Question

Three charges of equal magnitude $q$ is placed at the vertices of an equilateral triangle of side $l$. The force on a charge $Q$ placed at the centroid of the triangle is

• A. $\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$
• B. $\frac{2Qq}{4\pi {\epsilon }_0{l}^2}$
• C. $\frac{Qq}{2\pi {\epsilon }_0{l}^2}$
• D. zero

Answer 1

Answer: (D)

As shown in figure draw $AD\perp BC$.
$\therefore AD=ABcos30^{\circ }=\frac{l\sqrt{3}}{2}$
Distance $AO$ of the centroid $O$ from A
$=\frac{2}{3}AD=\frac{2l}{3}\frac{\sqrt{3}}{2}=\frac{l}{\sqrt{3}}$

$\therefore$ Force on $Q$ at $O$ due to charge $q_1=q$ at $A$,
${\vec{F}}_1=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{{\left(l/\sqrt{3}\right)}^2}=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$, along $AO$
Similarly, force on $O$ due to charge $q_2=q$ at $B$
${\vec{F}}_2=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$, along $BO$
and force on $Q$ due to charge $q_3=q$ at $C$
${\vec{F}}_3=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$, along $CO$
Angle between forces $F_2$ and $F_3=120^{\circ }$
By parallelogram law, resultant of ${\vec{F}}_2$ and ${\vec{F}}_2=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$,
along $OA$
$\therefore$ Total force on $Q=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}-\frac{3Qq}{4\pi {\epsilon }_0{l}^2}=0$