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)

$x = \frac{d}{2\sqrt{2}}$

Answer 2

Suppose third charge is similar to Q and it is q

So net force on it F_net = 2F cosθ

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 F_net to be maximum $\frac{dF_{net}}{dx} = 0$ i.e.

$$\frac{d}{dx}\left\lbrack \frac{2Qqx}{4\pi\varepsilon_{0}\left( x^{2} + \frac{d^{2}}{4} \right)^{3/2}} \right\rbrack = 0$$

or $\left\lbrack \left( x^{2} + \frac{d^{2}}{4} \right)^{- 3/2} - 3x^{2}\left( x^{2} + \frac{d^{2}}{4} \right)^{- 5/2} \right\rbrack = 0$ i.e. $x = \pm \frac{d}{2\sqrt{2}}$