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.

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.

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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}} $.

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Correct Answer: 1200

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