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:
(Assume, $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$)

We have a point charge $$q = 5 \times 10^{-9}$$ C, potential at P is $$V = 50$$ V, and $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$.

The electric potential due to a point charge is:

$$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{r}$$

Solving for $$r$$:

$$r = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q}{V}$$

$$r = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{50}$$

$$r = 9 \times 10^9 \times 10^{-10}$$

$$r = 0.9 \text{ m} = 90 \text{ cm}$$

So, the distance of P from the point charge is 90 cm. Hence, the correct answer is Option 4.