Question

Energy of the charged capacitor is $U$.Now it is removed from a battery and then connected to another identical uncharged capacitor in parallel. What will be the energy of each capacitor now?
$\text{A}\text{. }\dfrac{3U}{2}$
$\text{B}\text{. }U$
$\text{C}\text{. }\dfrac{U}{4}$
$\text{D}\text{. }\dfrac{U}{2}$

Answer 1

Hint: After capacitors are connected in parallel, we will find the final charge acquired by each capacitor and the voltage drop across each capacitor in parallel combination. Then we will calculate the relation between the initial energy of the capacitor and its final energy after being connected in parallel to another identical capacitor.
Formula used:
$Q=CV$
$U=\dfrac{{{Q}^{2}}}{2C}$

Completer step by step answer:
Capacitors are the devices that store electrical energy in an electric field. They are passive electronic components with two components. The effect of any capacitor is known as its Capacitance.
Capacitors can be connected in different combinations, series and parallel.
When capacitors are connected in series, the total capacitance is less than any one of the series capacitors’ individual capacitances.

Charge stored by each capacitor in series combination remains the same; however, the potential drop across each capacitor is different.
If $n$ capacitors are being connected in series, then, the Effective capacitance in series combination,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+.....\dfrac{1}{{{C}_{n}}}$
When capacitors are being connected in parallel, the total capacitance is the sum of the individual capacitors’ capacitances.

Voltage drop across each capacitor in parallel combination remains the same; however, the charge on each capacitor is different.
If $n$ capacitors are connected in parallel, then, the Effective capacitance in parallel combination,
$C={{C}_{1}}+{{C}_{2}}+.....{{C}_{n}}$
As given, Energy of charged capacitor $=U$
Let the energy of each capacitor after connected in parallel $=E$
Let the charge on capacitor be $Q$
Therefore, the initial energy stored in the capacitor will be, $\dfrac{{{Q}^{2}}}{2C}$
When the capacitors are connected in parallel, the flow of charge will be such that the potential difference and total charge$Q$across the capacitor remains the same.
Since, the capacitors connected are identical ones, therefore, potential difference across the capacitor,
$V=\dfrac{Q}{2C}$
Energy in each capacitor,
$E=\dfrac{1}{2}C{{V}^{2}}$
$E=\dfrac{1}{2}C{{\left( \dfrac{Q}{2C} \right)}^{2}}=\dfrac{1}{4}\times \dfrac{{{Q}^{2}}}{2C}$
$U=\dfrac{{{Q}^{2}}}{2C}$
We get,
$E=\dfrac{U}{4}$
The energy on each capacitor connected in parallel would be $\dfrac{U}{4}$
Hence, the correct option is C.
Note:
When capacitors are connected in parallel, the combination is such that the voltage drop across each capacitor remains the same and the charge gets divided between them. In the above case, the capacitors connected in parallel were identical, meaning they had equal value of capacitance, so the charge gets divided equally between two of them. However, when the capacitances are not equal, we will use the formula $Q=CV$ for finding the value of charge acquired by each capacitor.