Question

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

• A. $10^{-5}J$
• B. $10^{-6}J$
• C. $10^{-7}J$
• D. $10^{-4}J$

Answer 1

Answer: (B)

$C=400\times 10^{-12}F$
$V=100V$
Initially energy stored $=\frac{1}{2}C{V}^2$
$=\frac{1}{2}400\times 10^{-12}\times {\left(10^2\right)}^2$
$=\frac{1}{2}\times 4\times 10^{-10}\times 10^4$
$=2\times 10^{-6}J$
$\Rightarrow Q=CV=4\times 10^{-10}\times 100$
$=4\times 10^{-8}C$
$\Rightarrow$ By again connecting the charged capacitor with a uncharged ideal capacitor.
$\Rightarrow$ Now, the charge $Q$ will divide into two capacitors.
$\Rightarrow$ As $V$ and $C$ are same for both, $Q^{\prime }$ will be same for both,
$\Rightarrow$ So $Q^{\prime }=\frac{Q}{2}=2\times 10^{-8}C$
$\Rightarrow$ So, energy stored in both the capacitors $=\left(\frac{{\left(Q^{\prime }\right)}^2}{2C}\right)=\frac{{\left(Q^{\prime }\right)}^2}{2C}$
$=\frac{\left(2\times 10^{-8}\right)}{8\times 10^{-10}}=10^{-6}J$
So energy lost $=2\times 10^{-6}-0.25\times 10^{-6}$
$=0.75\times 10^{-6}J\approx 10^{-6}J$