91Ó°ÊÓ

Consider you have enough linear dielectric material, of dielectric constant to half-fill a parallel-plate capacitor

Write the formula of distribute material.

$\mathbf{D}={\mathrm{σ}}_{\mathbf{f}}$…… (1)

Here,${\mathrm{σ}}_{\mathbf{f}}$ is a free bound charge.

Write the formula of electric field of material.

${\mathbf{E}}_{\mathbf{material}}=\frac{\mathbf{D}}{{\mathrm{Î\mu }}_{\mathbf{0}}{\mathrm{Î\mu }}_{\mathbf{r}}}$ …… (2)

Here, $\mathbf{D}$is distribute material, ${\mathrm{Î\mu }}_{\mathbf{0}}$ is dielectric material and ${\mathrm{Î\mu }}_{\mathbf{r}}$ is dielectric constant.

Write the formula of electric field ofair.

${\mathbf{E}}_{\mathbf{air}}=\frac{\mathbf{D}}{{\mathrm{Î\mu }}_{\mathbf{0}}}$ …… (3)

Here, $\mathbf{D}$ is distribute material, ${\mathrm{Î\mu }}_{\mathbf{0}}$ is dielectric material.

Write the formula of polarization material.

$\mathbf{P}={\mathrm{Î\mu }}_{\mathbf{0}}\left({\mathrm{Î\mu }}_{\mathbf{r}}-\mathbf{1}\right){\mathbf{E}}_{\mathbf{material}}$ …… (4)

Here, ${\mathrm{Î\mu }}_{\mathbf{0}}$ is dielectric material, ${\mathrm{Î\mu }}_{\mathbf{r}}$ is dielectric constant and ${\mathbf{E}}_{\mathbf{material}}$is electric field of material.

Write the formula of free and bound charges on the lower.

${\mathrm{σ}}_{\mathbf{b}\mathbf{,}\mathbf{lower}}=-\mathbf{P}$ …… (5)

Here, $-\mathbf{P}$ is polarization material.

Write the formula of free and bound charges on the upper side.

${\mathrm{σ}}_{\mathbf{b}\mathbf{,}\mathbf{upper}}=\mathbf{P}$ …… (6)

Here,$\mathbf{P}$ is polarization material.

Draw the circuit diagram of figure 1 dielectric material of parallel plat capacitor.

Fig.1

Observe the first image. We shall first obtain the free charge density. To be on a greater potential, choose the bottom plate. I won't emphasise the vectorial nature of the vector quantities because they will all point in the z-direction.

$\begin{array}{rcl}D& =& \mathrm{Î\mu }E\\ & =& {\mathrm{σ}}_f\\ & & \end{array}$

Determine the output voltage between the plates.

$\begin{array}{rcl}{V}_0& =& -{∫}_{4\mathrm{δ}}^{0}Edx\\ & =& {∫}_0^{4\mathrm{δ}}\frac{D}{c}dx\\ & =& {∫}_0^{4\mathrm{δ}}\frac{{\mathrm{σ}}_f}{{\mathrm{Î\mu }}_0}dx\\ & =& {∫}_0^{\mathrm{δ}}\frac{{\mathrm{σ}}_f}{{\mathrm{Î\mu }}_0}dx+{∫}_{\mathrm{δ}}^{3\mathrm{δ}}\frac{{\mathrm{σ}}_f}{\mathrm{Î\mu }}dx+{∫}_{3\mathrm{δ}}^{4\mathrm{δ}}\frac{{\mathrm{σ}}_f}{\mathrm{Î\mu }}\\ & & \end{array}$

Solve further as

$\begin{array}{rcl}{V}_0& =& \frac{{\mathrm{σ}}_f}{{\mathrm{Î\mu }}_0}\left(\mathrm{δ}+\frac{2\mathrm{δ}}{{\mathrm{Î\mu }}_r}+\mathrm{δ}\right)\\ & =& \frac{2\mathrm{δ}{\mathrm{σ}}_f}{{\mathrm{Î\mu }}_0}\frac{{\mathrm{Î\mu }}_r+1}{{\mathrm{Î\mu }}_r}\\ {\mathrm{σ}}_f& =& \frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}\\ & & \end{array}$

Determine the value of distribute material.

Substitute $\frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$ for ${\mathrm{σ}}_f$into equation (1).

$\begin{array}{rcl}D& =& {\mathrm{σ}}_f\\ & =& \frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}\\ & & \end{array}$

Therefore, the value of distribute material is $D={\mathrm{σ}}_f=\frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$.

Determine the electric field of material and air.

Substitute $\frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$ for D into equation (2).

$\begin{array}{rcl}{E}_{material}& =& \frac{\frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}}{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}\\ & =& \frac{1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}\\ & & \end{array}$

Determine the electric field of air.

Substitute $\frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$ for D into equation (3).

$\begin{array}{rcl}{E}_{air}& =& \frac{\frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}}{{\mathrm{Î\mu }}_0}\\ & =& \frac{{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_0+1}\frac{V_0}{2\mathrm{δ}}\\ & & \end{array}$

Determine the polarization material.

Substitute $\frac{1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$ for $E_{material}$ into equation (4).

$P={\mathrm{Î\mu }}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$

Determine the free and negative bound charges on the lower side of the dielectric medium.

Substitute $-{\mathrm{Î\mu }}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$for $-P$into equation (5).

${\mathrm{σ}}_{b,lower}=-{\mathrm{Î\mu }}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$

Determine the free and positive bound charges on the upper side of the dielectric medium.

Substitute $-{\mathrm{Î\mu }}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$for P into equation (6).

${\mathrm{σ}}_{b,upper}={\mathrm{Î\mu }}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$

Therefore, the value of free and bound charges on the lower and upper side is ${\mathrm{σ}}_{b,lower}=-{\mathrm{Î\mu }}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$and${\mathrm{σ}}_{b,upper}={\mathrm{Î\mu }}_0\frac{{\mathrm{Î\mu }}_r-1}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}$

Determine the quotient factor of the two capacitances $\mathrm{η}$ is:

localid="1658912501962" $\mathrm{η}=\frac{C\text{'}}{C}$

Here, C is the capacitance with the dielectric, and C is capacitance without the dielectric.

localid="1658912510561" $\begin{array}{rcl}\mathrm{η}& =& \frac{\frac{{\mathrm{σ}}_{f}A}{V_0}}{{\mathrm{Î\mu }}_0\frac{A}{4\mathrm{δ}}}\\ & =& \frac{4\mathrm{δ}}{V_0{\mathrm{Î\mu }}_0}\frac{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\frac{V_0}{2\mathrm{δ}}\\ & =& \frac{2{\mathrm{Î\mu }}_r}{{\mathrm{Î\mu }}_r+1}\\ & & \end{array}$

Draw the circuit diagram of figure 2 dielectric material of parallel plat capacitor.

Fig.2

Now look at the second image. Once more, the voltage between the plates equals the integral of the electric field.

$\begin{array}{rcl}{V}_0& =& -{∫}_{\mathrm{δ}}^{0}Edx\\ & =& {∫}_0^{\mathrm{δ}}\frac{D}{\mathrm{Î\mu }}dx\\ & =& {∫}_0^{\mathrm{δ}}\frac{{\mathrm{σ}}_f}{\mathrm{Î\mu }}dx\\ & & \end{array}$

The two integrals have to be the same whether going trough material or air, so the free charge densities will not be the same:

${\mathrm{σ}}_{f,material}={\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r\frac{V_0}{\mathrm{δ}}$

And

${\mathrm{σ}}_{f,air}={\mathrm{Î\mu }}_0\frac{V_0}{\mathrm{δ}}$

Determine the distribute materialof second region.

Substitute ${\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r\frac{V_0}{\mathrm{δ}}$for ${\mathrm{σ}}_{f,material}$into equation (1).

$\begin{array}{rcl}{D}_{material}& =& {\mathrm{σ}}_{f,material}\\ & =& {\mathrm{Î\mu }}_0\frac{V_0}{\mathrm{δ}}\\ & & \end{array}$

Determine the distribute airof second region

Substitute ${\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r\frac{V_0}{\mathrm{δ}}$ for${\mathrm{σ}}_{f,air}$ into equation (1).

$\begin{array}{rcl}{D}_{air}& =& {\mathrm{σ}}_{f,air}\\ & =& {\mathrm{Î\mu }}_0\frac{V_0}{\mathrm{δ}}\end{array}$

Determine the electric field of second region is same both in material. .

Substitute $V_0$for D and $\mathrm{δ}$for $\mathrm{Î\mu }$into equation (3).

$\begin{array}{rcl}\text{E}& =& \frac{D}{\mathrm{Î\mu }}\\ & =& \frac{V_0}{\mathrm{δ}}\\ & & \end{array}$

Determine the polarization material of second region.

Substitute$E\frac{V_0}{\mathrm{δ}}$for E into equation (4).

$P={\mathrm{Î\mu }}_0\left({\mathrm{Î\mu }}_r-1\right)E\frac{V_0}{\mathrm{δ}}$

Determine the free and positive bound charges on the upper side of the dielectric medium.

Substitute $-{\mathrm{Î\mu }}_0\left({\mathrm{Î\mu }}_r-1\right)E\frac{V_0}{\mathrm{δ}}$for -P into equation (6).

${\mathrm{σ}}_{b,lower}=-{\mathrm{Î\mu }}_0\left({\mathrm{Î\mu }}_r-1\right)E\frac{V_0}{\mathrm{δ}}$

Determine the free and positive bound charges on the upper side of the dielectric medium.

Substitute${\mathrm{Î\mu }}_0\left({\mathrm{Î\mu }}_r-1\right)E\frac{V_0}{\mathrm{δ}}$ for P into equation (6).

${\mathrm{σ}}_{b,upper}={\mathrm{Î\mu }}_0\left({\mathrm{Î\mu }}_r-1\right)E\frac{V_0}{\mathrm{δ}}$

Therefore, the value of free and bound charges on the lower and upper side of second region is ${\mathrm{σ}}_{b,lower}=-{\mathrm{Î\mu }}_0\left({\mathrm{Î\mu }}_r-1\right)E\frac{V_0}{\mathrm{δ}}$and ${\mathrm{σ}}_{b,upper}={\mathrm{Î\mu }}_0\left({\mathrm{Î\mu }}_r-1\right)E\frac{V_0}{\mathrm{δ}}$

The two capacitances ratio, comes last. This configuration really consists of parallel connections of capacitors operating at the same voltage. Thus:

$C\text{'}=C_{material}+C_{air}$

Here, C is the capacitance with the dielectric,$C_{material}$is capacitance of material and $C_{air}$is capacitance of air.

Substitute $\frac{Q_{left}}{V_0}$ for $C_{material}$and $\frac{Q_{right}}{V_0}$for $C_{air}$ into above equation.

$\begin{array}{rcl}C\text{'}& =& \frac{Q_{left}}{V_0}+\frac{Q_{right}}{V_0}\\ & =& \frac{{\mathrm{σ}}_{f,material}\frac{A}{2}}{{\mathrm{σ}}_{f,material}\frac{\mathrm{δ}}{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}}+\frac{{\mathrm{σ}}_{f,air}\frac{A}{2}}{{\mathrm{σ}}_{f,air}\frac{\mathrm{δ}}{{\mathrm{Î\mu }}_0}}\\ & =& {\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r\frac{A}{2\mathrm{δ}}+{\mathrm{Î\mu }}_0\frac{A}{2\mathrm{δ}}\\ & =& {\mathrm{Î\mu }}_0\frac{A}{2\mathrm{δ}}\left(1+{\mathrm{Î\mu }}_r\right)\\ & & \end{array}$

Determine the quotient factor of the two capacitances $\mathrm{η}$ is:

$\mathrm{η}=\frac{C\text{'}}{C}$

Here, C is the capacitance with the dielectric, and C is capacitance without the dielectric.

Substitute ${\mathrm{Î\mu }}_0\frac{A}{2\mathrm{δ}}\left(1+{\mathrm{Î\mu }}_r\right)$for C and ${\mathrm{Î\mu }}_0\frac{A}{\mathrm{δ}}$ for C into above equation$\mathrm{η}$.

$\begin{array}{rcl}\mathrm{η}& =& \frac{{\mathrm{Î\mu }}_0\frac{A}{2\mathrm{δ}}\left(1+{\mathrm{Î\mu }}_r\right)}{{\mathrm{Î\mu }}_0\frac{A}{2\mathrm{δ}}}\\ & =& \frac{1}{2}\left(1+{\mathrm{Î\mu }}_r\right)\\ & =& \frac{1}{2}\left(1+{\mathrm{Î\mu }}_r\right)\\ & & \end{array}$