During the charging of a capacitor, the displacement current is

The correct answer is Option (1) → in the same direction as the conduction current.

Between the plates of a capacitor there is no conduction current, but the electric field there is changing. Maxwell’s displacement current is \( i_{d} = \varepsilon_{0}\,d\Phi_{E}/dt \).

While the capacitor charges, the charge \( q \) on the plates grows, so the field and the flux between them grow along the direction of the conduction current:

    \( \displaystyle i_{d} = \varepsilon_{0}\frac{d}{dt}\left(\frac{q}{\varepsilon_{0}}\right) = \frac{dq}{dt} = i_{c} \)

So the displacement current is equal to the conduction current and in the same direction; together they make the current continuous around the circuit.

NCERT (2026-27):
Displacement Current – “… This implies that for consistency,

    \( \displaystyle \varepsilon_0 \left(\frac{d\Phi_E}{dt}\right) = i \qquad (8.4) \)

This is the missing term in Ampere’s circuital law. If we generalise this law by adding to the total current carried by conductors through the surface, another term which is ε₀ times the rate of change of electric flux through the same surface, the total has the same value of current i for all surfaces.”