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

Answer: (C)

Suppose third charge is of same nature as that of $Q$ and let it magnitude be $q$.

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