How would you establish an instantaneous current 4A in the space between the two parallel plates of 2μF capacitor?
By changing the potential difference across the parallel plates of capacitor at the rate of 1×106 v/s
By changing the potential difference across the parallel plates of capacitor at the rate of 4×106 v/s
By changing the potential difference across the parallel plates of capacitor at the rate of 8×106 v/s
By changing the potential difference across the parallel plates of capacitor at the rate of 2×106 v/s
Displacement current ID=∈0dϕEdt ⇒ID=∈0AdEdt⇒ID=∈0AdVddt ⇒ID=∈oAddVdt=CdVdt ⇒dVdt=IDC=42×10−6=2×106Vs