A) decreases K times
B) increase K times
C) remains unchanged
D) becomes \[\frac{1}{{{K}^{2}}}\] times
Correct Answer: A
Solution :
According to Coulomb's law, force between two charges is directly proportional to product of charges and inversely proportional to square of distance between them. Thus, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}....(i)\] Here, \[\frac{1}{4\pi {{\varepsilon }_{0}}}=\]proportionality constant. If a dielectric medium of constant K is placed between them, then new force between them, \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}K}.\frac{{{q}_{2}}{{q}_{2}}}{{{r}^{2}}}...(ii)\] Dividing Eq. (ii) by Eq. (i), we have \[\frac{F'}{F}=\frac{1}{K}\] \[orF'=\frac{F}{K}\] Thus, new force decreases K times.
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