An isolated sphere has a capacitance of 50 pF. What would be the radius of the sphere? [UK PMT 2015]

As, potential of spherical charge (Q) having radius r is given by

                              $\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}·\frac{\mathrm{Q}}{\mathrm{r}}$

Also, capacitance, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{Q}}{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}}·{\displaystyle \frac{\mathrm{Q}}{\mathrm{r}}}}$

$\therefore$                          $\mathrm{C}=4{\mathrm{\pi \epsilon }}_0\mathrm{r}$

If $\mathrm{C}=50\times {10}^{-12}\mathrm{F},$ then the radius of the sphere,

                                $\mathrm{r}=\frac{50\times {10}^{-12}}{4{\mathrm{\pi \epsilon }}_0}$

                                  $=50\times {10}^{-12}\times 9\times {10}^9$

                                  $=50\times 9\times {10}^{-3}$   

                                   $=450\times {10}^{-3} \mathrm{m}$

                                   $=0.450 \mathrm{m} = 45 \mathrm{cm}$