A) 0.2
B) 2
C) 20
D) 200
Correct Answer: B
Solution :
Force of attraction between two point charges \[{{q}_{1}}\] and \[{{q}_{2}}\] at separation r is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] (Here: \[r=3cm=3\times {{10}^{-2}}m,\] \[F=40N,\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\] ) \[40=\frac{9\times {{10}^{9}}\times q\times q}{9\times {{10}^{-4}}}\] \[{{q}^{2}}=\frac{9\times 40\times {{10}^{-4}}}{9\times {{10}^{9}}}\] \[{{q}^{2}}=40\times {{40}^{-13}}=4\times {{10}^{-12}}\] =\[q=2\times {{10}^{-6}}=2\mu C\]You need to login to perform this action.
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