Question

A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current (\( I_c \)) and the displacement current (\( I_d \)) has the same value at all points of the circuit.

Hint:

In a capacitor, the displacement current compensates for the conduction current inside the plates. This ensures that the total current remains continuous in the circuit.

Answer 1

Sum of Conduction and Displacement Current in a Parallel Plate Capacitor  

When a parallel plate capacitor is charged by an AC source, both conduction current (\( I_c \)) and displacement current (\( I_d \)) are present in the circuit. To show that the sum of conduction and displacement current has the same value at all points in the circuit, let us analyze the situation:

Step 1: Conduction Current (\( I_c \)) in the Circuit

The conduction current (\( I_c \)) is the current that flows through the external circuit when the capacitor is charged by the AC source. According to Ohm's Law, the conduction current is related to the applied voltage (\( V \)) and the impedance (\( Z \)) of the circuit as:

\[ I_c = \frac{V}{Z} \]

Here, the applied voltage is alternating (AC), so the conduction current is also an alternating current with the same frequency as the source.

Step 2: Displacement Current (\( I_d \)) in the Capacitor

In the capacitor, a displacement current (\( I_d \)) arises due to the changing electric field between the plates. The displacement current is given by the time rate of change of the electric field in the capacitor:

\[ I_d = \epsilon_0 A \frac{dE}{dt} \]

Where: - \( \epsilon_0 \) is the permittivity of free space, - \( A \) is the area of the plates of the capacitor, - \( \frac{dE}{dt} \) is the rate of change of the electric field. The electric field \( E \) between the plates of the capacitor is related to the voltage across the plates by the relation: \[ E = \frac{V}{d} \] where \( d \) is the separation between the plates. Therefore, the displacement current can also be expressed as: \[ I_d = \epsilon_0 A \frac{d}{dt} \left( \frac{V}{d} \right) = \epsilon_0 A \frac{dV}{dt} \]

Step 3: Relationship Between Conduction and Displacement Currents

Now, let's consider the fact that the conduction current and displacement current are related through the behavior of the capacitor. Since the capacitor is being charged by an AC source, the changing voltage across the plates creates an electric field, and this results in a displacement current. The rate at which charge accumulates on the plates is equivalent to the displacement current.

In an ideal capacitor, the conduction current \( I_c \) and the displacement current \( I_d \) are related and are equal in magnitude, but opposite in direction. Hence, the total current entering and leaving the capacitor, when considered as a system, is the sum of the conduction current in the external circuit and the displacement current inside the capacitor. This means:

\[ I_c = I_d \]

Therefore, the sum of conduction current \( I_c \) and displacement current \( I_d \) has the same value at all points in the circuit.

Conclusion:

Thus, the sum of the conduction current (\( I_c \)) and the displacement current (\( I_d \)) is equal at all points in the circuit of a parallel plate capacitor being charged by an AC source.