Electrostatics Question 752
Three identical particles, each possessing the mass m and charge +q, are placed at the corners of an equilateral triangle with side $ r _{0}. $ The particles are simultaneously set free and start flying apart symmetrically due to Coulomb’s repulsion forces. The work performed by Coulomb’s forces acting on to a very large distance is $ (\text{where }k=1/4\pi {\varepsilon _{0}}.) $
Options:
A) $ \frac{3kq^{2}}{r _{0}} $
B) $ \frac{kq^{2}}{r _{0}} $
C) $ \frac{3kq^{2}}{2r _{0}} $
D) $ \frac{kq^{2}}{2r _{0}} $
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Answer:
Correct Answer: B
Solution:
[b] Since the given system is closed, the increase in KE is equal to decrease in P.E.
$ \Rightarrow \frac{3}{2}mv^{2}=\frac{2kq^{2}}{r _{0}}-\frac{3kq^{2}}{r} $
$ \Rightarrow v=\sqrt{\frac{2kq^{2}( r-r _{0} )}{m r r _{0}}}$
$ \Rightarrow {v _{\max }}=\sqrt{\frac{2kq^{2}}{mr _{0}}} $
The work performed by the interaction force during the variation of the system’s configuration is equal to the negative of the decrease in the potential energy
$ W=U _{1}-U _{2}=\frac{kq^{2}}{r _{0}} $
$ \therefore $ Work done per particle $ =\frac{kq^{2}}{r _{0}} $
BETA
