An electric charge $$10^{-6}$$ $$\mu$$C is placed at origin $$(0, 0)$$ m of $$X - Y$$ co-ordinate system. Two points $$P$$ and $$Q$$ are situated at $$(\sqrt{3}, \sqrt{3})$$ m and $$(\sqrt{6}, 0)$$ m respectively. The potential difference between the points $$P$$ and $$Q$$ will be :

A point charge $$q = 10^{-6}\mu C = 10^{-12}$$ C is placed at the origin, and we consider the points P $$(\sqrt{3},\sqrt{3})$$ and Q $$(\sqrt{6},0)$$.

The distance of point P from the origin is $$r_P = \sqrt{3+3} = \sqrt{6}$$ m, while the distance of point Q is $$r_Q = \sqrt{6}$$ m.

The electric potential due to a point charge is given by $$V = \frac{kq}{r}$$. Since $$r_P = r_Q = \sqrt{6}$$, it follows that

$$V_P = V_Q = \frac{kq}{\sqrt{6}}$$

Consequently, the potential difference between P and Q is

$$V_P - V_Q = 0$$

The correct answer is Option 3: 0 V.