Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is: ____.

We have two identical capacitors, each with capacitance

$C=200p^F=200{10}^{-12}F$

The first capacitor is charged to a voltage

$\mathrm{V\_0}=100V$

and then isolated, so its charge is

$Q=C\mathrm{V\_0}=200{10}^{-12}F100V=2{10}^{-8}C$

When this charged capacitor is connected in parallel with the second, uncharged capacitor, the total charge is conserved. The total capacitance becomes

$\mathrm{C\_\{tot\}}=2C$

Hence the final common voltage is

The initial electrostatic energy stored in the single charged capacitor was

$\mathrm{U\_i}=\frac{1}{2}C{\mathrm{V\_0}}^2=\frac{1}{2}200{10}^{-12}$ ${100}^2$

Evaluating:

$\mathrm{U\_i}=\frac{1}{2}200{10}^{-12}10000$

$\mathrm{U\_i}=1{10}^{-6}J$

After connection, the final energy is

$\mathrm{U\_f}=\frac{1}{2}\mathrm{C\_\{tot\}}{\mathrm{V\_f}}^2$

Substituting C_{tot}=2C and V_f=V_0/2:

$\mathrm{U\_f}=\frac{1}{2}2C{\mathrm{V\_0}2}^2$

which simplifies to

$\mathrm{U\_f}=\frac{1}{2}C{\mathrm{V\_0}}^2$

Thus

$\mathrm{U\_f}=\frac{1}{2}1{10}^{-6}J$

$\mathrm{U\_f}=0.5{10}^{-6}J$

The energy that is lost (converted to heat, radiation, etc.) during the charge redistribution is the difference between the initial and final energies:

$\mathrm{\backslash Delta\; U}=\mathrm{U\_i}-\mathrm{U\_f}=1{10}^{-6}J-0.5{10}^{-6}J=0.5{10}^{-6}J$

Therefore, the amount of electrostatic energy lost in the process is

$0.5{10}^{-6}J$