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Physics Question on Capacitors and Capacitance

A 20 μF20 \, \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 =45 V= 45 \,V
Hence, the final charge on the capacitor is
q=cvq=cv
Here, C=20 μF,V=45 VC =20\,\mu F, V=45\,V
∴q=20×10−6×45\therefore q=20\times10^{-6}\times45
or q=900×10−6q=900\times10^{-6}
or q=9×10−4 Cq=9\times10^{-4}\,C