Question

Two identical capacitors have the same capacitance $C$. One of them is charged to potential ${V_1}$ and the other to ${V_2}$. The negative ends of the capacitor are connected together. When the positive ends are also connected, the decrease in the energy of the combine system is
A. $\dfrac{C}{4}\left( {V_1^2 - V_2^2} \right)$
B. $\dfrac{C}{4}\left( {V_1^2 + V_2^2} \right)$
C. $\dfrac{C}{4}{\left( {{V_1} - {V_2}} \right)^2}$
D. $\dfrac{C}{4}{\left( {{V_1} + {V_2}} \right)^2}$

Answer 1

Use the formula for the energy stored in the capacitor. Using this formula determines the initial energy of the system of the capacitors. Then determine the potential of the system of two capacitors connected in parallel. Hence, calculate the energy of the system of capacitors connected in parallel. Take the difference of the initial and final energy of the capacitors to determine the decrease in the energy of the combined system.

Formula used:
The energy $U$ stored in the capacitor is given by
$U = \dfrac{1}{2}C{V^2}$ …… (1)
Here, $C$ is the capacitance and $V$ is the potential difference across the plates of the capacitor.

Complete step by step answer:
We have given that there is a system of two capacitors having the same capacitance $C$. One of them is charged to potential ${V_1}$ and the other to ${V_2}$.
The initial energy of the combination of the system of two capacitors is given by
${U_i} = \dfrac{1}{2}CV_1^2 + \dfrac{1}{2}CV_2^2$
When the two negative and positive ends of the capacitors are connected then the two capacitors behave as two capacitors connected in parallel.
Therefore, the resultant potential of these two capacitors is given by
$V = \dfrac{{{V_1} + {V_2}}}{2}$
When the two capacitors are connected in parallel, the energy of the combination of these two parallel capacitors is given by
${U_f} = \dfrac{1}{2}\left( {C + C} \right){V^2}$
Substitute $\dfrac{{{V_1} + {V_2}}}{2}$ for $V$ in the above equation.
${U_f} = \dfrac{1}{2}\left( {C + C} \right){\left( {\dfrac{{{V_1} + {V_2}}}{2}} \right)^2}$
$\Rightarrow {U_f} = \dfrac{C}{4}{\left( {{V_1} + {V_2}} \right)^2}$
The decrease in the energy of the system of capacitors is the difference of the initial energy ${U_i}$ of the system of capacitors and the final energy ${U_f}$ of the capacitors connected in parallel.
$\Delta U = {U_i} - {U_f}$
Substitute $\dfrac{1}{2}CV_1^2 + \dfrac{1}{2}CV_2^2$ for ${U_i}$ and $\dfrac{C}{4}{\left( {{V_1} + {V_2}} \right)^2}$ for ${U_f}$ in the above equation.
$\Delta U = \dfrac{1}{2}CV_1^2 + \dfrac{1}{2}CV_2^2 - \dfrac{C}{4}{\left( {{V_1} + {V_2}} \right)^2}$
$\Rightarrow \Delta U = \dfrac{1}{2}CV_1^2 + \dfrac{1}{2}CV_2^2 - \dfrac{C}{4}\left( {V_1^2 + 2{V_1}{V_2} + V_2^2} \right)$
$\Rightarrow \Delta U = \dfrac{1}{2}CV_1^2 + \dfrac{1}{2}CV_2^2 - \dfrac{C}{4}V_1^2 - \dfrac{C}{4}2{V_1}{V_2} - \dfrac{C}{4}V_2^2$
$\Rightarrow \Delta U = \dfrac{1}{4}CV_1^2 + \dfrac{1}{4}CV_2^2 - \dfrac{1}{2}C{V_1}{V_2}$
$\therefore \Delta U = \dfrac{C}{4}{\left( {{V_1} - {V_2}} \right)^2}$
Therefore, the decrease in the energy of combination of the system of the two capacitors is $\dfrac{C}{4}{\left( {{V_1} - {V_2}} \right)^2}$.

Hence, the correct option is C.

Note: The students should not forget to determine the resultant potential of the two capacitors connected in parallel. Also the students may think that the decrease in the energy of the combined system of capacitors is the energy of the capacitors connected in parallel. But the decrease in energy of the capacitors is the difference of the initial system of capacitors and final system of capacitors.