Two metallic spheres are placed in a liquid of dielectric constant 9, are place a such that their centers are nearly 1m apart. It is given that radius of each sphere is 2 cm and capacitance of system of spheres is 10^-N farads. The value of N is

As, distance between spheres is very large than the radius of sphere

System will be a capacitor, if we lace a charge + q over 1st and – q over 2nd sphere.

Potential on the surface of 1st sphere is ${\mathrm{V}}_1=\frac{\mathrm{Kq}}{{\mathrm{r}}_1}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{\epsilon }}_{\mathrm{r}}\mathrm{r}}$

and potential on the surface of 2nd sphere is ${\mathrm{V}}_2=\frac{\mathrm{Kq}}{{\mathrm{r}}_2}=\frac{-\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{\epsilon }}_{\mathrm{r}}\mathrm{r}}$

where C = capacitance of system= Charge/ Potential difference.

$\mathrm{C}=\frac{\mathrm{q}}{{\displaystyle \frac{\mathrm{q}}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{\epsilon }}_{\mathrm{r}}\mathrm{r}}}}$

$\begin{array}{l}\Rightarrow \mathrm{C}=2{\mathrm{\pi \epsilon }}_0{\mathrm{\epsilon }}_{\mathrm{r}}\mathrm{r}\Rightarrow \mathrm{C}=4{\mathrm{\pi \epsilon }}_0\frac{{\mathrm{\epsilon }}_{\mathrm{r}}}{2}\mathrm{r}\\ =\frac{9\times 2\times {10}^{-2}}{9\times {10}^9\times 2}={10}^{-7}\mathrm{F}\therefore \mathrm{N}=7\end{array}$