A) 2
B) 4
C) 5
D) 10
E) 7
Correct Answer: A
Solution :
From Coulambs law \[F=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}.\frac{{{q}_{1}}\,{{q}_{2}}}{{{r}^{2}}}\] \[3.7\times {{10}^{-9}}=9\times {{10}^{9}}\times \frac{{{q}^{2}}}{{{(5\times {{10}^{-10}})}^{2}}}\] \[q=3.2\times {{10}^{-19}}c\] \[q=nc\] \[n=\frac{q}{c}=\frac{3.2\times {{10}^{-19}}}{1.6\times {{10}^{-10}}}=2\]You need to login to perform this action.
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