Question

Three charges $q_1,q_2,q_3$ each equal to $q$ placed at the vertices's of an equilateral triangle of side $\ell .$ What will be the force on a charge $Q$ placed at the centroid of triangle?

• A. $4N$
• B. $5N$
• C. $10N$
• D. Zero

Answer 1

Answer: (D)

let the triangle be $ABC$ as follow:-

$\frac{AD}{AB}=\cos 30^{\circ }$
$AD=AB\cos 30^{\circ }$
$AD=\frac{\ell \sqrt{3}}{2}$
Distance $AO$ of centroid from $A$ is $\frac{2}{3}AD$
$=\frac{2}{3}\times \frac{\ell \sqrt{3}}{2}=\frac{\ell }{\sqrt{3}}$
Similarly $BO$ and $CO$ are equal to $\frac{\ell }{\sqrt{3}}$
Force on $Q$ at $O$ due to charge $q_1$ placed at $A$
$F_1=\frac{kQq}{(\ell /\sqrt{3}{)}^2}=\frac{3KQq}{{\ell }^2}$ along $AO$
Similarly $F_2=\frac{3KQq}{{\ell }^2}$ along $OB$
$F_3=\frac{3KQq}{{\ell }^2}$ along $CO$
Angle between $F_2$ and $F_3$ is $120^{\circ }$
According to parallelogram law of vector addition:
$F_r=\sqrt{F_2^2+F_3^2+2F_2{F}_3\cos 120^{\circ }}$
$F_1=F_2=F_3=F$
$F_r=\sqrt{F^2+F^2+2F^2\left(\frac{-1}{2}\right)}$
$F_r=\sqrt{F^2+F^2-F^2}$
$=\sqrt{F^2}=F$
Force due to charge $q$ at $A$ is equal and opposite to the resultant force $F_1$. So the force experienced is Zero.
Trick:- The charge is at the centre of the symmetrical system so the net force on it will be zero.