question_answer

Three charges of equal magnitude q are 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) \[\frac{Qq}{2\pi {{\varepsilon }_{0}}{{l}^{2}}}\]                     

D) zero

Correct Answer: D

Solution :

(d) zero As shown in figure, draw \[AD\bot BC\]. \[\therefore \,\,\,\,AD=AB\,\cos \,30{}^\circ =\frac{l\sqrt{3}}{2}\] 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 \[\overrightarrow{F}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{\left( l/\sqrt{3} \right)}^{2}}}=\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\], along AO Similarly, force on O due to charge, \[{{q}_{2}}=q\]at B, \[{{\overrightarrow{F}}_{2}}=\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]along BO and force on Q due to charge, \[{{q}_{3}}=q\]at C \[{{\overrightarrow{F}}_{3}}=\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\], along CO Angle between forces \[{{F}_{2}}\]and \[{{F}_{3}}=120{}^\circ \] By parallelogram law, resultant of forces \[{{\overrightarrow{F}}_{2}}\]and \[{{\overrightarrow{F}}_{3}}\] \[=\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}\]along OA \[\therefore \] Total force on \[Q=\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}-\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}=0\]