Electrostatics Question 752

Question: 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 foces. 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}} $

Show Answer

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} )}{mrr _{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 decrease in the potential energy

$ W=U _{1}-U _{2}=\frac{3kq^{2}}{r _{0}} $

$ \therefore $ Work done per particle $ =\frac{kq^{2}}{r _{0}} $



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