An electric charge 10^{-6} mu C is placed at | vedclass

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(C) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{kQ}{r}$.The distance of point $P(\sqrt{3}, \sqrt{3})$

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$0 	ext{ V}$

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(C) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{kQ}{r}$.The distance of point $P(\sqrt{3}, \sqrt{3})$ from the origin $(0,0)$ is $r_1 = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{3 + 3} = \sqrt{6} 	ext{ m}$.The distance of point $Q(\sqrt{6}, 0)$ from the origin $(0,0)$ is $r_2 = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6} 	ext{ m}$.Since $r_1 = r_2 = \sqrt{6} 	ext{ m}$,the potential at point $P$ is $V_P = \frac{kQ}{r_1}$ and the potential at point $Q$ is $V_Q = \frac{kQ}{r_2}$.Therefore,the potential difference $V_P - V_Q = \frac{kQ}{r_1} - \frac{kQ}{r_2} = 0 	ext{ V}$.

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

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