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At f = 100 MHz, show that silver (σ=6.1×107S/m,μr=1,εr=1)\left(\sigma=6.1 \times 10^{7} \mathrm{S} / \mathrm{m}, \mu_{r}=1, \varepsilon_{r}=1\right) is a good conductor, while rubber (σ=1015S/m,μr=1,εr=3.1)\left(\sigma=10^{-15} \mathrm{S} / \mathrm{m}, \mu_{r}=1, \varepsilon_{r}=3.1\right) is a good insulator.

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We are given a frequency of f=100 MHzf=100 ~\mathrm{MHz}, and the parameters of two materials:

  • Silver (σ=6.1107 Sm,ϵr=1,μr=1)(\sigma = 6.1 \cdot 10^7 \mathrm{~\frac{S}{m}}, \epsilon_r = 1, \mu_r = 1)

  • Rubber (σ=1015 Sm,ϵr=1,μr=3.1)(\sigma = \cdot 10^{-15} \mathrm{~\frac{S}{m}}, \epsilon_r = 1, \mu_r = 3.1)

We must show that silver is a good conductor at this frequency, and that rubber is a good insulator at this frequency.

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