Displacement current is produced between the plates of a capacitor, when the charge of the capacitor with time
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → increases or decreases
The displacement current is due to a changing electric flux between the plates. For a parallel plate capacitor the flux is \( \phi_E = Q/\varepsilon_0 \), so
\( \displaystyle I_d = \varepsilon_0\,\frac{d\phi_E}{dt} = \frac{dQ}{dt} \)
It is non-zero whenever the charge changes, whether the capacitor is charging or discharging; only its direction differs. If the charge stays constant, \( dQ/dt = 0 \) and there is no displacement current.
Why the other options are incorrect
Option (1) decreases only — Incorrect. A decreasing charge (discharging) does give a displacement current, but so does an increasing one, so “only” is wrong.
Option (2) remains same — Incorrect. A constant charge gives \( dQ/dt = 0 \), so no displacement current is produced.
Option (3) increases only — Incorrect. Charging gives a displacement current, but discharging does too, so “only” is wrong.
NCERT (2026-27):
Displacement Current – “… Now if the charge \( Q \) on the capacitor plates changes with time, there is a current \( i = (\mathrm{d}Q/\mathrm{d}t) \), so that using Eq. (8.3), we have
\( \displaystyle \frac{\mathrm{d}\Phi_E}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{Q}{\varepsilon_0}\right) = \frac{1}{\varepsilon_0}\frac{\mathrm{d}Q}{\mathrm{d}t} \)
This implies that for consistency,
\( \displaystyle \varepsilon_0\left(\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}\right) = i \qquad (8.4) \)
… The current, given by Eq. (8.4), is a new term, and is due to changing electric field (or electric displacement, an old term still used sometimes). It is, therefore, called displacement current or Maxwell’s displacement current.”