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

Answer 1

Answer: (D)

Zero

Answer 2

:

As shown in figure draw AD $\bot$BC.

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

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 $q_{1} = qatA$

$F_{1} = \frac{1}{4\pi\varepsilon_{0}}\frac{\text{Qq}}{(\text{l / }\sqrt{3})^{2}} = \frac{3Qd}{4\pi\varepsilon_{0}l^{2}},$ along AO

Similarly, force on O due to charge $q_{2} = q$ at B

$F_{2} = \frac{3Qq}{4\pi\varepsilon_{0}l^{2}}$ along BO

And force on Q due to charge $q_{3} = q$at C

$F_{3} = \frac{3Qq}{4\pi\varepsilon_{0}l^{2}}$ along CO

Angle between forces $F_{2}$and $F_{3}$= $120^{o}$

By parallelogram law, resultant of $F_{2}$ an

$F_{3} = \frac{3Qq}{4\pi\varepsilon_{o}l^{2}}$ along OA

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