JEE Main 2023 — Electrostatics Question with Solution

The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:

$V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$

In this case, we know (V) and (Q), and we're asked to solve for (r). We can rearrange the formula to solve for (r):

$r=\frac{1}{4\pi {ϵ}_0}\frac{Q}{V}$

Substituting the given values into this equation gives:

$r=\frac{9\times 10^9{\text{Nm}}^2{\text{C}}^{-2}}{1}\frac{5\times 10^{-9}\text{C}}{50\text{V}}=0.9\text{m}$

So, the distance of the point P from the point charge is 0.9 meters. This corresponds to 90 cm.