Electrostatics Question 693
Question: 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 any sphere due to the other two is
Options:
A) $ \frac{1}{4\pi {\varepsilon _{0}}}{{( \frac{q}{R} )}^{2}} $
B) $ \frac{\sqrt{3}}{4\pi {\varepsilon _{0}}}{{( \frac{q}{R} )}^{2}} $
C) $ \frac{\sqrt{3}}{16\pi {\varepsilon _{0}}}{{( \frac{q}{R} )}^{2}} $
D) $ \frac{\sqrt{5}}{16\pi {\varepsilon _{0}}}{{( \frac{q}{R} )}^{2}} $
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Answer:
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}}{{{( 2R )}^{2}}} $
$ =\frac{1}{4\pi {\varepsilon _{0}}}\frac{q^{2}}{4R^{2}}\text{along }\overrightarrow{BA} $
And force on A due to C,
$ F _{AB}=\frac{1}{4\pi {\varepsilon _{0}}}\frac{q^{2}}{{{( 2R )}^{2}}}==\frac{1}{4\pi {\varepsilon _{0}}}\frac{q^{2}}{4R^{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}{{( \frac{q}{R} )}^{2}} $
