Chapter 22, Problem 43GP

To determine

The value of rms of electric field that hits the Mars

Answer to Problem 43GP

The answer is $\text{469}\text{V/m}$

Explanation of Solution

Given data:

Power flux of the Earth, $S_{Earth}=1350W/m^2$

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

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: $S={\epsilon }_{0}c{E}_{rms}^2$

Where,

  $S$ is the power flux.

  $c$ is speed of the light.

  $E$ is the electric field.

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

Calculation:

It is been given that the radiation that comes from the Sun has the same intensity in all directions. Hence, the rate of passing will be the same, $P=S4\pi r^2$ .

The rate will be going to be the same for the Earth and the Sun, so, the ratio will be,

  $\frac{{\text{P}}_{\text{Mars}}}{{\text{P}}_{\text{Earth}}}\text{=}\left(\frac{{\text{S}}_{\text{Mars}}}{{\text{S}}_{\text{Earth}}}\right){\left(\frac{{\text{r}}_{\text{Mars}}}{{\text{r}}_{\text{Earth}}}\right)}^{\text{2}}\text{=1=}\left(\frac{{\text{S}}_{\text{Mars}}}{\text{1350}{\text{W/m}}^{\text{2}}}\right){\left(\text{1}\text{.52}\right)}^{\text{2}}$

Simplify further,

  ${\text{S}}_{\text{Mars}}\text{=}\frac{\text{1350}{\text{W/m}}^{\text{2}}}{{\left(\text{1}\text{.52}\right)}^{\text{2}}}\text{=584}\text{.3}{\text{W/m}}^{\text{2}}$

Substitute the given data into the formula $S={\epsilon }_{0}c{E}_{rms}^2$ ,

  $S={\epsilon }_{0}c{E}_{rms}^2=\text{584}\text{.3}{\text{W/m}}^{\text{2}}=8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\times 3\times {10}^8\text{m/s}\times E_{rms}^2$

Simplify more to get the rms value of the electric field,

  ${\text{E}}_{\text{rms}}\text{=}\sqrt{\frac{\text{584}\text{.3}{\text{W/m}}^{\text{2}}}{\text{8}{\text{.85×10}}^{\text{-12}}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}{\text{×3×10}}^{\text{8}}\text{m/s}}}=\text{469}\text{V/m}$

Conclusion: The value of the root mean squares of the electric field that hits the Mars is $\text{469}\text{V/m}$ .