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Boundary-Value Problems Involving Dielectrics

The differential from of Gauss’s law is
            ∇.D = ρ,                                … (1)
Here, ρ is the external charge density. ρ
If the dielectrics are linear, isotropic, and homogenous, then D = εE.
. :            ∇.E = 1/ερ.                                … (2)]

But the electrostatic field E is derivable from a scalar potential V, i.e.,
E = - ∇V

. :        ∇2V = 1/ερ.                                    … (3)

Thus the potential in the dielectric satisfies Poisson’s equation.
In most cases of interest the dielectric contains no charge distributed throughout its volume, that is r = 0 inside the dielectric material. The charge exists on the surfaces of conductors or is concentrated in the form of point charges, which may be embedded in the dielectric. In these circumstances, the potential satisfies Laplace’s equation throughout the body of the dielectric:
        ∇2V=0                                        … (4)

An electrostatic problem involving linear, isotropic and homogeneous dielectrics reduces, therefore, to finding solutions of Laplace’s equation in medium and joining the solutions in the various media by means of the boundary conditions of the preceding section.

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