When an A.C. supply is connected to a capacitor, the voltage across the capacitor is out of phase with the current. The voltage across the capacitor will be the peak voltage of the AC supply. The RMS value of the AC supply is 220 V. The peak voltage \( V_{\text{peak}} \) is related to the RMS value \( V_{\text{rms}} \) by the formula: \[ V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2} \] Substituting the given value of \( V_{\text{rms}} = 220 \, \text{V} \): \[ V_{\text{peak}} = 220 \times \sqrt{2} = 220\sqrt{2} \, \text{V} \] Thus, the potential difference across the capacitor is \( 220\sqrt{2} \, \text{V} \).
The correct answer is (D) : \(220\sqrt2\ V\).