91Ó°ÊÓ

Write the expression for the electric field of a point charge moving with constant velocity is,

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\frac{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{{\left(1-v^{2}si{n}^2{\displaystyle \frac{\mathrm{θ}}{c^2}}\right)}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}\frac{\hat{\mathbf{R}}}{{\mathbf{R}}^{\mathbf{2}}}$ …… (1)

Here, E is the electric field of a point charge moving, q is the point charge, v is the velocity, c is the speed of light and$\mathrm{θ}$ is the angle between R and v .

Write the equation for the flux associated with the point charge moving at constant velocity.

$\mathrm{ϕ}=∫E.da$

Here,$\mathrm{Ï•}$is the flux associated with the point charge moving at constant velocity.

Use the surface area of the sphere as$da=R^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}$.

Consider the expression and determine the flux integral as,

$$\begin{gathered} \mathrm{Ï•}=∫\frac{q}{4{\mathrm{Ï€Î\mu }}_0}\frac{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{R^2{\left(1-v^2{\sin }^2{\displaystyle \frac{\mathrm{θ}}{c^2}}\right)}^{{\displaystyle \frac{3}{2}}}}{R}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•} \\ =\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\mathrm{Ï€Î\mu }}_0}∫\frac{\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}}{{\left(1-v^2{\sin }^2{c}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\mathrm{Ï€Î\mu }}_0}2\mathrm{Ï€}{∫}_0^{\mathrm{Ï€}}\frac{\mathrm{²õ¾\pm ²Ôθ»åθ»åÏ•}}{{\left(1-{\mathrm{v}}^2{\sin }^2{\mathrm{c}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

Consider the substitution.

$$\begin{gathered} u=\cos \mathrm{θ} \\ du=-\sin \mathrm{θ}d\mathrm{θ} \\ 1-u^2={\sin }^2\mathrm{θ} \end{gathered}$$

Apply the values from the substitution method and then apply integration.

$$\begin{gathered} \mathrm{Ï•}=\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\mathrm{Î\mu }}_0}2\mathrm{Ï€}{∫}_0^{\mathrm{Ï€}}\frac{\mathrm{du}}{{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\left(1-{\mathrm{u}}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{\mathrm{q}\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)}{2{\mathrm{Î\mu }}_0}{∫}_{-1}^1\frac{\mathrm{du}}{{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\left(1-{\mathrm{u}}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{\mathrm{q}\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)}{2{\mathrm{Î\mu }}_0}\frac{2}{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)} \\ =\frac{\mathrm{q}}{{\mathrm{Î\mu }}_0} \end{gathered}$$

Therefore, the flux related to point charge that is moving with constant velocity is$\mathrm{φ}=\frac{q}{{\mathrm{Î\mu }}_0}$ .