A) \[9\times {{10}^{-3}}J\]
B) \[9\times {{10}^{-3}}eV\]
C) \[1J\]
D) zero
Correct Answer: A
Solution :
: Potential energy \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}\] \[=(9\times {{10}^{9}})\times \frac{{{10}^{-6}}\times {{10}^{-6}}}{1}J\] \[=9\times {{10}^{-3}}J\]You need to login to perform this action.
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