Chapter 22, Problem 41GP

(a)

To determine

The root mean square value of the electric field associated with radiation.

Answer to Problem 41GP

The answer is $0.07V/m$ .

Explanation of Solution

Given data:

Average energy density, $U=4\times {10}^{-14}J/m^3$

Permittivity of vacuum, ${\epsilon }_0=8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}$

Formula used:

It is known that: $U={\epsilon }_0{E}_{rms}^2$

Where,

  $\text{U}$ is the average energy density.

  ${\epsilon }_0$ is the permittivity of vacuum.

  $E_{rms}^2$$t$ is the root mean square value of the electric field.

Calculation:

Substitute the given data into the formula $U={\epsilon }_0{E}_{rms}^2$ ,

  $U={\epsilon }_0{E}_{rms}^2=4\times {10}^{-14}J/m^3=\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right)\times E_{rms}^2$

Now, simplify further,

  $E_{rms}^2=\frac{4\times {10}^{-14}J/m^3}{\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right)}=0.07V/m$

Conclusion: The root mean square value of the electric field associated with radiation is $0.07V/m$ .

(b)

To determine

The comparable value will be determined.

Answer to Problem 41GP

The answer is $8km$ .

Explanation of Solution

Given data:

Power, $P={10}^{5}W$

Speed of the light, $c=3\times {10}^8\text{m/s}$

Permittivity of vacuum, ${\in }_0=8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}$

The energy per meter length, $\text{E}=\text{0}\text{.1}\text{V/m}$

Formula used:

It is known that: $r^2=\frac{P}{\left(2\pi {\epsilon }_{0}c{E}^2\right)}$

Where,

  $P$ is the power.

  $c$ is speed of the light.

  $E_{rms}$ is root square means value of the electric field.

  ${\in }_0$ is the permittivity of vacuum.

  $\text{E}$ is the energy per meter length.

Calculation:

Substitute the given data into the formula $r^2=\frac{P}{\left(2\pi {\epsilon }_{0}c{E}^2\right)}$ ,

  $r^2=\frac{P}{\left(2\pi {\epsilon }_{0}c{E}^2\right)}=\frac{{10}^{5}W}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\right)\left(3\times {10}^8\text{m/s}\right){\left(\text{0}\text{.1}\text{V/m}\right)}^2}=5.995\times {10}^7{m}^2$

Simplify further,

  $r=\sqrt{5.995\times {10}^7{m}^2}=8km$

Conclusion: The comparable value will be $8km$ .