}$ .
Now to find the displacement current, which is given by the formula ${I_d} = {\varepsilon _0}\dfrac{{\partial {\Phi _E}}}{{\partial t}}$ . Substituting the known values in the formula we get
${I_d} = {\varepsilon _0}\dfrac{{d\left( {\dfrac{q}{{2{\varepsilon _0}}}} \right)}}{{dt}}$
${I_d} = \dfrac{{{\varepsilon _0}}}{{2{\varepsilon _0}}} \times \dfrac{{dq}}{{dt}}$
We know that current is nothing but the rate of change of flow of charges. \[\left( {i = \dfrac{{dq}}{{dt}}} \right)\] Substituting in the equation and solving we get,
$ \Rightarrow {I_d} = \dfrac{i}{2}$
Thus, we know that the current is halved when a plate of $\dfrac{A}{2}$ surface area is introduced in between the capacitor plates.

Note: We can see that when the area is halved the electric flux of the electric field is also halved and so is the displacement current. Thus, we can say that the displacement current is directly proportional to the electric flux which is directly proportional to the area of plates.