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(A) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{KQ}{r}$.Here,$Q = 10^{-8} \ C$.The position of the cha

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potential will be $18 \ V$

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(A) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{KQ}{r}$.Here,$Q = 10^{-8} \ C$.The position of the charge is $(4, 7, 2)$ and the observation point is $(1, 3, 2)$.The distance $r$ is calculated as $r = \sqrt{(1-4)^2 + (3-7)^2 + (2-2)^2} = \sqrt{(-3)^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \ m$.Substituting the values,$V = \frac{(9 	imes 10^9) 	imes 10^{-8}}{5} = \frac{90}{5} = 18 \ V$.Thus,the electric potential at the given point is $18 \ V$.

An electric charge $10^{-8} \ C$ is placed at the point $(4 \ m, 7 \ m, 2 \ m)$. At the point $(1 \ m, 3 \ m, 2 \ m)$,the electric

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