A charge Q = 10^-6 C is placed at origin. Find the potential difference between two points A and B whose position vectors are $(\sqrt{3}\stackrel{^<!--\; ^\; -->}{i}+\sqrt{3}\stackrel{^<!--\; ^\; -->}{j})m$ and $(\sqrt{6}\stackrel{^<!--\; ^\; -->}{j})m$ respectively.

The correct option is Zero.

To find the potential difference between two points A and B due to a point charge Q placed at the origin, we first need to determine the distances of these points from the origin.
The potential $V$ at a distance $r$ from a point charge $Q$ is given by the formula:
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$

Let the position vector of point A be:
${\overrightarrow{r}}_A=\sqrt{3}\hat{i}+\sqrt{3}\hat{j}$
The distance of point A from the origin, $r_A$, is the magnitude of this position vector:
$r_A=\left|{\overrightarrow{r}}_A\right|=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{3}\right)}^2}=\sqrt{3+3}=\sqrt{6}\text{m}$

Let the position vector of point B be:
${\overrightarrow{r}}_B=\sqrt{6}\hat{j}$
The distance of point B from the origin, $r_B$, is the magnitude of this position vector:
$r_B=\left|{\overrightarrow{r}}_B\right|=\sqrt{{\left(\sqrt{6}\right)}^2}=\sqrt{6}\text{m}$

Since both points A and B are at the same distance from the origin ($r_A=r_B=\sqrt{6}\text{m}$), the electric potential at point A is equal to the electric potential at point B:
$V_A=V_B$

Therefore, the potential difference between the two points A and B is:
$\Delta V=V_A-V_B=0$