Question

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

• A. $x=\frac{d}{\sqrt{2}}$
• B. $x=\frac{d}{2}$
• C. $x=\frac{d}{2\sqrt{2}}$
• D. $x=\frac{d}{2\sqrt{3}}$

Answer 1

The correct option is C $x=\frac{d}{2\sqrt{2}}$

Suppose third charge is similar to Q and it is q

So net force on it
$F_{net}=2Fcos\theta$

Where $F=\frac{1}{4\pi {ϵ}_0}.\frac{Qq}{(x^2+\frac{d^2}{4})}$ and $cos\theta =\frac{x}{\sqrt{x^2+\frac{d^2}{4}}}$
$\therefore F_{net}=2\times \frac{1}{4\pi {ϵ}_0}.\frac{Qq}{(x^2+\frac{d^2}{4})}\times \frac{x}{{(x^2+\frac{d^2}{4})}^{1/2}}$
$=\frac{2Qqx}{4\pi {ϵ}_0{(x^2+\frac{d^2}{4})}^{3/2}}$
For $F_{net}$ to be maximum $\frac{d{F}_{net}}{dx}=0$
i.e. $\frac{d}{dx}\left[\frac{2Qqx}{4\pi {ϵ}_0{(x^2+\frac{d^2}{4})}^{3/2}}\right]=0$
or $[{(x^2+\frac{d^2}{4})}^{-3/2}-3x^2{(x^2+\frac{d^2}{4})}^{-5/2}]=0$
i.e. $x=\pm \frac{d}{2\sqrt{2}}$