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. $\frac{3\,Qq}{4\pi\varepsilon_{0}l^{2}}$
• B. $\frac{2\,Qq}{4\pi\varepsilon_{0}l^{2}}$
• C. $\frac{Qq}{2\pi\varepsilon_{0}l^{2}}$
• D. zero

Answer 1

Answer: (D)

zero

Answer 2

As shown in figure draw $AD \bot BC$. $\therefore AD=AB\,cos30^{\circ}=\frac{l\sqrt{3}}{2}$ 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 $A$, $\vec{F}_{1}=\frac{1}{4\pi\varepsilon_{0}} \frac{Qq}{\left(l / \sqrt{3}\right)^{2}}=\frac{3Qq}{4\pi\varepsilon_{0}l^{2}}$, along $AO$ Similarly, force on $O$ due to charge $q_{2} = q$ at $B$ $\vec{F}_{2}=\frac{3Qq}{4\pi\varepsilon_{0}l^{2}}$, along $BO$ and force on $Q$ due to charge $q_{3}= q$ at $C$ $\vec{F}_{3}=\frac{3Qq}{4\pi\varepsilon_{0}l^{2}}$, along $CO$ Angle between forces $F_{2}$ and $F_{3}=120^{\circ}$ By parallelogram law, resultant of $\vec{F}_{2}$ and $\vec{F}_{2}=\frac{3Qq}{4\pi\varepsilon_{0}l^{2}}$, along $OA$ $\therefore$ Total force on $Q=\frac{3Qq}{4\pi\varepsilon_{0}l^{2}}-\frac{3Qq}{4\pi\varepsilon_{0}l^{2}}=0$