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Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric constant 4 is inserted between these two charges. If the thickness of slab is r/2 then the net force on a charge will become. Options a= 3F/5, b=4F/9 c=F/4 d=none of the above

Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric constant 4 is inserted between these two charges. If the thickness of slab is r/2 then the net force on a charge will become. Options a= 3F/5, b=4F/9 c=F/4 d=none of the above

Grade:12

3 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
6 years ago
As the dielectric constant increases the force between the two charges would decrease. Try to now solve the question.
Gauri
19 Points
4 years ago
The force between two point charges separated by distance ‘r’ when a dielectric of thickness ‘t’ is inserted between them is kq1q2/(r-t+tsqrt(K))^2. So the answer is 4F/9.
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the attached solution to your problem.
 
The electrostatic forces between two charges in a medium of di-electric constant k is
Here, kr2 can be taken as the effective depth square r’2
Hence, r’2 = kr2
or, r’ = r(k)1/2
Hence the effective thickness of slab is r/2 * k1/2
Hence the force between the charges is
Hence, option b) 4F/9 is correct.
 
Hope this helps.
Thanks and regards,
Kushagra

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