Question

At the interface between one linear dielectric and another, the electric field lines bend (see Fig. 4.34). Show that

${\mathbf{tan\theta }}_{\mathbf{2}}\mathbf{/}{\mathbf{tan\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{2}}\mathbf{/}{\mathbf{\epsilon }}_{\mathbf{1}}$

Assuming there is no free charge at the boundary. [Comment: Eq. 4.68 is reminiscent of Snell's law in optics. Would a convex "lens" of dielectric material tend to "focus’’ or "defocus," the electric field?]

Short answer

The relation is proved,$\frac{{\mathrm{tan\theta }}_2}{{\mathrm{tan\theta }}_1}=\frac{{\in }_2}{{\in }_1}$ .

Step-by-step solution

Define function

Write the expression for the parallel of electric fields.

${\mathbf{E}}_{\mathbf{above}}\mathbf{-}{\mathbf{E}}_{\mathbf{below}}\mathbf{=}\mathbf{0}$ …… (1)

Here,${\mathrm{E}}_{\mathrm{above}}$ is the parallel component of electric field above the interface and${\mathrm{E}}_{\mathrm{below}}$ is the parallel component of electric filed below the interface.

Write the expression for perpendicular components.

${\in }_1{\mathrm{E}}_{\mathrm{above}}^{\perp }-{\in }_2{\mathrm{E}}_{\mathrm{below}}^{\perp }={\mathrm{\sigma }}_{\mathrm{f}}$ …… (2)

Here,${\mathrm{E}}_{\mathrm{above}}^{\perp }$ is the perpendicular component of electric field above the interface and${\mathrm{E}}_{\mathrm{below}}^{\perp }$ is the perpendicular component of electric field blow the interface.

Determine perpendicular components

Across the material interface, electrical fields satisfied boundary requirements. There is no free boundary charge here. Hence,

${\mathrm{\sigma }}_{\mathrm{f}}=0$

Now, applying boundary conditions,

Write the expression for parallel components of electric fields.

role="math" localid="1657601144914"

Substitute$E_1$for$E_{above}$and $E_2$for$E_{below}$in above equation.

Therefore,

$E_1=E_2$

Hence, the parallel component of electric field that is, $E$ is continuous.

Write the expression for perpendicular components.

${\in }_1{\mathrm{E}}_{\mathrm{above}}^{\perp }-{\in }_2{\mathrm{E}}_{\mathrm{below}}^{\perp }={\mathrm{\sigma }}_{\mathrm{f}}$

Substitute$E_1$ forrole="math" localid="1657601336234" $E_{above}$ and$E_2$ for$E_{below}$ and$0$ for${\sigma }_f$ in above equation.

$\begin{array}{c}{\in }_1{E}_{above}^{\perp }-{\in }_2{E}_{below}^{\perp }={\sigma }_f\\ {\in }_1{E}_1-{\in }_2{E}_2=0\end{array}$

$\begin{array}{c}{\in }_1{E}_1={\in }_2{E}_2\\ \frac{E_1}{E_2}=\frac{{\in }_2}{{\in }_1}\end{array}$

Determine tangent angle

Write the expression for the tangent of angle ${\theta }_1$.

$\tan {\theta }_1=\frac{E_1}{E_1^{\perp }}$ …… (3)

Write the expression for the tangent of angle ${\theta }_2$.

$\tan {\theta }_2=\frac{E_2}{E_2^{\perp }}$ ……. (4)

Divide equation (4) by equation (3)

$\begin{array}{c}\frac{\tan {\theta }_2}{\tan {\theta }_1}=\frac{\left(\frac{E_2}{E_2^{\perp }}\right)}{\left(\frac{E_1}{E_1^{\perp }}\right)}\\ =\left(\frac{E_2}{E_1}\right)\left(\frac{E_1^{\perp }}{E_2^{\perp }}\right)\end{array}$

Substitute$E_1$ for$E_2$ and$\frac{{\in }_2}{{\in }_1}$ for$\left(\frac{E_1^{\perp }}{E_2^{\perp }}\right)$ in above equation

Therefore,

$\begin{array}{c}\frac{\tan {\theta }_2}{\tan {\theta }_1}=\left(\frac{E_2}{E_1}\right)\left(\frac{E_1^{\perp }}{E_2^{\perp }}\right)\\ =\left(\frac{E_1}{E_1}\right)\left(\frac{{\in }_2}{{\in }_1}\right)\\ =\frac{{\in }_2}{{\in }_1}\end{array}$

Hence, the relation is proved,$\frac{{\mathrm{tan\theta }}_2}{{\mathrm{tan\theta }}_1}=\frac{{\in }_2}{{\in }_1}$ .

If the medium 1 is air and medium 2 is dielectric then, ${\in }_2>{\in }_1$and filed lines bend away from the normal. Hence,

$\frac{\tan {\theta }_2}{\tan {\theta }_1}=\frac{{\in }_2}{{\in }_1}>1$

This is opposite of light rays. Thus, a convex lens would defocus the field line.

Hence, the convex lens of a dielectric material would defocus the electric field.