question_answer

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [NEET-2017]

A) Increases by a factor of 2

B) Increases by a factor of 4

C) Decreases by a factor of 2

D) Remains the same

Correct Answer: C

Solution :

[c]

Charge on capacitor

\[q=CV\]

when it is connected with another uncharged capacitor.

\[{{V}_{c}}=\frac{{{q}_{1}}+{{q}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{q+0}{C+C}\]

\[{{V}_{c}}=\frac{V}{2}\]

Initial energy

\[{{U}_{i}}=\frac{1}{2}C{{V}^{2}}\]

Final energy

\[{{U}_{f}}=\frac{1}{2}C{{\left( \frac{V}{2} \right)}^{2}}+\frac{1}{2}C{{\left( \frac{V}{2} \right)}^{2}}\]

\[=\frac{C{{V}^{2}}}{4}\]

Loss of energy \[={{U}_{i}}-{{U}_{f}}\]

\[=\frac{C{{V}^{2}}}{4}\]

i.e. decreases by a factor