Question

An electric field of $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{N}\mathbf{/}\mathit{C}$ acts on an electron, resulting in an acceleration of $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ for a short time. What is the magnitude of the radiative electric field observed at a location a distance of $\mathbf{2}\mathbf{}\mathbf{\text{cm}}$ away along a line perpendicular to the direction of the acceleration?

Short answer

The radiative electric field is $1.44\times {10}^{-7}N/C$.

Step-by-step solution

Given data

Applied electric field:

$E=1\times {10}^{6}N/C$

Acceleration of the electron:

$a=1.8\times {10}^{17}m/s^2$

Distance of measurement of radiative electric field

$$\begin{gathered} r=2\text{cm} \\ =2·\left(1\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right) \\ =0.02\text{m} \end{gathered}$$

Determine the concept of electric field

The radiative electric field of a particle of charge and acceleration is:

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{a}}{{\mathbf{c}}^{\mathbf{2}}\mathbf{r}}$ ……. (i)

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity of free space with value $\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{=}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathit{N}\mathbf{.}{\mathit{m}}^{\mathbf{2}}\mathbf{/}{\mathit{C}}^{\mathbf{2}}$

Here, $\mathit{E}$is the speed of light in vacuum with charge $\mathit{c}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.

Determine the radiative electric field

From equation (i), the radiative electric field for the electron with charge $1.6\times {10}^{-19}\text{C}$is calculated as:

$$\begin{gathered} E=9\times {10}^{9}N.m^2/C^2\times \frac{1.6\times {10}^{-19}\text{C}\times 1.8\times {10}^{17}m/s^2}{{\left(3\times {10}^{8}m/s\right)}^2\times 0.02\text{m}} \\ =1.44\times {10}^{-7}N/C \end{gathered}$$

Thus, the field is $1.44\times {10}^{-7}N/C$.