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  4. If the distance between the two charged particles are doubled, then the force will :

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Q. If the distance between the two charged particles are doubled, then the force will :

Delhi UMET/DPMTDelhi UMET/DPMT 2001

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.