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 \[Q=\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}-\frac{3Qq}{4\pi {{\varepsilon }_{0}}{{l}^{2}}}=0\]You need to login to perform this action.
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