Question 27 of 311

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$d$B$\dfrac{d}{2}$C$\dfrac{d}{\sqrt{2}}$D$\dfrac{d}{2\sqrt{2}}$

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Answer: (D) $\dfrac{d}{2\sqrt{2}}$

The distance between $q$ and $Q$ is $r=\sqrt{{x}^{2}+{\left(\dfrac{d}{2}\right)}^{2}}=\sqrt{{x}^{2}+\dfrac{{d}^{2}}{4}}$.
The force between $q$ and $Q$ will be $F=\dfrac{kqQ}{{r}^{2}}$.
From the given figure we can see that only vertical components of the forces will add, and the horizontal components will get cancelled out. Therefore,
${F}_{net}=\dfrac{2kqQ}{{r}^{2}}\cos \theta =\dfrac{2kqQ}{{r}^{2}}\times \dfrac{x}{r}=\dfrac{kqQx}{{\left({x}^{2}+\dfrac{{d}^{2}}{4}\right)}^{\dfrac{3}{2}}}$
For force to be maximum, $\dfrac{d{F}_{net}}{dx}=0.$
$\Rightarrow kqQ\left[\dfrac{{\left({x}^{2}+\dfrac{{d}^{2}}{4}\right)}^{\dfrac{3}{2}}-x\times \dfrac{3}{2}{\left({x}^{2}+\dfrac{{d}^{2}}{4}\right)}^{\dfrac{1}{2}}\times 2x}{{\left({\left({x}^{2}+\dfrac{{d}^{2}}{4}\right)}^{\dfrac{3}{2}}\right)}^{2}}\right]=0$
$\Rightarrow \left({x}^{2}+\dfrac{{d}^{2}}{4}\right)=3{x}^{2}\Rightarrow 2{x}^{2}=\dfrac{{d}^{2}}{4}\Rightarrow x=\dfrac{d}{2\sqrt{2}}$