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

Step 1: Definition of displacement current.
Displacement current arises due to a time-varying electric field, even where no physical charge flows. It 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: Displacement current in a charging capacitor.
- Conduction current flows in the connecting wires. - No real charge crosses the dielectric gap. - Electric field between plates changes with time: \[ \Phi_E = EA, \quad E = \frac{V}{d} \] - Rate of change of flux gives displacement current: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \] This ensures continuity of current: \[ i_c = i_d \]
Step 3: Displacement current for a conductor at constant voltage.
- Electric field is constant, so electric flux does not change with time: \[ \frac{d\Phi_E}{dt} = 0 \] - Therefore, displacement current: \[ i_d = 0 \]
Final Answers: \[ i_d = \varepsilon_0 \frac{d\Phi_E}{dt}, \quad \text{For a conductor at constant voltage: } i_d = 0 \]