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(B) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by the formula $V = \frac{kq}{r}$,where $k = 9 	imes 10^9 \, N \cdo

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(B) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by the formula $V = \frac{kq}{r}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.For point $A$ at $(\sqrt{2}, \sqrt{2})$,the distance from the origin $(0, 0)$ is $r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \, m$.For point $B$ at $(2, 0)$,the distance from the origin $(0, 0)$ is $r_B = \sqrt{2^2 + 0^2} = \sqrt{4} = 2 \, m$.Since the distances $r_A$ and $r_B$ are equal,the electric potentials at points $A$ and $B$ are equal: $V_A = V_B = \frac{kq}{2}$.The potential difference between points $A$ and $B$ is $\Delta V = V_A - V_B = 0 \, V$.

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

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