The electric foree between two electric charges in a medium is F … - Vidyadip

Question

The electric foree between two electric charges in a medium is F . The distance between them is $d$. At what distance were they placed from each other
so that the electric foree becomes (i) 3 F (ii) $\frac{ F }{4}$ ?

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Answer

From Coulomb's law$F \propto \frac{1}{r^2}$
$\Rightarrow$ $\frac{F_1}{F_2}=\left(\frac{r_2}{r_1}\right)^2 \Rightarrow r_2=r_1\left(\frac{F_1}{F_2}\right)^{\frac{1}{2}}$ $\therefore$
(i) $r_2=d\left(\frac{F}{3 F}\right)^{\frac{1}{2}}=\frac{d}{\sqrt{3}}$
(ii) $r_2=d\left(\frac{F}{ F / 4}\right)^{\frac{1}{2}}=d \sqrt{4}=2 d$

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