Question

A $60pF$ capacitor is fully charged by a $20V$supply. It is then disconnected from the supply and is connected to another uncharged $60pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is __(in $nJ$)

Answer 1

Step 1: Given data and drawing the diagram of situation

Value of capacitors, $C_1=60pF$

$=60\times {10}^{-12}F\left[As1F={10}^{12}pF\right]$

$C_2=60pF$

$=60\times {10}^{-12}F$

Potential difference, $V=20V$

Step 2: Find the resultant capacitor and draw the diagram after battery removal

Resultant capacitor, $C=C_1+C_2$

$$\begin{gathered} =C_1+C_1\left[As{C}_1=C_2\right] \\ =2C_1 \end{gathered}$$

Step 3: Find the lost energy

Let total amount of charge be $Q_0$.

Total charge, $Q_0=C_{1}V$ which will be always conserved.

Amount of lost energy, $∆E=\frac{{\left(Q_0\right)}^2}{2C_1}-\frac{{\left(Q_0\right)}^2}{2C}$

$$\begin{gathered} =\frac{{\left(C_{1}V\right)}^2}{2C_1}-\frac{{\left(C_{1}V\right)}^2}{2\times 2\times C_1} \\ =\frac{C_1{V}^2}{2}-\frac{C_1{V}^2}{4} \\ =\frac{C_1{V}^2}{4} \\ =\frac{60\times {10}^{-12}\times 20\times 20}{4} \\ =6000\times {10}^{-12}J \\ =6nJ \end{gathered}$$