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(B) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$.For point $A$ at $(\sqrt{2}, \sqrt{2})$,the di

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(B) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$.For point $A$ at $(\sqrt{2}, \sqrt{2})$,the distance from the origin $(0, 0)$ is $r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$.For point $B$ at $(2, 0)$,the distance from the origin $(0, 0)$ is $r_B = \sqrt{2^2 + 0^2} = 2$.Since the distances $r_A$ and $r_B$ are equal,the electric potentials at points $A$ and $B$ are equal:$V_A = \frac{kq}{r_A} = \frac{kq}{2}$$V_B = \frac{kq}{r_B} = \frac{kq}{2}$Therefore,the potential difference between points $A$ and $B$ is:$V_A - V_B = \frac{kq}{2} - \frac{kq}{2} = 0 \ V$.

$A$ charge of $10^{-3} \ \mu C$ is placed at the origin $(0, 0)$ of an $X-Y$ coordinate system. Two points $A$ and $B$ are located at $(\sqrt{2}, \sqrt{2})$ and $(2, 0)$ respectively. The potential difference between points $A$ and $B$ is ....... $V$.

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