question_answer

Three identical spheres, each having a charge q and radius R, are kept in such a way that each touches the other two. The magnitude of the electric force on any sphere due to the other two

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 \[|{{F}_{AB}}|=|{{F}_{AC}}|=F\] \[\therefore \,{{F}_{A}}=\sqrt{{{F}^{2}}+{{F}^{2}}+2FF\,\cos \,60}=\sqrt{3}\,F\] \[-\frac{1}{4\pi {{\varepsilon }_{0}}\,}\,\frac{\sqrt{3}}{4}\,{{\left( \frac{q}{R} \right)}^{2}}\]