Question

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

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

Answer 1

Answer: (D)

$F=\frac{KQq}{\left(x^2+\frac{d^2}{4}\right)}$
Net force on $g=2F\cos \theta$
$F_{\text{net }}=\frac{2KQqx}{{\left(x^2+\frac{d^2}{4}\right)}^{3/2}}$
For maximum $F_{net}$
$\frac{d{F}_{\text{net }}}{dx}=0$
we get $x=\frac{d}{2\sqrt{2}}$