Two point charges q_1 = 4 times 10^{-8} C an | vedclass

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(B) Let the point where the electric potential is zero be at a distance $x$ from point $A$ along the line $AB$.The potential $V$ due to a point charge

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$20$

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(B) Let the point where the electric potential is zero be at a distance $x$ from point $A$ along the line $AB$.The potential $V$ due to a point charge $q$ at a distance $r$ is given by $V = \frac{kq}{r}$.The total potential at the point is the algebraic sum of the potentials due to $q_1$ and $q_2$.$V_{total} = \frac{k q_1}{x} + \frac{k q_2}{50 - x} = 0$.Substituting the values: $\frac{4 	imes 10^{-8}}{x} + \frac{-6 	imes 10^{-8}}{50 - x} = 0$.$\frac{4}{x} = \frac{6}{50 - x}$.$4(50 - x) = 6x$.$200 - 4x = 6x$.$10x = 200$.$x = 20 \ cm$.Thus,the potential is zero at a distance of $20 \ cm$ from point $A$.

Two point charges $q_1 = 4 \times 10^{-8} \ C$ and $q_2 = -6 \times 10^{-8} \ C$ are placed at points $A$ and $B$ respectively,which are $50 \ cm$ apart. At what distance from point $A$ on the line $AB$ is the electric potential zero (in $cm$)?

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