91Ó°ÊÓ

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

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

Here, S is the poynting vector.

Write the expression for the Poynting vector.

$S=\frac{1}{{\mathrm{μ}}_0}\left(E×B\right)$ …… (2)

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

$B=\frac{1}{c^2}\left(v×E\right)$

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

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

Here,$\frac{1}{c^2}={\mathrm{μ}}_0{\mathrm{Î\mu }}_0$

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

$$\begin{gathered} S=\frac{\left({\mathrm{μ}}_0{\mathrm{Î\mu }}_0\right)}{{\mathrm{μ}}_0}\left(E×\left(v×E\right)\right) \\ S={\mathrm{Î\mu }}_0\left(E^{2}v-\left(v·E\right)E\right) \end{gathered}$$

Consider the figure,

The value of da is given as:

$da=\left(2\mathrm{Ï€}rdr\right)\hat{x}$

Substitute $S={\mathrm{Î\mu }}_0\left(E^{2}v-\left(v·E\right)E\right)$and $da=\left(2\mathrm{Ï€}rdr\right)\hat{x}$in equation (1).

$$\begin{gathered} P=∫{\mathrm{Î\mu }}_0\left(E^{2}v-\left(v·E\right)E\right)\left(2\mathrm{Ï€}rdr\right)\hat{x} \\ P={\mathrm{Î\mu }}_0∫\left(E^{2}v-\left(Ev\cos \mathrm{θ}\right)E_x\right)2\mathrm{Ï€}rdr \\ P={\mathrm{Î\mu }}_0∫\left(E^{2}v-E_x^{2}v\right)\left(2\mathrm{Ï€}r\right)dr \end{gathered}$$

Here, $E_x=E\cos \mathrm{θ}$

Hence, the above equation becomes,

role="math" localid="1653902580996" $$\begin{gathered} P={\mathrm{Î\mu }}_0∫\left(E^{2}v-{\left(E\cos \mathrm{θ}\right)}^{2}v\right)\left(2\mathrm{Ï€}r\right)dr \\ P=v{\mathrm{Î\mu }}_0∫\left(E^2-E^2{\cos }^2\mathrm{θ}\right)\left(2\mathrm{Ï€}r\right)dr \\ P=2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}v∫\left(E^2{\sin }^2\mathrm{θ}\right)rdr.......\left(4\right) \end{gathered}$$

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

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

$E=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{\mathrm{γ}}^2}\frac{1}{{\left(1-{\displaystyle \frac{v^2{\sin }^2\mathrm{θ}}{c^2}}\right)}^{3/2}}\frac{\hat{R}}{R^2}$

Substitute $E=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{\mathrm{γ}}^2}\frac{1}{{\left(1-{\displaystyle \frac{v^2{\sin }^2\mathrm{θ}}{c^2}}\right)}^{3/2}}\frac{\hat{R}}{R^2}$in equation (4).

$$\begin{gathered} P=2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}v\left({\left(\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{\mathrm{γ}}^2}\frac{1}{{\left(1-{\displaystyle \frac{v^2{\sin }^2\mathrm{θ}}{c^2}}\right)}^{3/2}}\frac{\hat{R}}{R^2}\right)}^2{\sin }^2\mathrm{θ}\right)rdr \\ P=2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}v{\left(\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)}^2\frac{1}{{\mathrm{γ}}^4}{∫}_0^{∞}\frac{r{\sin }^2\mathrm{θ}}{R^4{\left(1-{\left({\displaystyle \frac{v}{c}}\sin \mathrm{θ}\right)}^2\right)}^3}dr.....\left(5\right) \end{gathered}$$

Let,

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

From the figure, the data is observed as:

$\frac{1}{R}=\frac{\cos \mathrm{θ}}{a}$

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

$P=\frac{v}{2{\mathrm{γ}}^4}\frac{a^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{a^2}{∫}_0^{\frac{\mathrm{Ï€}}{2}}\frac{{\sin }^3\mathrm{θ}\cos \mathrm{θ}}{{\left(1-{\left({\displaystyle \frac{v}{c}}\sin \mathrm{θ}\right)}^2\right)}^3}d\mathrm{θ}$ …… (6)

Let,

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

Hence, the equation (6) becomes,

$$\begin{gathered} P=\frac{v{q}^2}{16\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2{\mathrm{γ}}^4}{∫}_0^1\frac{udu}{{\left(1-{\left({\displaystyle \frac{v}{c}}\right)}^{2}u\right)}^3} \\ P=\frac{v{q}^2}{16\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2{\mathrm{γ}}^4}×\frac{{\mathrm{γ}}^4}{2} \\ P=\frac{v{q}^2}{32\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2} \end{gathered}$$

Therefore, the total power passing through the plane is$P=\frac{v{q}^2}{32\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}$ .