A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2{{q}^{2}}}{a}\]
B) \[\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}a}\]
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{3{{q}^{2}}}{3R}\]
D) zero
Correct Answer: D
Solution :
As potential energy of the system initially and finally are given by \[{{U}_{i}}=\frac{(q)(-2q)+(q)(-2q)+(-2q)(-2q)}{4\pi {{\varepsilon }_{0}}a}=0\] and \[{{U}_{f}}=\frac{(q)(-2q)+(q)(-2q)+(-2q)(-2q)}{4\pi {{\varepsilon }_{0}}(2q)}=0\]As, \[\Delta U=0\] \[{{w}_{ext}}=0\]
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