$125$ identical drops,each charged to a potential of $50\;V$,are combined to form a single drop. The potential of the new drop will be......$V$.

Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is

In the given figure,three capacitors $C_1, C_2$ and $C_3$ are connected to a battery. With symbols having their usual meanings,the correct conditions are:

Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of radii $R_{1}$ and $R_{2},$ respectively. The electric fields $E_{1}$ (on $S_{1}$) and $E_{2}$ (on $S_{2}$) on their surfaces are such that $E_{1} / E_{2} = R_{1} / R_{2}.$ Then the ratio $V_{1} / V_{2}$ of the electrostatic potentials on each sphere is:

An electrician requires a capacitance of $6 \mu F$ in a circuit across a potential difference of $1.5 kV$. $A$ large number of $2 \mu F$ capacitors which can withstand a potential difference of not more than $500 V$ are available. The minimum number of capacitors required for the purpose is

$A$ spherical conductor of radius $2\,m$ is charged to a potential of $120\,V$. It is now placed inside another hollow spherical conductor of radius $6\,m$. Calculate the potential to which the bigger sphere would be raised.