Force Between Charges in a Dielectric Medium

Consider two small spherical charged conductors placed in an isotropic dielectric fluid of dielectric constant K. r is the distance between the spheres. Let Q₁ and Q₂ be the charges on the spheres. We calculate the field at a distance from Q1, taking into account the polarization charge at the dielectric surface surrounding Q₁.

We choose a spherical Gaussian surface of radius r around Q₁. By Gauss’s law, in terms of electric displacement D, we have
            Φ D. dS = Q₁

Over the spherical boundary of the dielectric, D has the same magnitude and is everywhere normal to the surface. Hence we can take D outside the integral. Thus,

            D (4π r²) = Q₁
                D = Q₁/ 4πr²

But D = εE = Eε0 E, where is the permittivity of the dielectric, and ε0 that of the free space.

            Kε₀E = Q₁/4πr²
            E = 1/4πε₀ Q₁/Kr²                             … (1)

The E (in the absence of the second charge Q₂) at a distance r is calculated.
The second charge Q₂, when placed in the dielectric medium, will polarize the fluid in its vicinity. But this polarizing effect produces a spherically symmetric polarization charge around Q₂ which does not affect the net force on Q₂. Therefore, the total force on Q₂ is

            F = Q₂E = (1/4πε₀ Q₁Q₂/Kr²)                        … (2)

Therefore, the force between point charges in a dielectric fluid is reduced by a factor 1/K, compared with vacuum.