91影视

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

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}}\frac{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{{\left(1-v^{2}si{n}^2{\displaystyle \frac{胃}{c^2}}\right)}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}\frac{\hat{\mathbf{R}}}{{\mathbf{R}}^{\mathbf{2}}}$ 鈥︹�� (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$胃$ is the angle between R and v .

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

$蠒=鈭�E.da$

Here,$蠒$is the flux associated with the point charge moving at constant velocity.

Use the surface area of the sphere as$da=R^2\sin 胃d胃d蠒$.

Consider the expression and determine the flux integral as,

$$\begin{gathered} 蠒=鈭�\frac{q}{4{\mathrm{蟺蔚}}_0}\frac{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{R^2{\left(1-v^2{\sin }^2{\displaystyle \frac{胃}{c^2}}\right)}^{{\displaystyle \frac{3}{2}}}}{R}^2\sin 胃d胃d蠒 \\ =\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\mathrm{蟺蔚}}_0}鈭�\frac{\sin 胃d胃d蠒}{{\left(1-v^2{\sin }^2{c}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\mathrm{蟺蔚}}_0}2\mathrm{蟺}{鈭�}_0^{\mathrm{蟺}}\frac{\mathrm{蝉颈苍胃诲胃诲蠒}}{{\left(1-{\mathrm{v}}^2{\sin }^2{\mathrm{c}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

Consider the substitution.

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

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

$$\begin{gathered} 蠒=\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{蔚}_0}2\mathrm{蟺}{鈭�}_0^{\mathrm{蟺}}\frac{\mathrm{du}}{{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\left(1-{\mathrm{u}}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{\mathrm{q}\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)}{2{\mathrm{蔚}}_0}{鈭�}_{-1}^1\frac{\mathrm{du}}{{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\left(1-{\mathrm{u}}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{\mathrm{q}\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)}{2{\mathrm{蔚}}_0}\frac{2}{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)} \\ =\frac{\mathrm{q}}{{\mathrm{蔚}}_0} \end{gathered}$$

Therefore, the flux related to point charge that is moving with constant velocity is$蠁=\frac{q}{{蔚}_0}$ .