Question

A particle of charge q is traveling at constant speed v along the x axis. Calculate the total power passing through the plane $\mathit{x}\mathbf{=}\mathit{a}$X, at the moment the particle itself is at the origin. [ Answer $\frac{{\mathbf{q}}^{\mathbf{2}}\mathbf{v}}{\mathbf{32}\mathbf{\Pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{a}}^{\mathbf{2}}}$]

Short answer

The total power passing through the plane is $P=\frac{v{q}^2}{32\pi {\epsilon }_0{a}^2}$.

Step-by-step solution

Expression for the total power passing through the plane:

Write the expression for the total power passing through the plane.

$\mathit{P}\mathbf{=}\mathbf{\int }\mathit{S}\mathbf{·}\mathit{d}\mathit{a}$ …… (1)

Here, S is the poynting vector.

Determine the Poynting vector:

Write the expression for the Poynting vector.

$S=\frac{1}{{\mu }_0}\left(E\times B\right)$ …… (2)

Here,${\mu }_0$ is the permeability of free space, E is the electric field, and B is the magnetic field which is given as:

$B=\frac{1}{c^2}\left(v\times E\right)$

Substitute $B=\frac{1}{c^2}\left(v\times E\right)$in equation (2).

role="math" localid="1653901917989" $$\begin{gathered} S=\frac{1}{{\mu }_0}\left(E\times \frac{1}{c^2}\left(v\times E\right)\right) \\ S=\frac{1}{{\mu }_0{c}^2}\left(E\times \left(v\times E\right)\right)......\left(3\right) \end{gathered}$$

Here,$\frac{1}{c^2}={\mu }_0{\epsilon }_0$

Substitute $\frac{1}{c^2}={\mu }_0{\epsilon }_0$in equation (3).

$$\begin{gathered} S=\frac{\left({\mu }_0{\epsilon }_0\right)}{{\mu }_0}\left(E\times \left(v\times E\right)\right) \\ S={\epsilon }_0\left(E^{2}v-\left(v·E\right)E\right) \end{gathered}$$

Determine the total power passing through the plane:

Consider the figure,

The value of da is given as:

$da=\left(2\pi rdr\right)\hat{x}$

Substitute $S={\epsilon }_0\left(E^{2}v-\left(v·E\right)E\right)$and $da=\left(2\pi rdr\right)\hat{x}$in equation (1).

$$\begin{gathered} P=\int {\epsilon }_0\left(E^{2}v-\left(v·E\right)E\right)\left(2\pi rdr\right)\hat{x} \\ P={\epsilon }_0\int \left(E^{2}v-\left(Ev\cos \theta \right)E_x\right)2\pi rdr \\ P={\epsilon }_0\int \left(E^{2}v-E_x^{2}v\right)\left(2\pi r\right)dr \end{gathered}$$

Here, $E_x=E\cos \theta$

Hence, the above equation becomes,

role="math" localid="1653902580996" $$\begin{gathered} P={\epsilon }_0\int \left(E^{2}v-{\left(E\cos \theta \right)}^{2}v\right)\left(2\pi r\right)dr \\ P=v{\epsilon }_0\int \left(E^2-E^2{\cos }^2\theta \right)\left(2\pi r\right)dr \\ P=2\pi {\epsilon }_{0}v\int \left(E^2{\sin }^2\theta \right)rdr.......\left(4\right) \end{gathered}$$

Here, E is the electric field due to a point charge.

Write the expression for an electric field due to a point charge.

$E=\frac{q}{4\pi {\epsilon }_0{\gamma }^2}\frac{1}{{\left(1-{\displaystyle \frac{v^2{\sin }^2\theta }{c^2}}\right)}^{3/2}}\frac{\hat{R}}{R^2}$

Substitute $E=\frac{q}{4\pi {\epsilon }_0{\gamma }^2}\frac{1}{{\left(1-{\displaystyle \frac{v^2{\sin }^2\theta }{c^2}}\right)}^{3/2}}\frac{\hat{R}}{R^2}$in equation (4).

$$\begin{gathered} P=2\pi {\epsilon }_{0}v\left({\left(\frac{q}{4\pi {\epsilon }_0{\gamma }^2}\frac{1}{{\left(1-{\displaystyle \frac{v^2{\sin }^2\theta }{c^2}}\right)}^{3/2}}\frac{\hat{R}}{R^2}\right)}^2{\sin }^2\theta \right)rdr \\ P=2\pi {\epsilon }_{0}v{\left(\frac{q}{4\pi {\epsilon }_0}\right)}^2\frac{1}{{\gamma }^4}{\int }_0^{\infty }\frac{r{\sin }^2\theta }{R^4{\left(1-{\left({\displaystyle \frac{v}{c}}\sin \theta \right)}^2\right)}^3}dr.....\left(5\right) \end{gathered}$$

Let,

$$\begin{gathered} r=a\tan \theta \\ dr=ase{c}^2\theta d\theta \\ dr=\frac{a}{{\cos }^2\theta }d\theta \end{gathered}$$

From the figure, the data is observed as:

$\frac{1}{R}=\frac{\cos \theta }{a}$

Substitute $r=a\tan \theta ,dr=\frac{a}{{\cos }^2\theta }d\theta$and $\frac{1}{R}=\frac{\cos \theta }{a}$in equation (5).

$P=\frac{v}{2{\gamma }^4}\frac{a^2}{4\pi {\epsilon }_0}\frac{1}{a^2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^3\theta \cos \theta }{{\left(1-{\left({\displaystyle \frac{v}{c}}\sin \theta \right)}^2\right)}^3}d\theta$ …… (6)

Let,

$$\begin{gathered} u={\sin }^2\theta \\ du=2\sin \theta \cos \theta \end{gathered}$$

Hence, the equation (6) becomes,

$$\begin{gathered} P=\frac{v{q}^2}{16\pi {\epsilon }_0{a}^2{\gamma }^4}{\int }_0^1\frac{udu}{{\left(1-{\left({\displaystyle \frac{v}{c}}\right)}^{2}u\right)}^3} \\ P=\frac{v{q}^2}{16\pi {\epsilon }_0{a}^2{\gamma }^4}\times \frac{{\gamma }^4}{2} \\ P=\frac{v{q}^2}{32\pi {\epsilon }_0{a}^2} \end{gathered}$$

Therefore, the total power passing through the plane is$P=\frac{v{q}^2}{32\pi {\epsilon }_0{a}^2}$ .