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

Dear Student, 
Please find below the solution to the asked query: 
F1=14πε0q1q2r2F2=14πεk0q1q2r2F2F1=1kF2=F1k=1005=20 N
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  • -15
let initial  electrostatic force of  attraction be F =KQ2/R2  where  K is 1/4piE0. 
final force F' =K/5. Q2/R
 taking ratio we get  F/F' =5  or F'=100/5 =20N
  • -8
20N
  • -3
Whenever a system of two charges is kept in a medium of dielectric constant k , the force of interaction ( note , here  this force is calculated when , they were kept in vacuum )  between them gets reduced by k ( now , they are kept in a medium of dielectric const. k ) .Thus , new force in this medium is ( 100 / 5 ) N or 20 N       
  • -9
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