Question

Two particles $A$ and $B$, each having a charge $Q$, are placed a distance $d$ apart. Where should a particle of charge $q$ be placed on the perpendicular bisector of $AB$ so that it experiences maximum force ? What is the magnitude of this maximum force ?

Answer 1

Let a point C at a distance $x$ on the perpendicular bisector.
Then force on C will be
$F=(F_1+F_2)\cos \theta$

Since both forces are equal in magnitude therefore horizontal component will cancel out each other

Therefore vertical component force is
$=\left(\frac{K.Q.q}{{\left(\frac{d}{2}\right)}^2+x^2}+\frac{K.Q.q}{{\left(\frac{d}{2}\right)}^2+x^2}\right)\frac{x}{\frac{d}{2}}$
$F=\frac{2K.Q.q}{{\left(\frac{d}{2}\right)}^2+x^2}.\frac{2x}{d}$
For maximum value of $F$ on differentiating we get
$\frac{dF}{dx}=\frac{d}{dx}[\frac{2K.Q.q.2x}{d({\frac{d}{4}}^2+x^2)}]$

On putting equal to zero for value of x we get
$\frac{4KQq}{d}\frac{d}{dx}\frac{x}{\frac{d^2}{4}+x^2}=0$
$\frac{d}{dx}\frac{x}{\frac{d^2}{4}+x^2}=0$
$\frac{(\frac{d^2}{4}+x^2).1-x.(0+2.x)}{{(\frac{d^2}{4}+x^2)}^2}=0$
$\frac{d^2}{4}+x^2-2x^2=0$
$\frac{d^2}{4}-x^2=0$

On simplifying we get
$\Rightarrow x=\frac{d}{2}$

On putting all the values we get the required force.

Hence the required maximum force is given by

$F=\frac{4KQq}{d^2}$