Question

Two point charges placed at a certain distance r in air exert a force F on each other. Then the distance R at which these charges will exert the same force in a medium of dielectric constant K is given

Answer 1

F (air) = (1/4 pi eo) [qQ/r^2] 

F(medium) = (1/4 pi eo*k) [qQ/R^2] 
==================== 
given> F(medium) = F(air) 
(1/4 pi eo*k) [qQ/R^2] = (1/4 pi eo) [qQ/r^2] 
(1/k) [1/R^2] = [1/r^2] 
kR^2 = r^2 
R^2 = r^2/k 
R = r/ [rootk] 

Answer 2

Given: the two point charges = q₁ and q₂

           distance between them = r

           Force exerted on each other = F

           the dielectric constant of the medium = K

To Find: the distance at which both charges will exert the same force, R.

Solution:

To calculate D, the formula used:

• F = k x ( q₁ .q₂ / r²)

        here k = 1 / 4π∈°

         ∈° is the vaccum permittivity

• F = q₁.q₂ / 4π∈° x r²

Applying the above formula, when the distance between the charges is r:

F₁ = q₁.q₂ / 4π∈° x r²                  ⇒ 1

When the dielectric constant of the medium is K and the distance between the two charges is R, then the force exerted is:

F₂ =  (q₁.q₂ / 4π∈° x R²) / K         ⇒2

On dividing equations 1 and 2:

F₁ / F₂ = [q₁.q₂ / 4π∈° x r² ] / [(q₁.q₂ / 4π∈° x R²) / K]

As F₁ = F₂ = F

∴ F / F = [q₁.q₂ / 4π∈° x r² ] / [(q₁.q₂ / 4π∈° x R²) / K]

        1 = ( 1/ 4π∈° x r² ) / [ (1 / 4π∈° x R²)/K]

   1 /r² = 1 / K x R²

Taking sqaure-root on both the side:

√ 1/ r² = √ 1 / K x R²

  1/ r   = 1 / R √K

Taking recirpocal on both the side:

     r  = R √K

     r / √K = R

    R = r / √K

Hence the the distance at which both charges will exert the same force is R = r / √K.