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(B) Let the two charges $q$ be placed at $(0, a)$ and $(0, -a)$. $A$ test charge $q_0$ is placed at $(x, 0)$.The force exerted by each charge on $q_0$

Vedclass · Accepted Answer

$2$

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(B) Let the two charges $q$ be placed at $(0, a)$ and $(0, -a)$. $A$ test charge $q_0$ is placed at $(x, 0)$.The force exerted by each charge on $q_0$ is $F' = \frac{K q q_0}{x^2 + a^2}$.The components of these forces along the $y$-axis cancel out,while the components along the $x$-axis add up.The net force is $F = 2 F' \cos 	heta = 2 \left( \frac{K q q_0}{x^2 + a^2} ight) \left( \frac{x}{\sqrt{x^2 + a^2}} ight) = \frac{2 K q q_0 x}{(x^2 + a^2)^{3/2}}$.To find the maximum force,we set $\frac{dF}{dx} = 0$.$\frac{d}{dx} [x(x^2 + a^2)^{-3/2}] = (x^2 + a^2)^{-3/2} + x \cdot (-\frac{3}{2}) (x^2 + a^2)^{-5/2} \cdot 2x = 0$.$(x^2 + a^2)^{-3/2} = 3x^2 (x^2 + a^2)^{-5/2}$.$x^2 + a^2 = 3x^2 \implies 2x^2 = a^2 \implies x = \frac{a}{\sqrt{2}}$.Comparing this with $\frac{a}{\sqrt{x}}$,we get $x = 2$.

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