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Physics Question on electrostatic potential and capacitance

A 400 pF400\,pF capacitor is charged by a 100 V100\,V supply. How much electrostatic energy is lost in the process of disconnecting from the supply and connecting another uncharged 400 pF400\,pF capacitor?

A

10−5  J10^{-5}\,\,J

B

10−6  J10^{-6}\,\,J

C

10−7  J10^{-7}\,\,J

D

10−4  J10^{-4}\,\,J

Answer

10−6  J10^{-6}\,\,J

Explanation

Solution

C=400×10−12FC=400 \times 10^{-12} F
V=100 VV=100\, V
Initially energy stored =12CV2=\frac{1}{2} C V^{2}
=12400×10−12×(102)2=\frac{1}{2} 400 \times 10^{-12} \times\left(10^{2}\right)^{2}
=12×4×10−10×104=\frac{1}{2} \times 4 \times 10^{-10} \times 10^{4}
=2×10−6J=2 \times 10^{-6} J
⇒Q=CV=4×10−10×100\Rightarrow Q=C V=4 \times 10^{-10} \times 100
=4×10−8C=4 \times 10^{-8} C
⇒\Rightarrow By again connecting the charged capacitor with a uncharged ideal capacitor.
⇒\Rightarrow Now, the charge QQ will divide into two capacitors.
⇒\Rightarrow As VV and CC are same for both, Q′Q' will be same for both,
⇒\Rightarrow So Q′=Q2=2×10−8CQ'=\frac{Q}{2}=2 \times 10^{-8} C
⇒\Rightarrow So, energy stored in both the capacitors =((Q′)22C)=(Q′)22C=\left(\frac{\left(Q'\right)^{2}}{2 C}\right)=\frac{\left(Q'\right)^{2}}{2 C}
=(2×10−8)8×10−10=10−6J=\frac{\left(2 \times 10^{-8}\right)}{8 \times 10^{-10}}=10^{-6} J
So energy lost =2×10−6−0.25×10−6=2 \times 10^{-6}-0.25 \times 10^{-6}
=0.75×10−6J≈10−6J=0.75 \times 10^{-6} J \approx 10^{-6} J