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(D) Let the two charges $Q$ be placed at points $A$ and $B$ separated by distance $d$. The mid-point is $O$. The charge $q$ is at point $P$ on the per

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

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(D) Let the two charges $Q$ be placed at points $A$ and $B$ separated by distance $d$. The mid-point is $O$. The charge $q$ is at point $P$ on the perpendicular bisector at distance $x$ from $O$.The distance between each charge $Q$ and charge $q$ is $r = \sqrt{x^2 + (d/2)^2}$.The magnitude of the Coulomb force exerted by each charge $Q$ on $q$ is $F = \frac{kQq}{x^2 + d^2/4}$.The horizontal components of the forces cancel out,and the net force $F_{	ext{net}}$ is directed along the perpendicular bisector:$F_{	ext{net}} = 2F \cos 	heta = 2 \left( \frac{kQq}{x^2 + d^2/4} ight) \left( \frac{x}{\sqrt{x^2 + d^2/4}} ight) = \frac{2kQqx}{(x^2 + d^2/4)^{3/2}}$.To find the maximum force,we set $\frac{dF_{	ext{net}}}{dx} = 0$:$\frac{d}{dx} \left[ 2kQqx (x^2 + d^2/4)^{-3/2} ight] = 0$.Using the product rule: $2kQq \left[ (x^2 + d^2/4)^{-3/2} + x(-3/2)(x^2 + d^2/4)^{-5/2}(2x) ight] = 0$.$(x^2 + d^2/4)^{-3/2} - 3x^2(x^2 + d^2/4)^{-5/2} = 0$.$(x^2 + d^2/4) - 3x^2 = 0 \implies d^2/4 = 2x^2 \implies x^2 = d^2/8$.Therefore,$x = \frac{d}{2\sqrt{2}}$.

Two point charges $Q$ each are placed at a distance $d$ apart. $A$ third point charge $q$ is placed at a distance $x$ from the mid-point on the perpendicular bisector. The value of $x$ at which charge $q$ will experience the maximum Coulomb force is ...............

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