Question

A parallel plate capacitor of area \( A = 16 \, \text{cm}^2 \) and separation between the plates \( 10 \, \text{cm} \), is charged by a DC current. Consider a hypothetical plane surface of area \( A_0 = 3.2 \, \text{cm}^2 \) inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through \( A_0 \) is \(\_\_\_\_\_ \)mA.

Hint:

The displacement current is proportional to the rate of change of the electric flux. For a parallel plate capacitor, the displacement current is given by the ratio of the area of the hypothetical surface inside the capacitor to the total area, multiplied by the current through the circuit.

Answer 1

Correct Answer: 1200

The displacement current \( I_{\text{displacement}} \) is defined as the rate of change of electric flux through a surface \( A_0 \). Maxwell's equations state that \( I_{\text{displacement}} = \epsilon_0 \frac{d\Phi_E}{dt} \), where \( \Phi_E = E \cdot A_0 \) is the electric flux, \( E \) is the electric field, and \( A_0 \) is the area of the hypothetical surface within the capacitor. The electric field \( E \) is related to the charge \( Q \) on the plates by \( E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A} \), where \( \sigma = \frac{Q}{A} \) is the surface charge density. The total current \( I \) in the circuit is the rate of change of charge: \( I = \frac{dQ}{dt} \). Therefore, the displacement current is proportional to the rate of change of the electric field, and \( I_{\text{displacement}} = \frac{A_0}{A} I \). Given \( A_0 = 3.2 \times 10^{-4} \, \text{m}^2 \), \( A = 16 \times 10^{-4} \, \text{m}^2 \), and \( I = 6 \, \text{A} \), we calculate \( I_{\text{displacement}} \) as follows:\[I_{\text{displacement}} = \frac{3.2 \times 10^{-4}}{16 \times 10^{-4}} \times 6 = \frac{3.2}{16} \times 6 = 1.2 \, \text{A}.\]Converting to milliamps, \( I_{\text{displacement}} = 1200 \, \text{mA} \). The displacement current through \( A_0 \) is \( \boxed{1200 \, \text{mA}} \).