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, εr = ?
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. 2Substituting 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, αe =?
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. εr = 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
εr 4.94, n²=2.69, α=αe+αi (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. 1We know εr=n² and at optical frequencies αi = 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 εr=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, εr = 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 εr =8.
Given
εr =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, εr =?
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 × 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}
10. Calculate the electric polarizability of Xenon. The radius of Xenon atom is 0.158 nm.
Solution :-
We know electric polarizability
αe = 4πε0R3
Given R = 0.158 × 10-9 m
Hence, αe = 4×π×8.85×10-12×(0.158×10-9)3
=4.388×10-40 Fm2
