An electric charge 10^{-3} mu C is placed at | vedclass

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(C) The distance of point $A(\sqrt{2}, \sqrt{2})$ from the origin is:$OA = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 	ext{ uni

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(C) The distance of point $A(\sqrt{2}, \sqrt{2})$ from the origin is:$OA = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 	ext{ units}$.The distance of point $B(2, 0)$ from the origin is:$OB = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 	ext{ units}$.The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.Since both points $A$ and $B$ are at the same distance $r = 2 	ext{ units}$ from the charge $Q = 10^{-3} \mu C$ placed at the origin,the potentials at these points are equal:$V_A = \frac{1}{4\pi\epsilon_0} \frac{Q}{OA}$ and $V_B = \frac{1}{4\pi\epsilon_0} \frac{Q}{OB}$.Since $OA = OB = 2$,it follows that $V_A = V_B$.Therefore,the potential difference between points $A$ and $B$ is:$V_A - V_B = 0 	ext{ V}$.

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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