A) 0.29 J
B) 0.72 J
C) 1.08 J
D) 1.44 J
Correct Answer: A
Solution :
Using the relation for potential energy \[U=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}\] or \[U=9\times {{10}^{9}}\times \frac{4\times {{10}^{-6}}\times 4\times {{10}^{-6}}}{0.5}\] \[U=0.29J\]
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