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)

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