An air-filled parallel plate capacitor is connected to an AC source of constant voltage amplitude, resulting in a peak displacement current I₀ between its plates. If the frequency of the AC source is halved and simultaneously a dielectric slab of dielectric constant K = 2 is inserted to completely fill the space between the capacitor plates, what will be the new peak displacement current?

Step-by-step solution

The initial peak displacement current is equal to the peak conduction current: I₀ = V_m X_C = V_m (2 f) C When the frequency is halved, the new frequency is f' = f 2 . When a dielectric of K = 2 is completely inserted, the new capacitance becomes C' = 2C . The new peak displacement current is: I' = V_m (2 f') C' = V_m (2 f 2 ) (2C) I' = V_m (2 f) C = I₀ Thus, the decrease in frequency is exactly compensated by the increase in capacitance, and the displacement current remains unchanged. Answer: I₀