Question

calculate the magnitude of the polarisation for a dielectric material in an electric field of 10^6 V/m, given that the susceptibility of the material is 4.38

Answer 1

Answer on Question #69568 – Physics – Field Theory

Calculate the magnitude of the polarization for a dielectric material in an electric field of $10^{\wedge}6$ V/m, given that the susceptibility of the material is 4.38.

Solution:

Electric susceptibility is defined as the constant of proportionality relating an electric field $E$ to the induced dielectric polarization density $P$ such that [1]:

$$P = \varepsilon_0 \chi_e E,$$

where $P$ is the magnitude of the polarization for a dielectric material (or polarization density), so

$$P = \varepsilon_0 \chi_e E \approx 8.85 \times 10^{-12} F \cdot m^{-1} \times 4.38 \times 10^6 V \cdot m^{-1} \approx 3.88 \times 10^{-5} C \cdot m^{-2},$$

Answer: $3.88 \times 10^{-5} C \cdot m^{-2}$.

[1] https://en.wikipedia.org/wiki/Electric_susceptibility

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