Question

A capacitor of capacitance $C =900 \,pF$ is charged fully by $100\, V$ battery $B$ as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance $C =900 \,pF$ as shown in figure (b). The electrostatic energy stored by the system (b) is :

• A. $1.5 \times 10^{-6} J$
• B. $4.5 \times 10^{-6} J$
• C. $3.25 \times 10^{-6} J$
• D. $2.25 \times 10^{-6} J$

Answer 1

Answer: (D)

We have the initial energy as $\frac{1}{2} CV ^2=\frac{1}{2} \times 9 \times 10^{-10} \times 100^2=4.5 \times 10^{-6} J$
Now the capacitor is disconnected and is connected to another $900 pF$ capacitor.
In the steady situation, the two capacitors have their positive plates at the same potential.
Let the common potential difference be $V$ '. The charge on each capacitor is $Q ^{\prime}= CV$ '.
By charge conservation, $Q^{\prime}=Q / 2$. This implies $V^{\prime}=V / 2$.
The total energy of the system is $=2 \times(1 / 2) Q ^{\prime} V '=(1 / 4) QV =2.25 \times 10^{-6} J$
Thus we get the electrostatic energy stored by the system as $\frac{1}{2}\left(C_1+C_2\right) V^2=$ $\frac{1}{2}\left(2 \times 9 \times 10^{-10}\right) \times 50^2=2.25 \times 10^{-6}$