JEE Main 2026 Physics question
A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 μF capacitor. The rate of change of potential difference across the plates of the capacitor is nearly α×10⁶ V/s. The value of α is .
Answer: (B)
Solution
= C(dV/dt), so dV/dt = 4/(6×10⁻⁶) = 6.67×10⁵ = 0.67×10⁶ V/s.