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(C) Let the two identical charges be $Q$ and the third charge be $q$. The distance between the two charges is $d$. The distance of the third charge fr

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

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(C) Let the two identical charges be $Q$ and the third charge be $q$. The distance between the two charges is $d$. The distance of the third charge from the midpoint of the line joining the two charges is $x$.The force exerted by each charge $Q$ on $q$ is $F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{x^2 + (d/2)^2}$.The resultant force $F_{res}$ on $q$ along the perpendicular bisector is $F_{res} = 2F \cos	heta$,where $\cos	heta = \frac{x}{\sqrt{x^2 + (d/2)^2}}$.Substituting $F$ and $\cos	heta$,we get $F_{res} = \frac{2Qqx}{4\pi\epsilon_0 (x^2 + d^2/4)^{3/2}}$.To find the maximum force,we differentiate $F_{res}$ with respect to $x$ and set it to zero: $\frac{dF_{res}}{dx} = 0$.Using the quotient rule or product rule,we get $(x^2 + d^2/4)^{3/2} - x \cdot \frac{3}{2}(x^2 + d^2/4)^{1/2} \cdot 2x = 0$.$(x^2 + d^2/4) - 3x^2 = 0 \implies 2x^2 = d^2/4 \implies x^2 = d^2/8$.Therefore,$x = \frac{d}{\sqrt{8}} = \frac{d}{2\sqrt{2}}$.

Two identical charges are separated by a distance $d$. $A$ third charge is placed on their perpendicular bisector at a distance $x$. For the third charge to experience maximum force,the value of $x$ is:

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