Given alternating emf

$E = 110\sqrt{2}\sin(100t)\ \text{V}$

Comparing with the standard form

$E = E_0\sin(\omega t)$

we get:

• Peak voltage: $E_0 = 110\sqrt{2}\ \text{V}$
• Angular frequency: $\omega = 100\ \text{rad/s}$

Find rms voltage

For a sinusoidal source,

$V_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{110\sqrt{2}}{\sqrt{2}} = 110\ \text{V}$

Given capacitance

$C = 2\,\mu\text{F} = 2\times 10^{-6}\ \text{F}$

Current in a purely capacitive AC circuit

The rms current is

$I_{\text{rms}} = V_{\text{rms}}\cdot \omega C$

Substituting values:

$I_{\text{rms}} = 110 \times 100 \times 2\times 10^{-6}$

$I_{\text{rms}} = 110 \times 2\times 10^{-4}$

$I_{\text{rms}} = 2.2\times 10^{-2}\ \text{A}$

Convert to mA

$2.2\times 10^{-2}\ \text{A} = 22\ \text{mA}$

Final answer

$\boxed{22}$