question_answer

A parallel place capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitor are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is:

A) zero                      

B)    \[\frac{3}{2}C{{V}^{2}}\]

C) \[\frac{25}{6}C{{V}^{2}}\]            

D)          \[\frac{9}{2}C{{V}^{2}}\]

Correct Answer: B

Solution :

Net charge shared by capacitor                  \[=Q'=4CV-CV=3CV\] \[{{q}_{1}}={{C}_{1}}{{V}_{1}}=CV\] \[{{Q}_{2}}={{C}_{2}}{{V}_{2}}\] \[=(2C)(2V)=4CV\] Common potential \[V'=\frac{Q'}{C'}=\frac{3CV}{3C}=V\] \[E'=\frac{1}{2}C'{{V}^{2}}=\frac{1}{2}(3C){{V}^{2}}=\frac{3}{2}C{{V}^{2}}\]