Sixty-four conducting drops,each of radius $0.02 \ m$ and each carrying a charge of $5 \ \mu C$,are combined to form a bigger drop. The ratio of the surface charge density of the bigger drop to that of the smaller drop will be ............

The four identical capacitors in the circuit shown below are initially uncharged. The switch is then thrown first to position $A$, and then to position $B$. After this is done:
Note: $V_{1,2,3,4}$ are the potential differences across $C_{1,2,3,4}$ and $Q_{1,2,3,4}$ are the final charges stored in $C_{1,2,3,4}$ respectively.

Seven capacitors each of capacitance $2 \mu F$ are to be connected in a configuration to obtain an effective capacitance $\left(\frac{10}{11}\right) \mu F$. The combination is

Initially,switch $S$ is connected to position $1$ for a long time as shown in the figure. The net amount of heat generated in the circuit after it is shifted to position $2$ is:

In the figure shown,after the switch $S$ is turned from position $A$ to position $B$,the energy dissipated in the circuit in terms of capacitance $C$ and total charge $Q$ is

$A$ hollow metal sphere has a radius $r$. The potential difference between a point on its surface and a point at a distance $3r$ from its center is $V$. The electric intensity at the distance $3r$ from the center of the sphere will be