A) \[1.08\times {{10}^{5}}N\]
B) \[1.08\times {{10}^{-6}}N\]
C) \[1.08\times {{10}^{-5}}N\]
D) None of these
Correct Answer: C
Solution :
\[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}_{2}}}\] \[=\frac{9\times {{10}^{9}}\times 3\times {{10}^{9}}\times {{10}^{-9}}}{{{(5\times {{10}^{-2}})}^{2}}}\] \[=1.08\times {{10}^{-5}}N\]
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