## Electrostatic Boundary Conditions

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# Electrostatic Boundary Conditions

Let us examine how the electric field changes at the boundary between two different media. Let us fine amount by which E changes at such a boundary.(A) Let us construct a small Gaussian pillbox, extending just barely over the edge in each direction. A is the area of the pillbox lid, h is the thickness of the pillbox. Applying Gauss’s law,

Φ E. da = 1/ε

_{0}Q

_{enc}= 1/ε

_{0}σA,

Now, the sides of the pillbox contribute nothing to the flux as h O. So

E

__|__

_{above}– E

__|__

_{below}= 1/ε

_{0}σ … (1)

Here, E

__|__

_{above}denotes the component of E that is perpendicular to the surface immediately above, and E

__|__

_{above}denotes the component of

**E**that is perpendicular to the surface immediately above, and E

__|__

_{below}is the same, only just below the surface. For consistency we let “upward” be the positive direction for both.

According to Eq. (1), the normal component of E is discontinuous by an amount σ/ε

_{0}at any boundary.

If the boundary is free from charge, σ = 0, then E

__|__is continuous, i.e., the normal component of E is continuous across a charge free boundary.

(B) Since the electrostatic field is conservative,

Φ E. d

__l__= 0

around the rectangular path EFGH. The lengths EF and GH will be taken equal to 1 and the segments FG and HE will be assumed to be negligibly small (h 0). Therefore

or E

_{||above}l – E

_{||below}l) = 0

E|

_{|above}= E

_{||below}… (2)

Here, E

_{||}represents the components of E which is parallel to the surface. Thus the tangential electric field is continuous across such a boundary. The boundary conditions on E [Eqs. (1) and (2)] can be combined into a single formula:

E

_{above}– E

_{below}= σ/ε

_{0}n, … (3)

Here, n is a unit vector perpendicular to the surface.

(C) The electrostatic potential must be continuous across any boundary, since

V

_{above}- V

_{below}= - ∫

^{b}

_{a}E.dl

As the path length shrinks to zero, so too does the integral.

. : V

_{above}= V

_{below}… (4)

From Eq. (3),

∇V

_{above}- ∇V

_{below}= - 1/ε

_{0}σn (. : E = - ∇V)

Or, ∂ V

_{above}/∂n - ∂ V

_{below}/∂n = - 1/ε

_{0}σ … (5)

Here, ∂V /∂n = ∇V. n denotes the normal derivative of V.

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