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                                           

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 :

                  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 [b]