Question

For a point charge moving at constant velocity, calculate the flux integral$\mathbf{\oint }\mathit{E}\mathbf{.}\mathit{d}\mathit{a}$ (using Eq. 10.75), over the surface of a sphere centered at the present location of the charge.

Short answer

The net flux associated with the point charge moving with constant velocity is$\phi =\frac{q}{{\epsilon }_0}$

Step-by-step solution

Expression for the electric field of a point charge moving with constant velocity:

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

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{{\left(1-v^{2}si{n}^2{\displaystyle \frac{\theta }{c^2}}\right)}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}\frac{\hat{\mathbf{R}}}{{\mathbf{R}}^{\mathbf{2}}}$ …… (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$\theta$ is the angle between R and v .

Determine the flux associated with the point charge moving with constant velocity:

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

$\varphi =\int E.da$

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

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

Consider the expression and determine the flux integral as,

$$\begin{gathered} \varphi =\int \frac{q}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{R^2{\left(1-v^2{\sin }^2{\displaystyle \frac{\theta }{c^2}}\right)}^{{\displaystyle \frac{3}{2}}}}{R}^2\sin \theta d\theta d\varphi \\ =\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\mathrm{\pi \epsilon }}_0}\int \frac{\sin \theta d\theta d\varphi }{{\left(1-v^2{\sin }^2{c}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\mathrm{\pi \epsilon }}_0}2\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\mathrm{sin\theta d\theta d\varphi }}{{\left(1-{\mathrm{v}}^2{\sin }^2{\mathrm{c}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

Consider the substitution.

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

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

$$\begin{gathered} \varphi =\frac{q\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}{4{\epsilon }_0}2\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\mathrm{du}}{{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\left(1-{\mathrm{u}}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{\mathrm{q}\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)}{2{\mathrm{\epsilon }}_0}{\int }_{-1}^1\frac{\mathrm{du}}{{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\left(1-{\mathrm{u}}^2\right)\right)}^{{\displaystyle \frac{3}{2}}}} \\ =\frac{\mathrm{q}\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)}{2{\mathrm{\epsilon }}_0}\frac{2}{\left(1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}\right)} \\ =\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0} \end{gathered}$$

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