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