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1. Electric Charges and Fields
hard
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
$x=\frac{d}{2 \sqrt{2}}$
(JEE MAIN-2022)
Solution
$F =\frac{ KQq }{\left( x ^{2}+\frac{ d ^{2}}{4}\right)}$
Net force on $g =2 F \cos \theta$
$F _{\text {act }}=\frac{2 KQqx }{\left( x ^{2}+\frac{ d ^{2}}{4}\right)^{3 / 2}}$
For maximum $F _{\text {act }}$
$\frac{ d F _{\text {net }}}{ dx }=0$
we get $x=\frac{d}{2 \sqrt{2}}$
Standard 12
Physics
