Read the following passage and answer the questions.
A parallel plate capacitor has two plates, each of area 0.10 m2, with air between them. A source charges the capacitor. The current in the connecting wires stays constant at 0.885 mA. Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\). A student draws a circular loop of radius \(r\) around one connecting wire, close to a plate. The plane of the loop is perpendicular to the wire. She marks a point P on the loop. She wants to find the magnetic field at P with the help of Ampere’s law. For this, she takes two surfaces that have the loop as their rim (edge). The first is a flat circular disc, and the wire passes through it. The second is a pot-shaped surface. Its bottom lies in the gap between the plates, and it does not touch the wire anywhere.
At an instant, the charge on the capacitor is \(Q\) and the area of each plate is \(A\). A flat surface between the plates has the same area \(A\) as a plate. The electric flux through this surface is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(Q/\varepsilon_0\)
The field between the plates is uniform and perpendicular to the plates. It vanishes outside the plate area, so the flux is the field multiplied by the plate area.
\( \begin{aligned} E &= \frac{\sigma}{\varepsilon_0} \\ &= \frac{Q}{\varepsilon_0A} \end{aligned} \)
\( \begin{aligned} \Phi_E &= EA \\ &= \frac{Q}{\varepsilon_0A}\times A \\ &= \frac{Q}{\varepsilon_0} \end{aligned} \)
Option (1) \( Q/\varepsilon_0A \) — Incorrect. This is the electric field \( E \) between the plates. The flux is this field multiplied by the area \( A \).
Option (2) \( Q\varepsilon_0 \) — Incorrect. In Gauss’s law the charge is divided by \( \varepsilon_0 \). It is not multiplied by it.
Option (3) \( Q/\varepsilon_0 \) — Correct. The area \( A \) comes once in the field and once in \( EA \). So the flux depends only on the charge \( Q \).
Option (4) \( QA/\varepsilon_0 \) — Incorrect. This takes the field as \( Q/\varepsilon_0 \) and then multiplies by \( A \). The field is \( Q/\varepsilon_0A \), so the area is used only once.
NCERT (2026-27):
Displacement Current – “… It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. … Using Gauss’s law, it is”
\( \displaystyle \Phi_E=EA=\frac{1}{\varepsilon_0}\frac{Q}{A}A= \)\( \displaystyle \frac{Q}{\varepsilon_0} \)\( \qquad (8.3) \)