An alternating emf $$E = 110\sqrt{2} \sin 100t$$ volt is applied to a capacitor of $$2\mu F$$, the rms value of current in the circuit is _____ mA.

We need to find the RMS value of current when an alternating EMF $$E = 110\sqrt{2} \sin 100t$$ volts is applied to a capacitor of $$2 \, \mu F$$. Since the general form of an AC voltage is $$E = E_0 \sin \omega t$$, comparing with the given expression yields the peak EMF $$E_0 = 110\sqrt{2}$$ V and the angular frequency $$\omega = 100$$ rad/s.

Substituting the peak EMF into the RMS relation gives the RMS voltage. From the equation $$E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{110\sqrt{2}}{\sqrt{2}} = 110 \text{ V}$$, we obtain an RMS voltage of 110 V.

Next, the capacitive reactance is given by $$X_C = \frac{1}{\omega C}$$. Substituting $$\omega = 100$$ rad/s and $$C = 2 \, \mu F = 2 \times 10^{-6}$$ F into this expression, we find $$X_C = \frac{1}{100 \times 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega$$.

Then the RMS current in a purely capacitive circuit follows from $$I_{rms} = \frac{E_{rms}}{X_C} = \frac{110}{5000} = 0.022 \text{ A} = 22 \text{ mA}$$. Therefore, the current is 22 mA.

The answer is 22 mA.