Question

Two equal point charges $A$ and $B$ are $R$ distance apart. An equal third charge is placed on the perpendicular bisector at a distance $d$ from the centre will experience maximum electrostatic force when

• A. $d=\frac{R}{2\sqrt{2}}$
• B. $d=\frac{R}{\sqrt{2}}$
• C. $d=R\sqrt{2}$
• D. $d=2\sqrt{2}R$

Answer 1

The correct option is A $d=\frac{R}{2\sqrt{2}}$


Let each charge be $Q$.
The third point charge $\left(Q\right)$ is placed on the perpendicular bisector, thus its distance from both charges at $A$ and $B$ will be equal to $r$
$r=\sqrt{d^2+\frac{R^2}{4}}$
From symmetry the forces exerted by $A$ and $B$ will be same.
$F=\frac{k\left(Q\right)\left(Q\right)}{r^2}$
$\Rightarrow F=\frac{k{Q}^2}{[d^2+\frac{R^2}{4}]}$
Net force on charge placed at perpendicular bisector,
$F_{\text{net}}=2F\cos \theta$ (Horizontal component of force will cancel each other)
and $\cos \theta =\frac{d}{\sqrt{d^2+\frac{R^2}{4}}}$
$\Rightarrow F_{\text{net}}=\frac{2k{Q}^2}{(d^2+\frac{R^2}{4})}.\frac{d}{\sqrt{d^2+\frac{R^2}{4}}}$
or, $F_{\text{net}}=\frac{2k{Q}^{2}d}{{(d^2+\frac{R^2}{4})}^{\frac{3}{2}}}$
For maximum force,
$\frac{d{F}_{\text{net}}}{d\left(d\right)}=0(\because d\text{is variable here})$
$\Rightarrow \frac{2k{Q}^2[{(d^2+\frac{R^2}{4})}^{\frac{3}{2}}.1-\frac{3}{2}.2d^2{(d^2+\frac{R^2}{4})}^{\frac{1}{2}}]}{{\left[{(d^2+\frac{R^2}{4})}^{\frac{3}{2}}\right]}^2}=0$
or, ${(d^2+\frac{R^2}{4})}^{\frac{3}{2}}-3d^2{(d^2+\frac{R^2}{4})}^{\frac{1}{2}}=0$
$\Rightarrow 3d^2=\frac{{[d^2+\frac{R^2}{4}]}^{\frac{3}{2}}}{{(d^2+\frac{R^2}{4})}^{\frac{1}{2}}}$
or, $3d^2=d^2+\frac{R^2}{4}$
$\Rightarrow 2d^2=\frac{R^2}{4}$
$\Rightarrow d={\left(\frac{R^2}{8}\right)}^{\frac{1}{2}}$
$\therefore d=\frac{R}{2\sqrt{2}}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Due to symmetry of forces exerted by charges at}A\text{and}B,\\ \text{the horizontal component of forces get cancelled}\\ \text{and net force will be along perpendicular bisector.}\end{array}$