the force between two charges at distance r is f what will be the force between these charges if the distance between them is doubled and charges are placed in a medium of dielectric constant 10 ?

Asked by aditya.loya9821 | 23rd Jul, 2021, 06:44: PM

Expert Answer:

Force between two charges q1 and q2 that are separated by a distance r is given as
 
begin mathsize 14px style f space equals space K space fraction numerator q subscript 1 space q subscript 2 over denominator r squared end fraction end style
where K = 1/(4begin mathsize 14px style pi epsilon subscript o end style) is Coulomb's constant and begin mathsize 14px style epsilon subscript o end style is permittivity of free space 
 
If distance is doubled and charges are placed in a medium of dielectric constant εr = 10 , then force F is given as
 
begin mathsize 14px style F thin space equals space K space fraction numerator q subscript 1 space q subscript 2 over denominator 10 space left parenthesis 2 r right parenthesis squared end fraction space equals space 1 over 40 K space fraction numerator q subscript 1 space q subscript 2 over denominator r squared end fraction space equals space 1 over 40 f end style
Hence Force F between same charges after doubling distance and placed in a medium of
dielectric constant 10 decreases by a factor (1/40) in comparison with initial force f
 

Answered by Thiyagarajan K | 23rd Jul, 2021, 07:13: PM