The rate of change of electric flux through the space between the plates of a charging capacitor is (\( i_d \) = displacement current)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \frac{i_d}{\varepsilon_0} \)
There is no conduction current between the plates of a charging capacitor. The current there is the displacement current, which is \( \varepsilon_0 \) times the rate of change of electric flux.
\( \displaystyle i_d=\varepsilon_0\,\frac{d\Phi_E}{dt} \)
\( \displaystyle \Rightarrow\; \frac{d\Phi_E}{dt}=\frac{i_d}{\varepsilon_0} \)
NCERT (2026-27):
Displacement Current – “The generalisation made by Maxwell then is the following. The source of a magnetic field is not just the conduction electric current due to flowing charges, but also the time rate of change of electric field. More precisely, the total current i is the sum of the conduction current denoted by ic, and the displacement current denoted by id (= ε0 (dΦE/dt)).”