Q.The charge on a parallel plate capacitor varies as $q = q_0\cos 2\pi\nu t$. The plates are very large and close together (area $= A$, separation $= d$). Neglecting the edge effects, find the displacement current through the capacitor.

The key idea is that a changing electric field between the plates produces a displacement current, which is equal to the conduction current in the wire.

Step 1: The displacement current is defined as

$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux.

Step 2: For a parallel plate capacitor (neglecting edge effects), the electric field between the plates is uniform:

$E = \frac{\sigma}{\varepsilon_0} = \frac{q}{\varepsilon_0 A}$.

The flux is $\Phi_E = EA = \frac{q}{\varepsilon_0}$.

Step 3: Differentiate: …

Why Displacement Current?

When a capacitor charges or discharges, no actual charge flows through the dielectric between the plates. Yet the magnetic field around the wires and between the plates behaves as if a current were flowing. Maxwell resolved this paradox by introducing the displacement current — a term that accounts for the changing electric field between the plates.

The key insight: wherever the electric flux $\Phi_E$ changes with time, there is an equivalent "current" given by:

For a parallel plate capacitor, the electric field between the plates is uniform (neglecting edge effects) and directly related to the charge on the plates. So the displacement current becomes simply the rate of change of charge — the same as the conduction current feeding the capacitor.

Let's work through it step by step.

1. The electric field between the plates

   For a parallel plate capacitor with plate area $A$ and charge $q$, the electric field between the plates (neglecting edge effects) is:

This field is uniform and perpendicular to the plates.

2. Electric flux through a surface between the plates

   Consider any surface parallel to the plates, of area $A$. The electric flux through it is:

Notice that the area cancels — the flux depends only on the charge, not on the plate area. This is a direct consequence of Gauss's law.

3. Displacement current definition

   The displacement current through any surface is:

Substituting $\Phi_E = q/\epsilon_0$:

This is the central result: the displacement current between the plates equals the conduction current in the wires.

4. Apply the given charge variation

   We are given $q = q_0 \cos(2\pi\nu t)$. Differentiate:

$= -2\pi\nu q_0 \sin(2\pi\nu t)$ …

Method: Computing Displacement Current from a Given Time-Varying Charge

Use this whenever a capacitor's charge (or voltage) is given as an explicit function of time and the question asks for the displacement current.

Steps

Step 1: Express the electric flux between the plates in terms of the charge

For a parallel-plate capacitor (edge effects neglected), $E = Q/(\varepsilon_0 A)$, so the flux through a surface of area $A$ between the plates is

Notice the plate area cancels out of this expression — the plate separation never enters either, so the final result below will not depend on the capacitor's physical size.

Step 2: Apply the displacement current definition

This confirms the general result: the displacement current between the plates always equals $dQ/dt$, the same as the conduction current charging the capacitor — true for any $Q(t)$, not just this one.

Step 3: Differentiate the given charge function …