Question #103176
A glass of relative permitivity 4 is kept in an external electric field of magnitude 10^2 Vm^–1
.
Calculate the polarisation vector, molecular/atomic polarisability and the refractive index
of the glass
1
Expert's answer
2020-02-20T09:31:58-0500

Polarization vector


P=χϵ0EP=\chi \epsilon_0 E


ϵ=1+χχ=ϵ1=41=3\epsilon=1+\chi \to \chi=\epsilon-1=4-1=3


P=χϵ0E=38.851012100=26.551010FVm2P=\chi \epsilon_0 E=3\cdot 8.85\cdot 10^{-12}\cdot 100=26.55\cdot 10^{-10} \frac{F\cdot V}{m^2}


Molecular/atomic polarisability



P=Nαϵ0Eα=PNϵ0E=26.5510106.0210238.851012100=0.51023m3P=N\alpha \epsilon_0 E \to \alpha=\frac{P}{N \epsilon_0 E}=\frac{26.55\cdot 10^{-10}}{6.02\cdot 10^{23}\cdot 8.85\cdot 10^{-12}\cdot 100}=0.5\cdot 10^{-23} m^3



Refractive index


n=ϵμ=40.99=1.99n=\sqrt{\epsilon\mu}=\sqrt{4\cdot 0.99}=1.99











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