(a) Suppose you imbedded some free charge in a piece of glass. About how long would it take for the charge to flow to the surface?

(b) Silver is an excellent conductor, but it’s expensive. Suppose you were designing a microwave experiment to operate at a frequency of${\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathit{H}\mathit{z}$. How thick would you make the silver coatings?

(c) Find the wavelength and propagation speed in copper for radio waves at role="math" localid="1655716459863" $\mathbf{1}\mathbf{}\mathit{M}\mathit{H}\mathit{z}$. Compare the corresponding values in air (or vacuum).

Write the expression for the characteristic time.

$\mathit{\zeta }\mathbf{=}\frac{\mathbf{\epsilon }}{\mathbf{\sigma }}$ …… (1)

Here, $\zeta$is the characteristic time, $\epsilon$is the permittivity of the medium, and $\sigma$is the conductivity.

Write the expression for the relativity permittivity.

$$\begin{gathered} {\mathit{\epsilon }}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{\epsilon }}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathit{\epsilon }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{r}}{\mathit{\epsilon }}_{\mathbf{0}} \end{gathered}$$ …… (2)

Here, ${\epsilon }_0$is the Permittivity of the free space.

(a)

Substitute the value of equation (2) in equation (1).

$\zeta =\frac{{\epsilon }_r{\epsilon }_0}{\sigma }$ …… (3)

Write the expression for the relation between relative permittivity and refractive index of the medium.

${\epsilon }_r=n^2$

Here, n is the refractive index of glass $\left(n=1.5\right)$.

Substitute $n=1.5$, ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/N.m$and $\sigma ={10}^{-12{\Omega }^{-1}}·m^{-1}$in equation (3).

$$\begin{gathered} \zeta =\frac{n^2{\epsilon }_0}{\sigma } \\ \zeta =\frac{{\left(1.5\right)}^2\times \left(8.85\times {10}^{-12}{C}^2/N·m\right)}{\left({10}^{-12}{\Omega }^{-1}·m^{-1}\right)} \\ \zeta =19.91s\approx 20s \\ \zeta =20s \end{gathered}$$

Therefore, the time in which the charge flows to the surface is $20s$.

(b)

Write the expression for the skin depth.

$d=\frac{1}{k}$ …… (4)

Here,kis the imaginary constant.

Write the expression for the value ofk.

$$\begin{gathered} d=\frac{1}{\sqrt{{\displaystyle \frac{\left(2\mathrm{\pi f}\right)\mu \left({\displaystyle \frac{1}{\rho }}\right)}{2}}}} \\ d=\sqrt{\frac{2\rho }{2\pi f\mu }} \\ d=\sqrt{\frac{\rho }{\pi f\mu }} \end{gathered}$$

For $\sigma >>>\omega {\epsilon }_0$, the above equation becomes,

$k=\sqrt{\frac{\omega \mu \sigma }{2}}$ …… (5)

Substitute the value of equation (5) in equation (4).

$$\begin{gathered} d=\frac{1}{\sqrt{\frac{\left(2\pi f\right)\mu \left(\frac{1}{\rho }\right)}{2}}} \\ d=\sqrt{\frac{2\rho }{2\pi f\mu }} \\ d=\sqrt{\frac{\rho }{\pi f\mu }} \end{gathered}$$

Here, $\rho$is the resistivity of silver $\left(1.59\times {10}^{-8}\Omega ·m^{-1}\right)$and f is the frequency.

Substitute $\rho =1.59\times {10}^{-8}\Omega ·m^{-1}$, $f={10}^{10}Hz$, and $\mu =4\pi \times {10}^{-7}H/m$in the above expression.

$$\begin{gathered} d=\sqrt{\frac{\left(1.59\times {10}^{-8}\Omega ·m^{-1}\right)}{\pi \times \left({10}^{10}Hz\right)\times \left(4\mathrm{\pi }\times {10}^{-7}\mathrm{H}/\mathrm{m}\right)}} \\ d=6.34\times {10}^{-7}m\times \left(\frac{{10}^{3}mm}{1m}\right) \\ d=6.34\times {10}^{4}mm \end{gathered}$$

Therefore, the thickness of the silver coating is $6.34\times {10}^{4}mm$.

(c)

Write the expression for the wavelength of a radio wave.

$\lambda =\frac{2\pi }{k}$

Here,kis the wavenumber.

Substitute the value of equation (5) in the above expression.

$$\begin{gathered} \lambda =\frac{2\pi }{\sqrt{{\displaystyle \frac{\omega \mu \sigma }{2}}}} \\ \lambda =2\pi \sqrt{\frac{2}{\left(2\pi f\right)\left({\displaystyle \frac{1}{\rho }}\right)\mu }} \end{gathered}$$

Substitute $f=1MHz$, $\rho =1.59\times {10}^{-8}\Omega ·m^{-1}$and $\mu =4\pi \times {10}^{-7}H/m$in the above expression.

$$\begin{gathered} \lambda =2\pi \sqrt{\frac{2}{2\pi \times \left(1MHz\times {\displaystyle \frac{{10}^{6}Hz}{1MHz}}\right)\times \left({\displaystyle \frac{1}{\left(1.59\times {10}^{-8}\Omega ·m^{-1}\right)}}\right)\left(4\pi \times {10}^{-7}H/m\right)}} \\ \lambda =4\times {10}^{-4}m\times \left(\frac{{10}^{3}mm}{1m}\right) \\ \lambda =0.4mm \end{gathered}$$

Write the expression for propagation speed.

$v=\lambda f$

Substitute $\lambda =4\times {10}^{-4}m$and $f=1MHz$in the above expression.

$$\begin{gathered} v=\left(4\times {10}^{-4}\right)\times \left(1MHz\times \frac{{10}^{6}Hz}{1MHz}\right) \\ v=400m/s \end{gathered}$$

Calculate the wavelength in a vacuum.

$$\begin{gathered} \lambda =\frac{c}{f} \\ \lambda =\frac{\left(3\times {10}^8{\displaystyle \frac{m}{s}}\right)}{\left(\left(1MHz\right)\times \frac{{10}^{6}Hz}{1MHz}\right)} \\ \lambda =300m \end{gathered}$$

The speed of radio waves is much lesser than the speed of light, which means the propagation speed in copper is losses gradually with respect to the penetration depth.

Therefore, the wavelength and propagation speed in copper for radio waves is $0.4mm$and $400\frac{m}{s}$respectively, and the corresponding values in the air is $300m$, and the propagation speed in copper is losses gradually with respect to the penetration depth.