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

The displacement current $$I_d$$ through a surface is given by $$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$$, where $$\Phi_E$$ is the electric flux through the surface.

For a parallel plate capacitor, the electric field $$E$$ between the plates is uniform. The electric field is $$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$$, where $$Q$$ is the charge on the plates and $$A$$ is the plate area.

The electric flux through the hypothetical surface of area $$A_0$$ (parallel to the plates and perpendicular to the electric field) is $$\Phi_E = E \cdot A_0$$. Substituting the expression for $$E$$:

$$\Phi_E = \left( \frac{Q}{\epsilon_0 A} \right) A_0$$

Now, the displacement current through $$A_0$$ is:

$$I_d = \epsilon_0 \frac{d}{dt} \left( \frac{Q}{\epsilon_0 A} A_0 \right) = \epsilon_0 \cdot \frac{A_0}{A} \cdot \frac{1}{\epsilon_0} \frac{dQ}{dt} = \frac{A_0}{A} \frac{dQ}{dt}$$

The rate of change of charge $$\frac{dQ}{dt}$$ is the conduction current $$I_c$$ in the circuit. Therefore:

$$I_d = \frac{A_0}{A} I_c$$

Given:

• Plate area $$A = 16 \text{cm}^2 = 16 \times 10^{-4} \text{m}^2$$
• Hypothetical surface area $$A_0 = 3.2 \text{cm}^2 = 3.2 \times 10^{-4} \text{m}^2$$
• Conduction current $$I_c = 6 \text{A}$$

Substitute the values:

$$I_d = \left( \frac{3.2 \times 10^{-4}}{16 \times 10^{-4}} \right) \times 6 = \frac{3.2}{16} \times 6$$

Simplify the fraction:

$$\frac{3.2}{16} = \frac{32}{160} = \frac{2}{10} = 0.2$$

So:

$$I_d = 0.2 \times 6 = 1.2 \text{A}$$

Convert to milliamperes (1 A = 1000 mA):

$$I_d = 1.2 \times 1000 = 1200 \text{mA}$$

The displacement current through $$A_0$$ is 1200 mA.