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Two charge particles A and B kept in air, electric force between them is 100 N. If they are placed in medium having dielectric constant k= 5, at same separation then magnitude of electric force on one of the charge particle due to medium is.

(A) 20 N (B) 100 N (C) 80 N (D) 500 N

Please find below the solution to the asked query:

${\mathrm{F}}_{1}=\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{2}=\frac{1}{4\mathrm{\pi \epsilon k}0}\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{{\mathrm{r}}^{2}}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{F}}_{2}}{{\mathrm{F}}_{1}}=\frac{1}{\mathrm{k}}\phantom{\rule{0ex}{0ex}}{\mathrm{F}}_{2}=\frac{{\mathrm{F}}_{1}}{\mathrm{k}}\phantom{\rule{0ex}{0ex}}=\frac{100}{5}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{20}\mathbf{}\mathbf{N}$

Hope this information will clear your doubts about

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