A) \[\frac{F}{8}\]
B) \[\frac{F}{4}\]
C) \[4F\]
D) \[16\,F\]
Correct Answer: D
Solution :
\[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2{{q}_{1}}\times 2{{q}_{2}}}{{{\left( \frac{r}{2} \right)}^{2}}}=\frac{4}{1/4}F\] \[\Rightarrow \] \[F'=15F\Rightarrow nF=16F\]
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