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(C) Let the two equal charges be $Q$ and the third charge be $q$. The distance between the two charges $Q$ is $d$. The third charge $q$ is placed at a

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$x = \frac{d}{2\sqrt{2}}$

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(C) Let the two equal charges be $Q$ and the third charge be $q$. The distance between the two charges $Q$ is $d$. The third charge $q$ is placed at a distance $x$ on the perpendicular bisector.The force exerted by each charge $Q$ on $q$ is $F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qq}{x^2 + (d/2)^2}$.The net force $F_{net}$ on $q$ is the sum of the components of these forces along the perpendicular bisector:$F_{net} = 2F \cos 	heta$,where $\cos 	heta = \frac{x}{\sqrt{x^2 + (d/2)^2}}$.Substituting the values:$F_{net} = 2 \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{Qq}{x^2 + d^2/4} \cdot \frac{x}{(x^2 + d^2/4)^{1/2}} = \frac{2Qqx}{4\pi\varepsilon_0 (x^2 + d^2/4)^{3/2}}$.For $F_{net}$ to be maximum,$\frac{dF_{net}}{dx} = 0$.Using the quotient rule or differentiating $f(x) = x(x^2 + a^2)^{-3/2}$ where $a = d/2$:$f'(x) = (x^2 + a^2)^{-3/2} + x \cdot (-3/2)(x^2 + a^2)^{-5/2} \cdot 2x = 0$.$(x^2 + a^2)^{-5/2} [ (x^2 + a^2) - 3x^2 ] = 0$.$a^2 - 2x^2 = 0 \implies x^2 = a^2/2 \implies x = a/\sqrt{2}$.Since $a = d/2$,we get $x = \frac{d/2}{\sqrt{2}} = \frac{d}{2\sqrt{2}}$.

Two equal charges are separated by a distance $d$. $A$ third charge placed on a perpendicular bisector at $x$ distance will experience maximum Coulomb force when

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