Question

What is meant by displacement current? What is the difference between displacement current and conduction current? Write down their formulae.

Hint:

Displacement current comes from a changing electric flux, not moving charge: \(I_d = \varepsilon_0\, d\Phi_E/dt\), while \(I_c = dq/dt\).

Answer 1

Step 1 (Meaning of displacement current): Maxwell noticed that Ampere's law failed in the gap of a charging capacitor, where no charges cross the gap yet a magnetic field still exists. He proposed that a changing electric field in that region acts like a current. This is the displacement current \(I_d\). It is not a flow of charge; it is produced by a time-varying electric flux and, like a real current, it creates a magnetic field.

Step 2 (Formula for displacement current): If \(\Phi_E\) is the electric flux through a surface,
\[ I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
where \(\varepsilon_0\) is the permittivity of free space.

Step 3 (Meaning of conduction current): The conduction current \(I_c\) is the ordinary current caused by the actual motion (drift) of free charges, such as electrons in a metal wire.

Step 4 (Formula for conduction current):
\[ I_c = \frac{dq}{dt} \]
the rate of flow of charge; equivalently \(I_c = \dfrac{V}{R}\) in a resistive conductor.

Step 5 (Key differences):
• Conduction current needs a real flow of charges through a conductor; displacement current needs no charge flow, only a changing electric field.
• Conduction current dominates inside wires; displacement current dominates in the gap of a capacitor or in free space (electromagnetic waves).
• Both produce magnetic fields, so the generalised Ampere-Maxwell law uses their sum: \(I = I_c + I_d\).
\[\boxed{I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}, \qquad I_c = \dfrac{dq}{dt}}\]