Question

Two equal point charges $A$ and $B$ are $R$ distance apart. A third point charge 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

Answer: (A)

$F_1=F_2=\frac{K{Q}^2}{\left[d^2+\frac{R^2}{4}\right]}$
$F_N=F_1\cos \theta +F_2\cos \theta$
$=2F_1\cos \theta =$
$F_N=2\cdot \frac{K{Q}^2}{\left(d^2+\frac{R^2}{4}\right)}\cdot \frac{d}{\left[d^2+\frac{R^2}{4}\right]}\frac{1}{2}$
If $F=$ Maximum than $\frac{dF}{d^{\prime \prime }{d}^{\prime }}=0$
So we get $\alpha =\frac{R}{2\sqrt{2}}$