Point charges q_1 = 2,mu C and q_2 = -1,mu C | vedclass

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(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$. The total potent

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$x = 4$ and $x = 12$

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(C) The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$. The total potential is the algebraic sum of potentials due to individual charges.Let the point where potential is zero be at position $x$. The potential is given by:$V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q_1}{x} + \frac{q_2}{|x - 6|} ight] = 0$Case $1$: Point between the charges $(0 $\frac{2 	imes 10^{-6}}{x} + \frac{-1 	imes 10^{-6}}{6 - x} = 0$$\frac{2}{x} = \frac{1}{6 - x} \implies 12 - 2x = x \implies 3x = 12 \implies x = 4$.Case $2$: Point outside the charges $(x > 6)$:$\frac{2 	imes 10^{-6}}{x} + \frac{-1 	imes 10^{-6}}{x - 6} = 0$$\frac{2}{x} = \frac{1}{x - 6} \implies 2x - 12 = x \implies x = 12$.Thus,the potential is zero at $x = 4$ and $x = 12$.

Point charges $q_1 = 2\,\mu C$ and $q_2 = -1\,\mu C$ are kept at points $x = 0$ and $x = 6$ respectively. Electrical potential will be zero at points

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