A parallel plate capacitor of area $A=16 \mathrm{~cm}^2$ and separation between the plates 10 cm , is charged by a DC current. Consider a hypothetical plane surface of area $\mathrm{A}_0=3.2 \mathrm{~cm}^2$ 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 $\mathrm{A}_0$ is __________ mA .

<p>To determine the displacement current through a hypothetical plane surface within a parallel plate capacitor, follow these steps:</p> <p><strong>Current Density Calculation</strong>:</p> <p><p>The current density ($ J_d $) is the current ($ I $) divided by the area ($ A $) of the capacitor plates.</p></p> <p><p>Given: $ I = 6 \, \text{A} $ and $ A = 16 \, \text{cm}^2 $.</p> <p>$ J_d = \frac{I}{A} = \frac{6 \, \text{A}}{16 \, \text{cm}^2} $</p></p> <p><strong>Displacement Current Through the Hypothetical Surface</strong>:</p> <p><p>The hypothetical surface has an area $ A_0 = 3.2 \, \text{cm}^2 $.</p></p> <p><p>The displacement current through this smaller area ($ I_{small} $) is the current density multiplied by the area of the hypothetical surface.</p> <p>$ I_{small} = J_d \times A_0 = \frac{6 \, \text{A}}{16 \, \text{cm}^2} \times 3.2 \, \text{cm}^2 $</p></p> <p><strong>Calculation</strong>:</p> <p><p>Simplify the expression to find the displacement current:</p> <p>$ I_{small} = \left(\frac{6}{16}\right) \times 3.2 = 1.2 \, \text{A} = 1200 \, \text{mA} $</p></p> <p>Thus, the displacement current through the hypothetical plane surface is 1200 mA.</p>