Electric potential at a point $P$ due to a point charge of $5 \times 10^{-9}\; C$ is $50 \;V$. The distance of $P$ from the point charge is ......... $cm$
(Assume, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^{+9}\; Nm ^2 C ^{-2}$)
JEE MAIN 2023, Easy
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$V _{ P }=\frac{K Q}{ r }$
$50=\frac{9 \times 10^9 \times 5 \times 10^{-9}}{ r }$
$r =\frac{45}{50}=\frac{9}{10}=0.9\,m =90\,cm$
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