Boundary conditions static electric fields

Boundary conditions static Electric fields:-

So far, we have considered the existence of the electric field in a Homogeneous medium. If the field exists in a region consisting of two different media, the conditions that the field must satisfy at the interface separating the media are called “Boundary conditions”.

These conditions are helpful in determining the field on one side of the boundary if the field on the other side is known.

The conditions will be dictated by the types of material the media are made of.

We shall consider the Boundary conditions at an interface separating

• Di-electric $(\epsilon&space;_{1})$ and Di-electric $(\epsilon&space;_{2})$.
• Conductor $(\sigma&space;)$ and Di-electric$(\epsilon&space;)$.
• Conductor and free space.

To determine the boundary conditions, we need to use two Maxwell’s equations.

1. $\oint_{s}&space;\overrightarrow{D}&space;.\overrightarrow{ds}&space;=Q_{enclosed}$ –  Gauss’s law in Electrostatics.
2. $\oint_{l}&space;\overrightarrow{E}&space;.\overrightarrow{dl}&space;=&space;0$.

The Electric field intensity could be considered as result of two components tangential and normal components.

$\overrightarrow{E}=\overrightarrow{E_{t}}+\overrightarrow{E_{n}}$.

Di-electric $(\epsilon&space;_{1})$ and Di-electric $(\epsilon&space;_{2})$:-

In order to find out the $\overrightarrow{E}$ , $\overrightarrow{D}$   at the interface between two different magnetic materials boundary conditions are required.

Consider two Di-electric materials having permeabilities  $\epsilon&space;_{1}$   and  $\epsilon&space;_{2}$ as shown in the figure

To apply    $\oint_{l}&space;\overrightarrow{E}&space;.\overrightarrow{dl}&space;=&space;0$  a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda  on the surface of the boundary

Then $\oint_{l}&space;\overrightarrow{E}&space;.\overrightarrow{dl}&space;=&space;0$ .

$E_{1t}\Delta&space;w&space;-E_{1n}\frac{\Delta&space;h}{2}-E_{2n}\frac{\Delta&space;h}{2}-E_{2t}\Delta&space;w+E_{2n}\frac{\Delta&space;h}{2}+E_{1n}\frac{\Delta&space;h}{2}=0$.

(since $\Delta&space;h=0$   On the boundary)

$E_{1t}\Delta&space;w-E_{2t}\Delta&space;w=0$ .

$E_{1t}=E_{2t}$.

$\frac{D_{1t}}{\epsilon&space;_{1}}=\frac{D_{2t}}{\epsilon&space;_{2}}$ , So the tangential components of  $\overrightarrow{D}$    are discontinuous where as $\overrightarrow{E}$   are continuous.

In order to apply $\oint_{s}&space;\overrightarrow{D}&space;.\overrightarrow{ds}&space;=Q$ , a surface is required which is nothing but Gaussian surface (or) a Pillbox enclosing some charge.

$\oint_{s}&space;\overrightarrow{D}&space;.\overrightarrow{ds}&space;=&space;\oint_{side}&space;\overrightarrow{D}&space;.\overrightarrow{ds}&space;+&space;\oint_{top}&space;\overrightarrow{D}&space;.\overrightarrow{ds}&space;+&space;\oint_{bottom}&space;\overrightarrow{D}&space;.\overrightarrow{ds}$ .

($\oint_{s}&space;\overrightarrow{D}&space;.\overrightarrow{ds}&space;=Q$  -Because on the boundary  $\Delta&space;h=0$, so there exists no side of surface).

$D_{1n}\Delta&space;S&space;-D_{2n}\Delta&space;S&space;=\Delta&space;Q$.

$D_{1n}-D_{2n}=\frac{\Delta&space;Q}{\Delta&space;S}$ .

$D_{1n}-D_{2n}=\rho&space;_{s}-----(1)$.

since $\overrightarrow{D}&space;=&space;\epsilon&space;\overrightarrow{E}$ ,

$\epsilon&space;_{1}E_{1n}-\epsilon&space;_{2}E_{2n}=\rho&space;_{s}-----(2)$

From equations (1) and (2)  the normal components of $\overrightarrow{D}$ and $\overrightarrow{E}$   are  discontinuous at the boundary .

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