A) \[2.4\times {{10}^{-3}}N\](repulsive)
B) \[2.4\times {{10}^{-3}}N\](attractive)
C) \[1.5\times {{10}^{-3}}N\] (repulsive)
D) \[1.5\times {{10}^{-3}}N\] (attractive)
Correct Answer: D
Solution :
Key Idea: Like charges repel each other while unlike charges attract each other. From Coulombs law, the force of attraction/repulsion between two point charges \[{{q}_{1}}\] and \[{{q}_{2}}\] placed a distance r apart is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}N\] When similar charges are taken \[{{q}_{1}}=3\times {{10}^{-6}}C,\,{{q}_{2}}=8\times {{10}^{-6}}C\] \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(3\times {{10}^{-6}})\times (8\times {{10}^{-6}})}{{{r}^{2}}}\] ?. (i) (repulsive) When additional charge \[-6\times {{10}^{-6}}C\] is given to each charge, then \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(3-6)\times {{10}^{-6}}\times (8-6)\times {{10}^{-6}}}{{{r}^{2}}}\] (attractive) \[\therefore \] \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-3)\times {{10}^{-6}}\times 2\times {{10}^{-6}}}{{{r}^{2}}}N\]? (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{F}{F}=-\frac{6}{24}\] \[\Rightarrow \]\[F=-\frac{F}{4}\] \[=-\frac{6\times {{10}^{-3}}}{4}=-1.5\times {{10}^{-3}}N\] Negative sign indicates, force is attractive.
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