A parallel plate capacitor with plate area A and separation d between the plates is charged by a constant current I. Consider a plane surface of area A/2 parallel to the plates and drawn between the plates. The displacement current through this area is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( I/2 \)
Between the plates there is no conduction current. The current there is the displacement current \( i_d=\varepsilon_0\,d\Phi_E/dt \). The electric field between the plates is uniform, with the same magnitude over the whole plate area. So a surface of area \( A/2 \) receives half of the electric flux.
With charge \( Q \) on the plates, the field between them is \( E=Q/\varepsilon_0 A \). The flux through the surface of area \( A/2 \) is:
\( \begin{aligned} \Phi_E &= E\times\frac{A}{2} \\ &= \frac{Q}{\varepsilon_0 A}\times\frac{A}{2} \\ &= \frac{Q}{2\varepsilon_0} \end{aligned} \)
The charging current is \( I=dQ/dt \), so the displacement current through this surface is:
\( \begin{aligned} i_d &= \varepsilon_0\,\frac{d\Phi_E}{dt} \\ &= \varepsilon_0\times\frac{1}{2\varepsilon_0}\,\frac{dQ}{dt} \\ &= \frac{I}{2} \end{aligned} \)
NCERT (2026-27):
Displacement Current – “… If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q/A)/ε0 … It has the same magnitude over the area A of the capacitor plates, and vanishes outside it.”
Displacement Current – “… the total current i is the sum of the conduction current denoted by ic, and the displacement current denoted by id (= ε0 (dΦE/dt)).”