A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}{{\left( \frac{q}{R} \right)}^{2}}\]
B) \[\frac{\sqrt{3}}{4\pi {{\varepsilon }_{0}}}{{\left( \frac{q}{R} \right)}^{2}}\]
C) \[\frac{\sqrt{3}}{16\pi {{\varepsilon }_{0}}}{{\left( \frac{q}{R} \right)}^{2}}\]
D) \[\frac{\sqrt{5}}{16\pi {{\varepsilon }_{0}}}{{\left( \frac{q}{R} \right)}^{2}}\]
Correct Answer: C
Solution :
For external points, a charged sphere behaves as if the whole of its charge is concentrated at its center. \[{{F}_{AB}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{(2R)}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{4{{R}^{2}}}\,\,along\,\,\overline{BA}\] And force on A due to C, \[{{F}_{AC}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{{{q}^{2}}}{{{(2R)}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{q}^{2}}}{4{{R}^{2}}}\,\,along\,\overline{CA}\] Now as angle between BA and CA is\[~60{}^\circ \]and \[|{{F}_{AB}}|=|{{F}_{AC}}|=F\] \[\therefore \,{{F}_{A}}=\sqrt{{{F}^{2}}+{{F}^{2}}+2FF\,\cos \,60}=\sqrt{3}\,F\] \[-\frac{1}{4\pi {{\varepsilon }_{0}}\,}\,\frac{\sqrt{3}}{4}\,{{\left( \frac{q}{R} \right)}^{2}}\]You need to login to perform this action.
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