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

Hint: We will find initial charge on initial capacitor of $60pF$ and total energy of circuit. Then after adding a second capacitor, we will do voltage balance in the circuit. From there we will find the final charge in both capacitors. Then we will find the final energy of the circuit. The difference between final and initial energies will be the energy loss in the process.

Complete step-by-step solution:Let initial charge on capacitor be ${Q_0}$
Then, ${Q_0} = {C_1}V$
Where ${C_1} = 60pF$ and $V = 20V$

Therefore, ${Q_0} = 60 \times {10^{ - 10}} \times 20$
So we get,
${Q_0} = 120 \times {10^{ - 10}}C$
Therefore, ${Q_0} = 1200pC$
Initial energy ${U_i}$ is given by $\dfrac{1}{2}{C_1}{V^2}$
On putting values, we get,
${U_i} = \dfrac{1}{2} \times 60 \times {10^{ - 10}} \times {(20)^2}$
So we get,
\[{U_i} = 120 \times {10^{ - 10}}J\]
When battery is disconnected and new capacitor is added, then,
Let final charge on first capacitor be ${Q_f}$ and charge on second capacitor be $q$

Therefore, ${Q_f} = {Q_0} - q$









Doing force balance on above circuit, we get the equation,
${V_B} - \dfrac{q}{{{C_2}}} + \dfrac{{{Q_0} - q}}{{{C_1}}} = {V_B}$ ,
On simplifying we get,
\[\dfrac{{{Q_0} - q}}{{{C_1}}} = \dfrac{q}{{{C_2}}}\] ,
Putting values we get,
$\dfrac{{1200 - q}}{{60}} = \dfrac{q}{{60}}$ ,
On simplifying we get,
$1200 = 2q$ ,
On further simplification, we get,
$q = 600pC$
So, ${Q_f} = {Q_0} - q$
So we get,
${Q_f} = 1200 - 600 = 600$
Final energy ${U_f}$ is the sum of energies in both capacitors so,
${U_f} = \dfrac{{{Q_f}^2}}{{2 \times {C_1}}} + \dfrac{{{q^2}}}{{2 \times {C_2}}}$
On putting values we get,
${U_f} = \dfrac{{{{\left( {600} \right)}^2}}}{{2 \times 60}} + \dfrac{{{{\left( {600} \right)}^2}}}{{2 \times 60}}$
On solving we get,
\[{U_f} = 60 \times {10^{ - 10}}J\]
Energy lost is ${U_i} - {U_f}$
So, ${\text{energy loss = 120}} \times {\text{1}}{{\text{0}}^{ - 10}}J - 60 \times {10^{ - 10}}J$
On solving, ${\text{energy loss = 12}} \times {\text{1}}{{\text{0}}^{ - 9}}J - 6 \times {10^{ - 9}}J$
On solving we get,
${\text{energy loss = }}6 \times {10^{ - 9}}J{\text{ = 6nJ}}$
So, answer is $6$

Note:- We subtracted final energy from initial energy in solution because initial energy will be greater than final energy. Energy loss in the process is due to heat generated while connecting the initial capacitor with the second capacitor. That’s why energy loss is in the form of heat.