Dielectric Materials - Solved Problems

1. If an ionic crystal is subjected to an electric field of 1000 V/m and the resulting polarization is 4.3 × 10^-8 cm², Calculate the relative permittivity of the crystal.

Given

E = 1000 V/m, ε₀ =8.85×10^-12 F/m, P = 4.3 × 10^-8 C/m², ε_r = ?

Solution :-

$\begin{aligned} P & =\varepsilon_{\mathrm{o}}\left(\varepsilon_{\mathrm{r}}-1\right) \mathrm{E} \\ \varepsilon_{\mathrm{r}} & =1+\frac{\mathrm{P}}{\varepsilon_{0} \mathrm{E}} \\ & =1+\frac{4.3 \times 10^{-8}}{\left(8.85 \times 10^{-12} \times 1000\right)} \\ \varepsilon_{\mathrm{r}} & =5.86 . \end{aligned}$

2. A solid material contains 5 × 10²⁸ atoms/m² each with a polarisability of 2 × 10^-40 Fm². Calculate the ratio of internal field to the applied field.

Given

$\mathrm{N}=5 \times 10^{28} \text { atoms } / \mathrm{m}^{3}, \alpha=2 \times 10^{-40} \mathrm{Fm}^{2}, \frac{\mathrm{E}_{\mathrm{i}}}{\mathrm{E}}=?$

Solution :-

eqn. 1

$\begin{aligned} E_{i} & =E+\frac{P}{3 \varepsilon_{0}} \\ P & =N \alpha E_{i} \\ & =5 \times 10^{28} \times 2 \times 10^{-40} \times E_{i} \\ P & =1 \times 10^{-11} \times E_{i} \end{aligned}$ eqn. 2

Substituting eqn. (2) in (1) we get,

$\begin{aligned} \mathrm{E}_{\mathrm{i}} & =\mathrm{E}+\frac{1 \times 10^{-11} \times \mathrm{E}_{\mathrm{i}}}{3 \times 8.85 \times 10^{-12}} \\ & =\mathrm{E}_{\mathrm{i}}-\frac{1 \times 10^{-11} \times \mathrm{E}_{\mathrm{i}}}{3 \times 8.85 \times 10^{-12}} \\ & =\mathrm{E}_{\mathrm{i}}\left[1-\frac{1 \times 10^{-11}}{3 \times 8.85 \times 10^{-12}}\right] \\ & =\mathrm{E}_{\mathrm{i}}(0.62352) \\ \frac{\mathrm{E}_{\mathrm{i}}}{\mathrm{E}} & =\frac{1}{0.6235} \\ \frac{\mathrm{E}_{\mathrm{i}}}{\mathrm{E}} & =1.6038 \end{aligned}$

3. Calculate the electronic polarizability of argon atom. Given ∈_r = 1.0024 at NTP and N = 2.7 × 10²⁵ atoms/m³.

Given

∈_r = 1.0024, ∈_o = 8.85 × 10^-12 F/m, N=2.7 × 10²⁵ atoms/m³, α_e =?

Solution:-

$\begin{aligned} \mathrm{P} & =\mathrm{N} \alpha_{\mathrm{e}} \mathrm{E} \\ \alpha_{\mathrm{e}} & =\frac{\mathrm{P}}{\mathrm{NE}} \\ \mathrm{P} & =\varepsilon_{\mathrm{o}}\left(\varepsilon_{\mathrm{r}}-1\right) \mathrm{E} \\ \alpha_{\mathrm{e}} & =\frac{\varepsilon_{\mathrm{o}}\left(\varepsilon_{\mathrm{r}}-1\right) \mathrm{E}}{\mathrm{NE}}=\frac{\varepsilon_{\mathrm{o}}\left(\varepsilon_{\mathrm{r}}-1\right)}{\mathrm{N}} \\ & =\frac{8.85 \times 10^{-12} \times(1.0024-1)}{2.7 \times 10^{25}} \\ \alpha_{\mathrm{e}} & =7.9 \times 10^{-40} \mathrm{Fm}^{2} \end{aligned}$

4. The following data refers to a dielectric material. ε_r = 4.94 and n² = 2.69, where n is the index of refraction. Calculate the ratio between electron and ionic Polarizability for this material.

Given

ε_r4.94, n²=2.69, α=α_e+α_i (α₀ is negligibly small)

Solution :-

Clausius – Mossotti relation is

$\begin{aligned} \frac{\varepsilon_{\mathrm{r}}-1}{\varepsilon_{\mathrm{r}}+2} & =\frac{N \alpha}{3 \varepsilon_{0}} \\ \frac{\varepsilon_{\mathrm{r}}-1}{\varepsilon_{\mathrm{r}}+2} & =\frac{N\left(\alpha_{\mathrm{e}}+\alpha_{\mathrm{i}}\right)}{3 \varepsilon_{0}} \\ \frac{4.94-1}{4.94+2} & =\frac{\mathrm{N}\left(\alpha_{\mathrm{e}}+\alpha_{\mathrm{i}}\right)}{3 \varepsilon_{0}} \\ \frac{\mathrm{N}\left(\alpha_{\mathrm{e}}+\alpha_{\mathrm{i}}\right)}{3 \varepsilon_{0}} & =0.5677 \end{aligned}$ eqn. 1

We know ε_r $=n²$ and at optical frequencies α_i = 0

Hence $\begin{aligned} \frac{\mathrm{n}^{2}-1}{\mathrm{n}^{2}+2} & =\frac{\mathrm{N} \alpha_{\mathrm{e}}}{3 \varepsilon_{0}} \\ \frac{\mathrm{N} \alpha_{\mathrm{e}}}{3 \varepsilon_{0}} & =\frac{2.69-1}{2.69+2} \\ \frac{\mathrm{N} \alpha_{\mathrm{e}}}{3 \varepsilon_{0}} & =0.36034 \end{aligned}$ eqn. 2

Dividing equation (1) by equation (2)

$\begin{aligned} \frac{N\left(\alpha_{e}+\alpha_{i}\right)}{N \alpha_{e}} & =\frac{0.5677}{0.36034} \\ 1+\alpha_{i} / \alpha_{e} & =1.575 \\ \alpha_{i} / \alpha_{e} & =0.575 \\ \alpha_{e} / \alpha_{i} & =1.738 \end{aligned}$

5. A parallel plate condenser has a capacitance of $2μF$ . The dielectric has permittivity ε_r $=100$ . For an applied voltage of $1000 V$ , find the energy stored in the condenser as well as the energy stored in polarizing the dielectric.

Given

C = 2 × 10^-6 F, V=10³ V, ε_r= 100, W₀ = ?

Solution :-

$\begin{aligned} & \mathrm{E}=\frac{1}{2} \mathrm{CV}^{2} \\ & \mathrm{E}=\frac{2 \times 10^{-6} \times\left(10^{3}\right)^{2}}{2}=1 \text { joules } \end{aligned}$

To calculate the energy stored in the dielectric material which is in between the parallel plates of the condenser, capacitance has to be calculated removing the dielectric material.

$\mathrm{C}_{0}=\frac{\mathrm{C}}{\varepsilon_{\mathrm{r}}}=\frac{2 \times 10^{-6}}{100}=0.02 \mu \mathrm{F}$

Energy stored without the dielectric

$\mathrm{E}_{\mathrm{o}}=\frac{\mathrm{C}_{\mathrm{o}} \mathrm{V}^{2}}{2}=\frac{0.02 \times 10^{-6} \times 10^{6}}{2}=0.01 \text { joules }$

Hence energy stored in the dielectric

$\begin{aligned} & \mathrm{E}_{1}=\mathrm{E}-\mathrm{E}_{\mathrm{o}}=1-0.01 \\ & \mathrm{E}_{1}=0.99 \mathrm{Joule} \end{aligned}$

6. Find the capacitance of layer of Al₂O₃ having thickness 0.5μm and area 2500 mm² with ε_r =8.

Given

ε_r =8, d=0.5 × 10^-6 m, A = 2500 × 10^-6 m², C =?

Solution:-

$\begin{aligned} \mathrm{C} & =\frac{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mathrm{A}}{\mathrm{d}} \\ & =\frac{8.85 \times 10^{-12} \times 8 \times 2500 \times 10^{-6}}{0.5 \times 10^{-6}} \\ \mathrm{C} & =0.354 \times 10^{-6} \mathrm{~F} \end{aligned}$

7. Calculate the dielectric constant of a material which when inserted in a parallel plate condenser of area $100 mm²$ with a distance of separation of $1 mm$ gives a capacitance of $10 -9 F$ .

Given

A = 100 × 10^-6 m², d=1×10^-3 m, C = 10^-9 F, ε_r =?

Solution:-

$\begin{aligned} C & =\frac{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mathrm{A}}{\mathrm{d}} \\ \varepsilon_{\mathrm{r}} & =\frac{\mathrm{Cd}}{\varepsilon_{\mathrm{o}} \mathrm{A}} \end{aligned}$

$\begin{aligned} & =\frac{10^{-9} \times 1 \times 10^{-3}}{8.85 \times 10^{-12} \times 100 \times 10^{-6}} \\ & =\frac{10^{-12}}{8.85 \times 10^{-18}} \\ & =1.129 \times 10^{-3} \times 10^{6} \\ & \varepsilon_{\mathrm{r}}=1129.9 \end{aligned}$

8. Calculate the relative dielectric constant of a barium titanate crystal, which when inserted in a parallel plate condenser of area 10 $mm \times 10 mm$ and distance of separation of $2 mm$ , gives a capacitance of $10 -9 F$ .

Solution :-

Given

$\begin{aligned} \mathrm{C} & =\frac{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mathrm{A}}{\mathrm{d}} \\ \mathrm{C} & =10^{-9} \mathrm{~F} \\ \mathrm{~d} & =2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m} \\ \varepsilon_{\mathrm{o}} & =8.854 \times 10^{-12} \mathrm{Fm}^{-1} \\ \varepsilon_{\mathrm{r}} & =\frac{\mathrm{Cd}}{\varepsilon_{\mathrm{o}} \mathrm{A}}=\frac{10^{-9} \times 2 \times 10^{-3}}{8.854 \times 10^{-12} \times 100 \times 10^{-6}} \\ & =2259 \mathrm{~F} . \end{aligned}$

9. Calculate the polarization produced in a dielectric medium of dielectric constant 6 when it is subjected to an electric field of 100 $V/m$ .

Solution :-

Formula $\begin{aligned} \mathrm{p} & =\varepsilon_{\mathrm{o}} \chi_{\mathrm{o}} \mathrm{E} \\ & =\mathrm{E} \varepsilon_{\mathrm{o}}\left(\varepsilon_{\mathrm{r}}-1\right) \\ \mathrm{E} & =100 \mathrm{~V} / \mathrm{m} \\ \varepsilon_{\mathrm{o}} & =8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m} \\ \varepsilon_{\mathrm{r}} & =6 \end{aligned}$

Given

Hence $\begin{aligned} \mathrm{P} & =100 \times 8.85 \times 10^{-12}(6-1) \\ & =100 \times 8.85 \times 5 \times 10^{-12} \\ & =4.425 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2} \end{aligned}$