Question

Considering the case of charging of a capacitor, show that i_d = `(d phi_E)/dt`. What is the value of i_d for a conductor across which a constant voltage is applied?

Answer 1

Displacement current is defined as:

i_d = `(d phi_E)/dt`

Where:

ε₀ = permittivity of free space

Φ_E = electric flux

When a capacitor is charging, conduction current flows in wires; no real charge flows across the dielectric gap, but the electric field between the plates changes with time.

Electric flux between plates (Φ_E) = EA

As the voltage increases, the electric field (Φ_E) changes with time

E = `V/d`

Thus,

i_d = `epsilon_0 (d phi_E)/dt`

This ensures continuity:

i_c = i_d

For a conductor with a constant applied voltage, the electric field is constant, and the electric flux does not change with time.

So:

`(d phi_E)/dt` = 0

Hence, i_d = 0