Electrostatics> Force between two identical charges place...

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

shivam agarwal , 9 Years ago

Grade 12

3 Answers

Sumit Majumdar

Last Activity: 9 Years ago

As the dielectric constant increases the force between the two charges would decrease. Try to now solve the question.

Gauri

Last Activity: 8 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

Last Activity: 4 Years 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 * k^1/2

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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Electrostatics> Force between two identical charges place...

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

shivam agarwal , 9 Years ago

Sumit Majumdar

Gauri

Kushagra Madhukar

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