Question

A glass of relative permittivity 4 is kept in an external electric field of magnitude
102 Vm−1. Calculate the polarisation vector, molecular/atomic polarisability and the
refractive index of the glass.

Answer 1

Answer on Question #69083, Physics / Electromagnetism

A glass of relative permittivity 4 is kept in an external electric field of magnitude 102 Vm⁻¹. Calculate the polarisation vector, molecular/atomic polarisability and the refractive index of the glass.

Solution:

Polarisation vector:

$\vec{P} = \chi \varepsilon_0 \vec{E}$ (1), where $\chi$ is the electric susceptibility, $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$, $E$ is the magnitude of electric field

The relative permittivity of a medium $\varepsilon_r$ is related to its electric susceptibility, $\chi$:

$$\varepsilon_r = 1 + \chi$$ $$\text{Of (2)} \Rightarrow \chi = \varepsilon_r - 1$$ $$\text{Of (3)} \Rightarrow \chi = 4 - 1 = 3$$ $$\text{Of (1)} \Rightarrow \vec{P} = 3 \times 8.85 \times 10^{-12} \, \frac{\text{F}}{\text{m}} \times 10^{2} \, \frac{\text{V}}{\text{m}} = 26.55 \times 10^{-10} \, \frac{\text{F} \times \text{V}}{\text{m}^{3}}$$

Polarisation vector:

$\vec{P} = N \vec{p} = N \alpha \varepsilon_0 \vec{E}$ (5), where $N = 6.02 \times 10^{23} \, \text{mol}^{-1}$, $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$, $\alpha$ is atomic polarisability, $E$ is the magnitude of electric field

$$\text{Of (5)} \Rightarrow \alpha = \frac{p}{N \varepsilon_0 E}$$ $$\text{Of (6)} \Rightarrow \alpha = 0.5 \times 10^{-23} \, \text{m}^{3}$$

Refractive index:

$n = \sqrt{\varepsilon_r \mu_r}$ (7), where $\varepsilon_r$ is the relative permittivity, $\mu_r$ is the relative permeability

$$\text{Of (7)} \Rightarrow n = \sqrt{4 \times 0.99} = 1.99$$

Answer:

polarisation vector: $26.55 \times 10^{-10} \, \frac{\text{F} \times \text{V}}{\text{m}^{3}}$

atomic polarisability: $0.5 \times 10^{-23} \, \text{m}^{3}$

refractive index: 1.99

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