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

As $\cos$ components cancel each other $\sin$ components of force add up. We can find the maxima of the equation to find the place of maximum value of force.

$2F\sin \theta =2\frac{k{q}^2}{r^2+\frac{d^2}{4}}\sin \theta$

Put $\frac{d\left(2F\sin \theta \right)}{dx}=0$

We get,

$\frac{d}{dx}\left(\frac{2k{q}^2}{x^2+\frac{d^2}{4}}\times \frac{x}{\sqrt{x^2+\frac{d^2}{4}}}\right)=0$

$=\frac{2k{q}^2}{{\left(x^2+\frac{d^2}{4}\right)}^{\frac{3}{2}}}+\frac{2k{q}^2}{{\left(x^2+\frac{d^2}{4}\right)}^{\frac{3}{2}}}\times \left(\frac{-3}{2}\right)\times x$

On solving we get,

$x^2+\frac{d^2}{4}=3x^2$

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