The displacement current due to time varying electric field is given by:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle \varepsilon_0\frac{d\phi_E}{dt} \)
Maxwell added a term to Ampere's circuital law so that it holds for a capacitor being charged: a changing electric flux behaves like a current, called the displacement current.
\( \displaystyle i_d = \varepsilon_0\frac{d\phi_E}{dt} \)
Shortcut: A displacement current must have the unit of current. \( \varepsilon_0\,\Phi_E \) has the unit of charge (Gauss's law gives \( \Phi_E = q/\varepsilon_0 \)), so \( \varepsilon_0\,d\Phi_E/dt \) is charge per second, an ampere. The forms with \( \mu_0 \) or with no constant have other units.
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 \( i_c \), and the displacement current denoted by \( i_d \) (= \( \varepsilon_0 (\mathrm{d}\Phi_E/\mathrm{d}t) \)). So we have
\( \displaystyle i = i_c + i_d = i_c + \varepsilon_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t} \qquad (8.5) \)”