Question

A 900 pF capacitor is charged by 100 V battery as shown in figure. The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?

• A. $1.25\times {10}^{-6}J$
• B. $5.25\times {10}^{-6}J$
• C. $2.25\times {10}^{-6}J$
• D. $8.24\times {10}^{-6}J$

Answer 1

The correct option is C $2.25\times {10}^{-6}J$

$\text{Step 1: Initial energy stored in the capacitor}\left(U_i\right)$

$U_i=\frac{1}{2}{C}_1{V}^2=\frac{1}{2}\times 900\times {10}^{-12}F\times (100{)}^2$ $=4.5\times {10}^{-6}J$

$\text{Step 2: Charge on each capacitor}$

$C_1=900pF$ is connected with another $C_2=900pF$ capacitor.

In steady-state, the two capacitors will have the same potential, say $V^{\prime }$. Let say $Q_1$ and $Q_2$ are the charges on the capacitor.

$\therefore \frac{Q_1}{900pF}=\frac{Q_2}{900pF}\Rightarrow Q_1=Q_2$

Initial charge $=Q$

Charge conservation : $Q_1+Q_2=Q$

$\therefore Q_1=\frac{Q}{2},Q_2=\frac{Q}{2}$

$\text{Step 3: New potential on the capacitors.}$

$V^{\prime }=\frac{Q_1}{C_1}=\frac{Q}{2C_1}=\frac{V}{2}=\frac{100}{2}=50V$.

$\text{Step 4: Final electrostatic energy stored}$

$U_f=\frac{1}{2}{C}_1{V}^{\prime 2}+\frac{1}{2}{C}_2{V}^{\prime 2}$ $[\because C_1=C_2]$

$=C_1{V}^{\prime 2}$

$=900\times {10}^{-12}F\times (50{)}^2$$=2.25\times {10}^{-6}J$

Hence, Option (C) is correct.

$\text{Note:}$ $U_i-U_f=2.25\times {10}^{-6}J$

This much energy is lost in the form of heat.