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Physics Question on Capacitors and Capacitance

The energy stored in the capacitor as shown in Fig. (a) is 4.5×10−6 J4.5 \times 10^{-6} \,J. If the battery is replaced by another capacitor of 900 pF900 \,pF as shown in Fig. (b), then the total energy of system is

A

4.5×10−6 J4.5 \times 10^{-6}\, J

B

2.25×10−6 J2.25 \times 10^{-6}\, J

C

zero

D

9×10−6 J9 \times 10^{-6} \,J

Answer

2.25×10−6 J2.25 \times 10^{-6}\, J

Explanation

Solution

Energy stored in the capacitor in Fig (a) 12Q2C=4.5×10−6J\frac{1}{2} \frac{Q^{2}}{C}=4.5 \times 10^{-6} J If battery in Fig. (a) is replaced by capacitor in Fig. (b), total energy stored =12(12Q2C)=\frac{1}{2}\left(\frac{1}{2} \frac{Q^{2}}{C}\right) =12×4.5×10−6= \frac{1}{2} \times 4.5 \times 10^{-6} =2.25×10−6 J= 2.25 \times 10^{-6}\, J