A parallel plate capacitor has a capacitance $$C = 200$$ pF. It is connected to $$230$$ V ac supply with an angular frequency $$300$$ rad s$$^{-1}$$. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :

The problem asks to find the rms values of conduction and displacement currents for a capacitor connected to an AC supply.

For a pure capacitor in an AC circuit, the rms conduction current is given by $$I_{\text{rms}} = \omega C V_{\text{rms}}$$.

Substituting the values $$\omega = 300$$ rad/s, $$C = 200\text{ pF} = 200\times10^{-12}\text{ F}$$, and $$V_{\text{rms}} = 230$$ V into this expression yields

$$I_{\text{rms}} = 300 \times 200 \times 10^{-12} \times 230$$

$$ = 300 \times 200 \times 230 \times 10^{-12} = 13{,}800{,}000 \times 10^{-12} = 13.8 \times 10^{-6}\text{ A} = 13.8\ \mu\text{A}$$

According to Maxwell's equations, the displacement current in a capacitor equals the conduction current in the external circuit because the changing electric field between the plates creates a displacement current $$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$$ that exactly matches the conduction current, ensuring continuity of current in the circuit.

Therefore, the rms displacement current in the capacitor is also $$13.8\ \mu\text{A}$$.

The correct answer is Option D: 13.8 $$\mu$$A and 13.8 $$\mu$$A.