The RMS value of conduction current in a parallel plate capacitor is $$6.9 \mu A$$. The capacity of this capacitor, if it is connected to $$230 \text{ V}$$ AC supply with an angular frequency of $$600 \text{ rad s}^{-1}$$, will be

The RMS value of conduction current in a parallel plate capacitor is $$I_{rms} = 6.9 \text{ } \mu A$$. The supply voltage is $$V_{rms} = 230 \text{ V}$$ and the angular frequency is $$\omega = 600 \text{ rad/s}$$.

For a purely capacitive circuit, the rms current and voltage are related by $$I_{rms} = V_{rms} \times \omega C$$.

Rearranging this relation to solve for capacitance gives $$C = \frac{I_{rms}}{V_{rms} \times \omega}$$.

Substituting the given values, we get $$C = \frac{6.9 \times 10^{-6}}{230 \times 600}$$, which simplifies to $$C = \frac{6.9 \times 10^{-6}}{138000}$$ and can also be expressed as $$C = \frac{6.9 \times 10^{-6}}{1.38 \times 10^5}$$.

Thus, the capacitance is $$C = 5 \times 10^{-11} \text{ F}$$, or $$C = 50 \text{ pF}$$.

Hence, the correct answer is Option B.