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Physics Question on Capacitors and Capacitance

A 20 muF20\,mu F capacitor is connected to 45 V45\,V battery through a circuit whose resistance is 2000 Ω2000\, \Omega . What is the final charge on the capacitor ?

A

9×10−4 C9 \times 10^{-4}\,C

B

9.154×10−4 C9.154 \times 10^{-4}\,C

C

9.8×10−4 C9.8 \times 10^{-4}\,C

D

None of these

Answer

9×10−4 C9 \times 10^{-4}\,C

Explanation

Solution

We know that in steady state the capacitor behaves like as an open circuit i.e., capacitor will not pass the current.

So, the potential difference across the capacitor
=45V=45 V
Hence, the final charge on the capacitor is
q=CVq=C V
Here, C=20μF,V=45VC=20 \mu F , V=45 V
∴q=20×10−6×45\therefore q=20 \times 10^{-6} \times 45
or q=900×10−6q=900 \times 10^{-6}
or q=9×10−4Cq=9 \times 10^{-4} C