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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.

Force Between Charges
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