Q 12-08-032JEE MainJEE Main 2026 (5 Apr, Shift 1)Easy
A displacement current of $4.0$ A can be set up in the space between two parallel plates of $6\,\mu\text{F}$ capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\alpha \times 10^6$ V/s. The value of $\alpha$ is ______.
Answer: (B) $0.67$
Concept. Between the plates no charge flows, but the changing field carries a displacement current equal to the conduction current in the wires: $I_d=\varepsilon_0\dfrac{d\Phi_E}{dt}=C\dfrac{dV}{dt}$.
Step 1. Rate of change of voltage. $\dfrac{dV}{dt}=\dfrac{I_d}{C}=\dfrac{4.0}{6\times10^{-6}}=0.67\times10^6$ V/s.
Answer. $\alpha\approx0.67$ (Option B).
Solution by Sreeraj P, M.Sc Physics
