# Calculate the potential at a point P due to a charge of $4 \times 10^{–7}C$ located $9\; cm$ away.

$q= 4 \times 10^{-7}\;C$
$r= 9\;cm=9 \times 10^{-2}\;m$
$V=?$
$V=\large\frac{1}{4 \pi \in_0 }\frac{q}{r}$$=9 \times 10^{9} NM^2 C^{-2} \times \large\frac{4 \times 10^{-7}C}{9 \times 10^{-2} m}$
$\qquad=4 \times 10^{-9}\;C$
Hence A is the correct answer.