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Physics Question on electrostatic potential and capacitance

An 8 μF8\, \mu F capacitor is connected across 220 V,50 Hz220\,V, 50 \,Hz line. What is the peak value of the charge through capacitor?

A

2.5×10−3 C2.5 \times 10^{-3}\,C

B

2.5×10−4 C2.5 \times 10^{-4}\,C

C

5×10−5 C5 \times 10^{-5}\,C

D

7.5×10−2 C7.5 \times 10^{-2}\,C

Answer

2.5×10−3 C2.5 \times 10^{-3}\,C

Explanation

Solution

Capacitive reactance
Xc=1ωC=12π×50×8×10−6X_{c}=\frac{1}{\omega C}=\frac{1}{2 \pi \times 50 \times 8 \times 10^{-6}}
=398 Ω=398\, \Omega
Crms=220V,Vpeak=2×220C_{\text{rms}}=220 V, V_{\text{peak}}=\sqrt{2} \times 220
=311 V=311 \,V
qrms=VrmsC=1.76×10−3 Cq_{\text{rms}}=V_{\text{rms}} C=1.76 \times 10^{-3}\, C
qrms=VrmsC=1.76×10−3 Cq_{\text{rms}}=V_{\text{rms}} C=1.76 \times 10^{-3} \,C
Peak vaiue of charge
qpeak=2qrms=1.76×10−3×2q_{\text{peak}}=\sqrt{2} q_{\text{rms}}=1.76 \times 10^{-3} \times \sqrt{2}
=2.5×10−3 C=2.5 \times 10^{-3} \,C