`0.2 F` capacitor is charged to `600 V` by a battery. On removing the battery it is connected with another parallel plate condenser of `1 F`. The potential decreases to

STATEMENT-1: A parallel plate capacitor is charged by a battery. The battery is then disconnected. If the distance the plates is increased, the enegry strored in the capacitor will decrease. STATEMENT-2: Work has to be done to increase the separation between the plates of a charged capacitor.

A 2 muF condenser is charged upto 200 V and then battery is removed. On combining this with another uncharged condenser in parallel, the potential differences between two plates are found to be 40 V. The capacity of second condenser is

A 2 muF condenser is charged upto 200 V and then battery is removed. On combining this with another uncharged condenser in parallel, the potential differences between two plates are found to be 40 V. The capacity of second condenser is

A capacitor C is charged to a potential V by a battery. It is then disconnected from the battery and again connected with its polarity reversed to the battery

A capacitor C is charged to a potential V by a battery. It is then disconnected from the battery and again connected with its polarity reversed to the battery

A 2 mu F condenser is charged upto 200 volt and then battery is removed. On combining this with another uncharged condenser in parallel, the potential difference is found to be 40 volt. Find the capacity of second condenser (in mu F )

2mu F capacitor is connected with 50V supply & 3mu F capacitor is connected with 100V supply. Now after removing battery if two plates of same type of charges are placed to form new capacitor then potential difference is ____ V.