The lower plate of a parallel plate capacitor is supported on a rigid rod. The upper plate is suspended from one end of a balance. The two plates are joined together by a thin wire and subsequently disconnected. The balance is then counterpoised. Now a voltage $V = 5000\, volt$ is applied between the plates. The distance between the plates is $d =5\, mm$ and the area of each plate is $A = 100 cm^2.$ Then find out the additional mass placed to maintain balance.......$g$ [All the elements other than plates are massless and nonconducting] :-

Two spherical conductors each of capacity $C$ are charged to potentials $V$ and $ - V$. These are then connected by means of a fine wire. The loss of energy will be

Two identical capacitors are connected in parallel across a potenial difference $V$. After they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be

The energy of a charged capacitor is given by the expression ($q$= charge on the conductor and $C$ = its capacity)

Two small spheres each carrying a charge $q$ are placed $r$ metre apart. If one of the spheres is taken around the other one in a circular path of radius $r$, the work done will be equal to

A capacitor $4\,\mu F$ charged to $50\, V$ is connected to another capacitor of $2\,\mu F$ charged to $100 \,V$ with plates of like charges connected together. The total energy before and after connection in multiples of $({10^{ - 2}}\,J)$ is