Question

What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?

Hint:

Displacement current exists only when electric field changes with time. No change in electric field → no displacement current.

Answer 1

Concept: Displacement current was introduced by Maxwell to explain continuity of current in circuits containing capacitors. It arises due to time-varying electric field, even where no charge flows physically.
Step 1: Definition of displacement current. Displacement current is defined as: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \] Where:

\( \varepsilon_0 \) = permittivity of free space
\( \Phi_E \) = electric flux

Step 2: Charging capacitor case. When a capacitor is charging:

Conduction current flows in wires
No real charge flows across dielectric gap
But electric field between plates changes with time
Electric flux between plates: \[ \Phi_E = EA \] As voltage increases, electric field changes: \[ E = \frac{V}{d} \Rightarrow \Phi_E \text{ changes with time} \] Thus: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \] This ensures continuity: \[ i_c = i_d \]
Step 3: Conductor with constant voltage. For a conductor with constant applied voltage:

Electric field is constant
Electric flux does not change with time
So: \[ \frac{d\Phi_E}{dt} = 0 \] Hence: \[ i_d = 0 \] Final Answers:

Displacement current: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
For a conductor at constant voltage: \[ i_d = 0 \]