Physics - Electrostatics Question with Solution | TestHub

For charged spherical conductor, potential inside the sphere is same as that on its surface.

$\therefore$Potential inside the surface,$V_{in}=V_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}$

$\frac{\mathrm{ \; \; 1}}{4\pi {\epsilon }_0}\frac{q}{10}=V$

and potential outside the surface,$\frac{1}{4\pi {\epsilon }_0}\frac{q}{15}$

$\therefore$The ratio$\frac{V_{15}}{V}=\frac{q}{15}\times \frac{10}{q}=\frac{2}{3}$

$\Rightarrow V_{15}=\frac{2}{3}V$

$\Rightarrow V_{out}=\frac{2}{3}V$