Question

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

Answer 1

In the first condition, electrostatic energy is
$U_i=\frac{1}{2}C{V}_0^2$

$U_i=\frac{1}{2}\times 60\times {10}^{-12}\times 400$

$U_i=12\times {10}^{-9}\text{J}$

In the second condition,
$U_f=\frac{1}{2}{C}^{\prime }{V}^{\prime 2}$

As the capacitors are connected in parallel, and have the same capacitance,
$C^{\prime }=2C$

$V^{\prime }=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

Here, $C_1=C_2=C,V_1=V_0,V_2=0$

$\Rightarrow V^{\prime }=\frac{V_0}{2}$

$\Rightarrow U_f=\frac{1}{2}2C{\left(\frac{V_0}{2}\right)}^2$

$\Rightarrow U_f=\frac{1}{4}\times 60\times {10}^{-12}\times (20{)}^2$

$\Rightarrow U_f=6\times {10}^{-9}\text{J}$

Energy lost $\Delta U=U_i-U_f$

$\Rightarrow \Delta U=12\times {10}^{-9}\text{J}-6\times {10}^{-9}\text{J}$

$\Rightarrow \Delta U=6\text{nJ}$

Hence, $6$ is the correct answer.