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 :
[c] For external points, a charged sphere behaves as if he whole of its charge is concentrated at its center. Force on A due to B. \[{{F}_{AB}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{\left( 2R \right)}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{4{{R}^{2}}}\text{along }\overrightarrow{BA}\] And force on A due to C, \[{{F}_{AB}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{\left( 2R \right)}^{2}}}==\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{4{{R}^{2}}}\text{along }\overrightarrow{\text{CA}}\] Now as angle between BA and CA is \[60{}^\circ \]and \[|{{F}_{AB}}|=|{{F}_{AC}}|=F\] \[\therefore {{F}_{A}}=\sqrt{{{F}^{2}}+{{F}^{2}}+2FF\cos 60}=\sqrt{3}F\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sqrt{3}}{4}{{\left( \frac{q}{R} \right)}^{2}}\]
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