A parallel plate air capacitor is being charged. The potential difference across it rises at a steady rate, and the displacement current between its plates is \(i\). A second capacitor has the same plate area but half the plate separation. Its potential difference rises at the same rate. The displacement current between the plates of the second capacitor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2i\)
No charge crosses the gap between the plates. The changing electric field in the gap gives a displacement current, and it is equal to the charging current \( dQ/dt \). Since \( Q=CV \), this current is \( C\,dV/dt \).
For a parallel plate capacitor in air, \( C=\varepsilon_0A/d \). So the displacement current is inversely proportional to \( d \) when \( A \) and \( dV/dt \) stay the same.
\( i=\frac{\varepsilon_0A}{d}\frac{dV}{dt} \)
\( \begin{aligned} i_2 &= \frac{\varepsilon_0A}{d/2}\frac{dV}{dt} \\ &= 2\times\frac{\varepsilon_0A}{d}\frac{dV}{dt} \\ &= 2i \end{aligned} \)
NCERT (2026-27):
Displacement Current – “… Now if the charge Q on the capacitor plates changes with time, there is a current i = (dQ/dt) …”
Displacement Current – “In explicit terms, this means that outside the capacitor plates, we have only conduction current ic = i, and no displacement current, i.e., id = 0. On the other hand, inside the capacitor, there is no conduction current, i.e., ic = 0, and there is only displacement current, so that id = i.”