What is the electric potential at a distance of 9 cm from 3 nC? [KCET 2014]

Given, $\mathrm{q}=3 \mathrm{nC}=3\times {10}^{-9} \mathrm{C}$

and      $\mathrm{r}=9 \mathrm{cm}=9\times {10}^{-2} \mathrm{m}$

We know that, potential due to point charge is given by

           $\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\mathrm{r}}=\frac{9\times {10}^9\times 3\times {10}^{-9}}{9\times {10}^{-2}}=3\times {10}^2 \mathrm{V}$

$\therefore$ Electric potential, V = 300 V