A) \[43.2\times {{10}^{-6}}m\]
B) \[42.4\times {{10}^{-3}}m\]
C) \[18.1\times {{10}^{-3}}m\]
D) \[19.2\times {{10}^{-6}}m\]
Correct Answer: B
Solution :
(b) \[42.4\times {{10}^{-3}}m\] Here, \[{{q}_{1}}=0.2\,\mu C=0.2\times {{10}^{-6}}C\] \[{{q}_{2}}=-0.4\,\mu C=-0.4\times {{10}^{-6}}C,\,F=-0.4\,N\] As \[F=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[\therefore \,\,\,\,\,\,\,{{r}^{2}}=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}F}\] \[=\frac{0.2\times {{10}^{-6}}\times 0.4\times {{10}^{-6}}\times 9\times {{10}^{9}}}{0.4}\] \[\Rightarrow {{r}^{2}}=1.8\times {{10}^{-3}}\] \[\therefore \,\,\,\,\,\,r={{\left( 1.8\times {{10}^{-3}} \right)}^{{\scriptstyle{}^{1}/{}_{2}}}}\] \[=0.0424\,m=42.4\times {{10}^{-3}}\,m\]
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