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, we first revisit the displacement current formula within a capacitor. The displacement current (\(I_d\)) is given by:

\(I_d = \varepsilon_0 \frac{d\phi_E}{dt}\)

where \(\varepsilon_0\) is the permittivity of free space and \(\frac{d\phi_E}{dt}\) is the rate of change of electric flux.

The electric flux (\(\phi_E\)) 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. For a capacitor, \(I = \frac{dQ}{dt}\), thus \(\frac{d\phi_E}{dt} = \frac{1}{\varepsilon_0} \cdot \frac{dQ}{dt} \cdot A\).

Substituting in the equation for \(I_d\):

\(I_d = A \cdot \frac{1}{\varepsilon_0} \cdot I\)

Given \(A_0 = 3.2 \, \text{cm}^2 = 3.2 \times 10^{-4} \, \text{m}^2\) and \(I = 6 \, \text{A}\), with \(\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\), plug in the values:

\(I_d = 3.2 \times 10^{-4} \cdot \frac{6}{8.854 \times 10^{-12}}\)

Simplifying:

\(I_d \approx 2.17 \times 10^{-3} \, \text{A} = 2170 \, \text{mA}\)

The displacement current through \(A_0\) is therefore correctly calculated as 2170 mA, which is within the range provided (1200 mA).

Answer 2

The displacement current \( I_{\text{displacement}} \) is related to the rate of change of the electric flux through the surface \( A_0 \). According to Maxwell's equations, the displacement current is given by: \[ I_{\text{displacement}} = \epsilon_0 \frac{d\Phi_E}{dt}, \] where \( \Phi_E = E \cdot A_0 \) is the electric flux through the surface \( A_0 \), \( E \) is the electric field between the plates, and \( A_0 \) is the area of the hypothetical surface inside 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. For the displacement current, we use the fact that the total current \( I \) through the circuit is related to the rate of change of the charge: \[ I = \frac{dQ}{dt}. \] Thus, the rate of change of the charge \( \frac{dQ}{dt} \) is the current through the circuit. Since the displacement current is proportional to the rate of change of the electric field, we have: \[ I_{\text{displacement}} = \frac{A_0}{A} I. \] Given: - \( A_0 = 3.2 \, \text{cm}^2 = 3.2 \times 10^{-4} \, \text{m}^2 \), - \( A = 16 \, \text{cm}^2 = 16 \times 10^{-4} \, \text{m}^2 \), - \( I = 6 \, \text{A} \). Substitute these values into the formula: \[ I_{\text{displacement}} = \frac{3.2 \times 10^{-4}}{16 \times 10^{-4}} \times 6 = \frac{3.2}{16} \times 6 = 1.2 \, \text{A}. \] Converting this to milliamps: \[ I_{\text{displacement}} = 1200 \, \text{mA}. \] Thus, the displacement current through \( A_0 \) is \( \boxed{1200 \, \text{mA}} \).