A) zero
B) 9 N
C) 18 N
D) 2 N
Correct Answer: A
Solution :
Using the formula, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] Here, \[F=+27\text{ }N\] \[{{q}_{1}}=+3C\] and \[{{q}_{2}}=+9C\] \[\therefore \] \[27=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{3\times 9}{{{r}^{2}}}\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}=1\] ?(i) When\[-3C\]charge is given to each charge, then \[{{q}_{1}}'=(+3-3)=0\,C\] and \[{{q}_{2}}'=(+9-3)=6\,C\] \[\therefore \] New force, \[F'=\left( \frac{1}{4\pi {{\varepsilon }_{0}}{{r}^{2}}} \right)({{q}_{1}}'\times {{q}_{2}}')\] \[=1(0\times 6)\] \[=0\] \[\therefore \]Force of interaction is zero.
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