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

Answer 1

Answer: (D)

let the triangle be $ABC $ as follow:-

$\frac{A D}{A B}=\cos 30^{\circ} $
$A D=A B \cos 30^{\circ} $
$A D=\frac{\ell \sqrt{3}}{2}$
Distance $A O$ of centroid from $A$ is $\frac{2}{3} A D$
$=\frac{2}{3} \times \frac{\ell \sqrt{3}}{2}=\frac{\ell}{\sqrt{3}}$
Similarly $BO$ and $CO$ are equal to $\frac{\ell}{\sqrt{3}}$
Force on $Q$ at $O$ due to charge $q_{1}$ placed at $A$
$F_{1}=\frac{k Q q}{(\ell / \sqrt{3})^{2}}=\frac{3 K Q q}{\ell^{2}}$ along $A O$
Similarly $F _{2}=\frac{3 K Qq }{\ell^{2}}$ along $OB$
$F _{3}=\frac{3 K Qq }{\ell^{2}}$ along $CO$
Angle between $F_{2}$ and $F _{3}$ is $120^{\circ}$
According to parallelogram law of vector addition:
$F_{ r }=\sqrt{ F _{2}^{2}+ F _{3}^{2}+2 F _{2} F _{3} \cos 120^{\circ}}$
$F _{ 1 }= F _{2}= F _{3}= F$
$F_{r}=\sqrt{F^{2}+F^{2}+2 F^{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.