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

To calculate the displacement current in a capacitor, we begin with the formula: \(I_d = \varepsilon_0 \frac{d\phi_E}{dt}\). Here, \(\varepsilon_0\) is the permittivity of free space, and \(\frac{d\phi_E}{dt}\) represents the rate of change of electric flux. The electric flux is defined as \(\phi_E = E \cdot A = \frac{Q}{\varepsilon_0} \cdot A\), where \(Q\) is the charge and \(E\) is the electric field. Since the current \(I = \frac{dQ}{dt}\), we can express the rate of change of electric flux as \(\frac{d\phi_E}{dt} = \frac{1}{\varepsilon_0} \cdot \frac{dQ}{dt} \cdot A\). Substituting this into the displacement current equation yields: \(I_d = A \cdot \frac{1}{\varepsilon_0} \cdot I\).

Given the area \(A_0 = 3.2 \, \text{cm}^2 = 3.2 \times 10^{-4} \, \text{m}^2\), current \(I = 6 \, \text{A}\), and \(\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\), we substitute these values into the formula: \(I_d = 3.2 \times 10^{-4} \cdot \frac{6}{8.854 \times 10^{-12}}\). The result of this calculation is approximately \(I_d \approx 2.17 \times 10^{-3} \, \text{A}\), which is equivalent to 2170 mA. Therefore, the displacement current through \(A_0\) is 2170 mA, which falls within the provided range of 1200 mA.