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
$F_{\text {net }}=2 F \cos \theta$
where $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q q}{\left(x^{2}+\frac{d^{2}}{4}\right)}$ and $\cos \theta=\frac{x}{\sqrt{x^{2}+\frac{d^{2}}{4}}}$
$\therefore F_{\text {net }}=2 \times \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q q}{\left(x^{2}+\frac{d^{2}}{4}\right)} \times \frac{x}{\left(x^{2}+\frac{d^{2}}{4}\right)^{1 / 2}}$
$=\frac{2 Q q x}{4 \pi \varepsilon_{0}\left(x^{2}+\frac{d^{2}}{4}\right)^{3 / 2}}$
For $F_{\text {net }}$ to be maximum, $\frac{d F_{n e t}}{d x}=0$
$\frac{d}{d x}\left[\frac{2 Q q x}{4 \pi \varepsilon_{0}\left(x^{2}+\frac{d^{2}}{4}\right)^{3 / 2}}\right]=0$
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., $x=\pm \frac{d}{2 \sqrt{2}}$