A parallel plate capacitor has plate area A and separation between the plates equal to d. A material of dielectric constant K and resistivity ρ is filled between the plates. The switch is closed to connect the capacitor to a cell of emf V.
(a) Write the steady state current in the circuit and charge on the capacitor.
(b) When the circuit is in steady state, switch S is opened (at t = 0). Write charge on the capacitor as function of time (t) after this.

Question

G1 Physics · Accepted Answer

Image Leakage resistance of dielectric R = ρd/A and capacitance C = Kε₀A/d. With cell connected, steady current I = V/R = VA/(ρd), while capacitor charge q₀ = CV = Kε₀AV/d. After opening, discharge time constant RC = ρKε₀, so q(t)=q₀ e ( − t / ( ρ K ε 0 ) ) . e^(-t/(ρKε₀)). e ( − t / ( ρ K ε 0 ​ )) .