When a $0.75 \mu F$ capacitor is charged to a voltage of $20 \ V$,then the magnitude of charge on each plate is (in $\mu C$)

Two spherical conductors of capacities $3 \mu F$ and $2 \mu F$ are charged to the same potential,having radii $3 \ cm$ and $2 \ cm$ respectively. If $\sigma_1$ and $\sigma_2$ represent the surface charge density on the respective conductors,then $\frac{\sigma_1}{\sigma_2}$ is:

Eight drops of mercury, each of same radius and same charge, combine to form a bigger drop. The ratio of the capacitance of the bigger drop to that of each smaller drop is: (in $ : 1$)

The capacitance of a spherical capacitor is $100 \ pF$. If the spacing between the two spheres is $1 \ cm$,then the radius of the inner sphere of the capacitor is: (in $cm$)

For the shown situation,in the steady state condition,the ratio of the charge stored in the first capacitor to the charge stored in the $n^{th}$ capacitor is:

The radii of the inner and outer spheres of a spherical capacitor are $9\,cm$ and $10\,cm$ respectively. If the dielectric constant of the medium between the two spheres is $6$ and the charge on the inner sphere is $18 \times 10^{-9}\,C$,calculate the potential of the inner sphere,given that the outer sphere is earthed.