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Two equal charges are separated by a distance $d$. A third charge placed on a perpendicular bisector at $x$ distance will experience maximum coulomb force when
$x = \frac{d}{{\sqrt 2 }}$
$x = \frac{d}{2}$
$x = \frac{d}{{2\sqrt 2 }}$
$x = \frac{d}{{2\sqrt 3 }}$
Solution
(c) Suppose third charge is similar to $Q$ and it is $q$
So net force on it
$F_{net} = 2F cos \theta$
Where $F = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Qq}}{{\left( {{x^2} + \frac{{{d^2}}}{4}} \right)}}$and $\cos \theta = \frac{x}{{\sqrt {{x^2} + \frac{{{d^2}}}{4}} }}$
${F_{net}} = 2 \times \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Qq}}{{\left( {{x^2} + \frac{{{d^2}}}{4}} \right)}} \times \frac{x}{{{{\left( {{x^2} + \frac{{{d^2}}}{4}} \right)}^{1/2}}}}$
$ = \frac{{2Qqx}}{{4\pi {\varepsilon _0}{{\left( {{x^2} + \frac{{{d^2}}}{4}} \right)}^{3/2}}}}$
For Fnet to be maximum $\frac{{d{F_{net}}}}{{dx}} = 0$
i.e. $\frac{d}{{dx}}\left[ {\frac{{2Qqx}}{{4\pi {\varepsilon _0}{{\left( {{x^2} + \frac{{{d^2}}}{4}} \right)}^{3/2}}}}} \right] = 0$
or $\left[ {{{\left( {{x^2} + \frac{{{d^2}}}{4}} \right)}^{ – 3/2}} – 3{x^2}{{\left( {{x^2} + \frac{{{d^2}}}{4}} \right)}^{ – 5/2}}} \right] = 0$
i.e. $x = \pm \frac{d}{{2\sqrt 2 }}$
