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In vaccume two forces placed at distance r have force F when theae charges are placed in a uniform dielectric medium (Dielectric constant K) the distance between them for the force will be
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Asked by Raghavendrakumarsir | 23 May, 2021, 11:43: AM
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Let two charges q1 and q2 are at a distance r in vacuum
 
Then electrostatic force F between two charges is given as
 
begin mathsize 14px style F space equals space fraction numerator q subscript 1 space q subscript 2 over denominator 4 pi epsilon subscript o r squared end fraction end style   ......................(1)
where εo is the permittivity of free space

If the same charges q1 and q2 are placed in medium of dielectric constant K , to get force of same magnitude,
required distance r1 is determined as
 
begin mathsize 14px style F space equals space fraction numerator q subscript 1 space q subscript 2 over denominator 4 pi epsilon space r subscript 1 squared end fraction end style .............................(2)
 
where ε is the permittivity of the medium with dielectric constant K so that ε = K εo
 
If we equate , equations (1) and (2) , we get
 
εo r2 = ε r12
 
begin mathsize 14px style r subscript 1 space equals space square root of epsilon subscript o over epsilon r squared end root space equals space fraction numerator r over denominator square root of K end fraction end style

 
Answered by Thiyagarajan K | 23 May, 2021, 01:15: PM
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