Electric potential at a point P due to a point charge of 5 × 10 ^ -...

The electric potential for a point charge is given by

$V = \dfrac{Q}{4\pi{ε}_{0}r}$

The data given is

$V = 50VQ = 5 \times {10}^{- 9}C$

So, the distance of the point is

$r = \dfrac{1}{4\pi{ε}_{0}}\dfrac{Q}{V}$
$⇒ r = 9 \times {10}^{9} \times \dfrac{5 \times {10}^{- 9}}{50}$
$⇒ r = 0 . 9m = 90cm$