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)

zero

As shown in figure draw $AD\perp BC$

$\therefore AD=AB\cos {30}^o=\frac{l\sqrt{3}}{2}$

Distance $AO$ of the centroid $O$ from $A$

$=\frac{2}{3}AD=\frac{2l}{3}\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{3}}$

$\therefore$ Force on $Q$ at $O$ due to charge $q_1=q$ at $A$

$\overrightarrow{F_1}=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{{\left(l\sqrt{3}\right)}^2}=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$,along $AO$

Similarly, force on $O$ due to charge $q_2=q$ at $B$

$\overrightarrow{F_2}=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$, along $BO$

and force on $Q$ due to charge $q_3=q$ at $C$

$\overrightarrow{F_3}=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$ along $CO$

Angle between forces $F_2$ and $F_3={120}^o$

By parallelogram law, resultant of $\overrightarrow{F_2}$ and $\overrightarrow{F_3}$ $\frac{3Qq}{4\pi {\epsilon }_0{l}^2}$ along $OA$

$\therefore$ force on $Q=\frac{3Qq}{4\pi {\epsilon }_0{l}^2}-\frac{3Qq}{4\pi {\epsilon }_0{l}^2}=0$