91影视

Write the expression for the scalar potential of a point charge.

$\mathit{V}\left(r,t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{R}\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{胃}}}$

Here, q is the point charge, v is the velocity, c is the speed of light, and R is the radius.

Write the expression for the scalar potential in terms of retarded time.

$\mathit{V}\left(r,t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{c}}{\sqrt{{\left(c^{2}t-r脳v\right)}^{\mathbf{2}}\mathbf{+}\left(c^2-v^2\right)\left(r^2-c^2{t}^2\right)}}$ 鈥︹�� (1)

Solve the denominator value ${\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\left(r^2-c^2{t}^2\right)$from equation (1).

${\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\left(r^2-c^2{t}^2\right)=\left[\begin{array}{c}{c}^4{t}^2+{\left(r路v\right)}^2-2c^{2}t\left(r路v\right)+\\ c^2{r}^2-c^4{t}^2-v^2{r}^2+v^2{c}^2{t}^2\end{array}\right]$

${\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\left(r^2-c^2{t}^2\right)=\left[\begin{array}{c}{\left(r路v\right)}^2+\left(c^2-v^2\right)r^2+\\ c^2{\left(vt\right)}^2-2c^2\left(r路vt\right)\end{array}\right]$ .......(2)

Using a given problem, it is given that:

$$\begin{gathered} R=r-vt \\ vt=r-R \end{gathered}$$

Hence, equation (2) becomes,

$$\begin{gathered} \left[\begin{array}{c}{\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\\ \left(r^2-c^2{t}^2\right)\end{array}\right]=\left[\begin{array}{c}{\left(r路v\right)}^2+\left(c^2-v^2\right)r^2+\\ c^2{\left(r-R\right)}^2-2c^2\left(r路\left(r-R\right)\right)\end{array}\right] \\ \left[\begin{array}{c}{\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\\ \left(r^2-c^2{t}^2\right)\end{array}\right]=\left[\begin{array}{c}{\left(r路v\right)}^2+c^2{r}^2-v^2{r}^2+c^2\left(r^2+R^2-2rR\right)-\\ 2c^2\left(r^2-rR\right)\end{array}\right] \\ \left[\begin{array}{c}{\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\\ \left(r^2-c^2{t}^2\right)\end{array}\right]=\left[\begin{array}{c}{\left(r路v\right)}^2+2c^2{r}^2-v^2{r}^2+c^2{R}^2-\\ 2c^{2}rR-2c^2{r}^2+2c^{2}rR\end{array}\right] \\ \left[\begin{array}{c}{\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\\ \left(r^2-c^2{t}^2\right)\end{array}\right]={\left(r路v\right)}^2-v^2{r}^2+c^2{R}^2.....\left(3\right) \end{gathered}$$

Solve the value${\left(r路v\right)}^2-r^2{v}^2$from the above expression.

$$\begin{gathered} {\left(r路v\right)}^2-r^2{v}^2={\left(\left(R+vt\right)v\right)}^2-{\left(R+vt\right)}^2{v}^2 \\ {\left(r路v\right)}^2-r^2{v}^2={\left(R路v\right)}^2+v^4{t}^2+2\left[\left(R路v\right)v^{2}t\right]-{\left(R^2{v}^2\right)}^2-v^4{t}^2-2\left(R路v\right)t{v}^2 \\ {\left(r路v\right)}^2-r^2{v}^2={\left(R路v\right)}^2-R^2{v}^2 \\ {\left(r路v\right)}^2-r^2{v}^2=R^2{v}^2{\cos }^2胃-R^2{v}^2 \end{gathered}$$

On further solving,

role="math" localid="1653889596337" $$\begin{gathered} {\left(r路v\right)}^2-r^2{v}^2=R^2{v}^2\left({\cos }^2胃-1\right) \\ {\left(r路v\right)}^2-r^2{v}^2=-R^2{v}^2{\sin }^2胃.....\left(4\right) \end{gathered}$$

Substitute the value of equation (4) in equation (3).

${\left(c^{2}t-r路v\right)}^2+\left(c^2-v^2\right)\left(r^2-c^2{t}^2\right)=-R^2{v}^2{\sin }^2胃+c^2{R}^2$

Substitute the above value in equation (1).

$$\begin{gathered} V\left(r,t\right)=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{qc}{\sqrt{-R^2{v}^2{\sin }^2胃+c^2{R}^2}} \\ V\left(r,t\right)=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{qc}{cR\sqrt{1-{\displaystyle \frac{v^2}{c^2}}{\sin }^2胃}} \end{gathered}$$

Therefore, the scalar potential of a point charge moving with constant velocity is derived as $V\left(r,t\right)=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{qc}{cR\sqrt{1-{\displaystyle \frac{v^2}{c^2}}{\sin }^2胃}}$.