A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{l}\]
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2{{q}^{2}}}{l}\]
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{3{{q}^{2}}}{l}\]
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4{{q}^{2}}}{l}\]
Correct Answer: C
Solution :
\[U=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{Q}_{1}}{{Q}_{2}}}{r};\] net potential energy \[{{U}_{net}}=3\times \frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}^{2}}}{l}\]
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