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(C) Let the two charges $Q$ be at points $B$ and $C$,separated by distance $d$. The third charge $q$ is at distance $x$ on the perpendicular bisector.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the two charges $Q$ be at points $B$ and $C$,separated by distance $d$. The third charge $q$ is at distance $x$ on the perpendicular bisector.The net force $F_{net}$ on charge $q$ is the sum of the components of the electrostatic forces along the perpendicular bisector:$F_{net} = 2F \cos 	heta$Here,$F = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Qq}{x^2 + (d/2)^2}$ and $\cos 	heta = \frac{x}{\sqrt{x^2 + (d/2)^2}}$.Substituting these into the expression for $F_{net}$:$F_{net} = 2 \cdot \frac{1}{4\pi \varepsilon_0} \cdot \frac{Qq}{x^2 + d^2/4} \cdot \frac{x}{(x^2 + d^2/4)^{1/2}} = \frac{2Qqx}{4\pi \varepsilon_0 (x^2 + d^2/4)^{3/2}}$To find the maximum force,we set the derivative with respect to $x$ to zero:$\frac{dF_{net}}{dx} = 0$Using the quotient rule or product rule,we differentiate the expression:$\frac{d}{dx} [x(x^2 + d^2/4)^{-3/2}] = 0$$(x^2 + d^2/4)^{-3/2} + x \cdot (-3/2)(x^2 + d^2/4)^{-5/2} \cdot (2x) = 0$$(x^2 + d^2/4)^{-3/2} - 3x^2(x^2 + d^2/4)^{-5/2} = 0$Multiplying by $(x^2 + d^2/4)^{5/2}$:$(x^2 + d^2/4) - 3x^2 = 0$$d^2/4 = 2x^2$$x^2 = d^2/8$$x = \frac{d}{2\sqrt{2}}$

List-$I$	List-$II$
$P. Q_1, Q_2, Q_3, Q_4$ all positive	$1. +x$
$Q. Q_1, Q_2$ positive; $Q_3, Q_4$ negative	$2. -x$
$R. Q_1, Q_4$ positive; $Q_2, Q_3$ negative	$3. +y$
$S. Q_1, Q_3$ positive; $Q_2, Q_4$ negative	$4. -y$

Two point positive charges are placed at a distance $d$ apart. $A$ third positive charge is placed at a distance $x$ on the perpendicular bisector. For what value of $x$ is the force on the third charge maximum?

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