A) remain same
B) be one-fourth
C) be halved
D) be doubled
Correct Answer: B
Solution :
From Coulombs law \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] where \[{{q}_{1}},{{q}_{2}}\]are charges separated at a distance r. Given, \[{{F}_{1}}=F,\,\,{{r}_{1}}=r,\,\,{{r}_{2}}=2\,r\] \[\therefore \] \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}=\frac{{{(2\,r)}^{2}}}{{{r}^{2}}}=4\] \[\Rightarrow \] \[{{F}_{2}}=\frac{{{F}_{1}}}{4}\] Hence, force becomes one-fourth.
You need to login to perform this action.
You will be redirected in
3 sec
