A capacitor of capacity `20microF` is charged upto `50V` and and disconnected from cell. Now this charged capacitor is connected to another capacitor of capacitance `C` If final common potential is `20V` then find the capacitance `C`.

A capacitor of capacitance of 2muF is charged to a potential difference of 200V , after disconnecting from the battery, it is connected in parallel with another uncharged capacitor. The final common potential is 20V then the capacitance of second capacitor is :

A capacitor of capacity C charged to potential "V" and " then connected to an identical uncharged capacitor then loss of energy is

A capacitor of capacity C_(0) is charged to certain potential and enrgy stored in it is U_(0) . Now this capacitor is connected in parrallel to uncharged capacitor of capacitance. C . Find the loss of energy.

A capacitor of capacitance 4muF is charged to a potential of 100 V. it is then disconnected from the battery and connected in parallel with another capacitor C_(2) . If their common potential is 40 volts, then the value of C_(2) is

A capacitor of capacitance 10muF charged to 100V is connected to an uncharged capacitor. The effective potential now is 40 V. The capacitance on uncharged capacitor is

An isolated capacitor of unknown capacitance has been charged to potential V_(0) . This capacitor is then connected to an uncharged capacitor of capacity C . If common potential is V(ltV_(0) , find unknow capacitance.

A capacitor of capacitance C is charged to a potential differece V_(0) . The terminals of the charged capacitor are then connected to those of an uncharged capacitor of capacitance C//2 . a. Compute the original charge of the system. b. Find the final potential differce across each capacitor. c. Find the final energy of the system. d. Calculate the decrease in energy when the capacitors are connected. e. Where did the ''lost'' energy go?