Question

The electric field between the plates of a parallel plate capacitor changes at the rate of \(4.5 \times 10^{7} \, {V/m/s}\). If the plates of the capacitor are circular with radius 2 cm, then the displacement current inside the capacitor is:

• A. 0.2 \(\mu A\)
• B. 0.3 \(\mu A\)
• C. 0.4 \(\mu A\)
• D. 0.5 \(\mu A\)

Hint:

Displacement current in capacitor: \( I_d = \epsilon_0 A \frac{dE}{dt} \). Use plate area and rate of change of electric field.

Answer 1

The Correct Option is D

Given: \[ \frac{dE}{dt} = 4.5 \times 10^{7} \, {V/m/s}, r = 2 \, {cm} = 0.02 \, m \] Displacement current \( I_d \) is related to the rate of change of electric flux: \[ I_d = \epsilon_0 A \frac{dE}{dt} \] where - \( \epsilon_0 = 8.85 \times 10^{-12} \, {F/m} \), - \( A = \pi r^2 = \pi (0.02)^2 = 1.2566 \times 10^{-3} \, m^2 \). 
Calculate displacement current: \[ I_d = 8.85 \times 10^{-12} \times 1.2566 \times 10^{-3} \times 4.5 \times 10^{7} \] \[ I_d = 8.85 \times 1.2566 \times 4.5 \times 10^{-8} = 5.0 \times 10^{-7} \, A = 0.5 \, \mu A \]