91影视

The function is given as:

Here,${蔚}_0$is the permittivity of free space, q is the charge, and r is the distance between two charged particles.

(a)

Take the partial derivative of equation (1).

$$\begin{gathered} \frac{鈭�A}{鈭�t}\frac{鈭�}{鈭�t}\left(-\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{qt}{r^2}\stackrel{铃�}{r}\right) \\ \frac{鈭�A}{鈭�t}=-\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{q}{r^2}\stackrel{铃�}{r} \end{gathered}$$

Write the expression for the electric field strength.

$$\begin{gathered} E=-鈭�V-\frac{鈭�A}{鈭�t}......\left(2\right) \\ \mathrm{Here},\mathrm{V}\mathrm{and}\mathrm{A}\mathrm{are}\mathrm{the}\mathrm{scalar}\mathrm{and}\mathrm{vector}\mathrm{potential}. \\ \mathrm{Substitute}\mathrm{v}=0\mathrm{and}\frac{鈭�\mathrm{A}}{鈭�\mathrm{t}}=-\frac{1}{4{\mathrm{蟺蔚}}_0}\stackrel{铃�}{\mathrm{r}}\mathrm{in}\mathrm{equation}\left(2\right). \\ \mathrm{E}=0-\left(-\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}}\right) \\ \mathrm{E}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}} \\ \mathrm{Write}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{strength}. \\ \stackrel{\mathbf{鈬赌}}{\mathbf{B}\mathbf{=}}\mathbf{鈭�}\mathbf{脳}\stackrel{\mathbf{鈬赌}}{\mathbf{A}}\mathbf{}\mathbf{}\mathbf{}鈥�鈥�\left(3\right) \\ \mathrm{Find}\mathrm{the}\mathrm{curl}\mathrm{of}\mathrm{A}. \\ 鈭�脳\mathrm{A}=\left[\begin{array}{c}\frac{1}{\mathrm{r}\sin 胃}\left[\frac{鈭�}{鈭�\mathrm{胃}}\left(\sin {\mathrm{A}}_{\mathrm{蠒}}\right)-\frac{鈭�{\mathrm{A}}_{\mathrm{胃}}}{鈭�\mathrm{蠒}}\right]\stackrel{铃�}{\mathrm{r}}+\frac{1}{\mathrm{r}}\left[\frac{1}{\mathrm{蝉颈苍胃}}\frac{鈭�{\mathrm{A}}_{\mathrm{r}}}{鈭�\mathrm{蠒}}-\frac{鈭�}{鈭�\mathrm{r}}\left({\mathrm{rA}}_{\mathrm{蠒}}\right)\right]\stackrel{铃�}{\mathrm{胃}}\\ \frac{1}{\mathrm{r}}\left[\frac{鈭�}{鈭�\mathrm{r}}\left({\mathrm{rA}}_{\mathrm{蠒}}\right)-\frac{鈭�{\mathrm{A}}_{\mathrm{r}}}{鈭�\mathrm{蠒}}\right]\stackrel{铃�}{\mathrm{蠒}}\end{array}\right] \\ \mathrm{Substitute}{\mathrm{A}}_{\mathrm{蠒}}={\mathrm{A}}_{\mathrm{胃}}=0\mathrm{and}{\mathrm{A}}_{\mathrm{r}}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \end{gathered}$$ 鈥︹�� (2)

$$\begin{gathered} 鈭�脳A=\left[\begin{array}{cccc}\frac{1}{r\sin 胃}\left[\frac{鈭�}{鈭�胃}\left(\sin 胃\left(0\right)\right)-\frac{鈭�\left(0\right)}{鈭�蠒}\right]& \stackrel{铃�}{r}+\frac{1}{r}\left[\frac{1}{\sin 胃}\frac{鈭�\left({\displaystyle \frac{1}{4{\mathrm{蟺蔚}}_0}}\frac{qt}{r^2}\right)}{鈭�蠒}\frac{鈭�}{鈭�胃}\left(r脳0\right)\right]蠒& & \\ \frac{1}{r}\left[\frac{鈭�}{鈭�r}\left(r脳0\right)-\frac{鈭�\left(-{\displaystyle \frac{1}{4{\mathrm{蟺蔚}}_0}}{\displaystyle \frac{qt}{r^2}}\right)}{鈭�蠒}\right]\stackrel{铃�}{蠒}& & & \end{array}\right] \\ 鈭�脳A=\left[\begin{array}{c}0+\frac{1}{2}\left[\frac{1}{\sin 胃}\frac{鈭�\left(-{\displaystyle \frac{1}{4{\mathrm{蟺蔚}}_0}}\frac{qt}{r^2}\right)}{鈭�蠒}-0\right]\\ \frac{1}{2}\left[0-\frac{鈭�\left(-{\displaystyle \frac{1}{4{\mathrm{蟺蔚}}_0}}\frac{qt}{r^2}\right)}{鈭�蠒}\right]\stackrel{铃�}{蠒}+\end{array}\stackrel{铃�}{蠒}+\right] \\ 鈭�脳A=0 \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}鈭�脳\stackrel{鈬赌}{\mathrm{A}}=0\mathrm{in}\mathrm{equation}\left(3\right). \\ \stackrel{鈬赌}{\mathrm{B}}=0 \\ \mathrm{Write}\mathrm{the}\mathrm{divergence}\mathrm{of}\mathrm{an}\mathrm{electric}\mathrm{field}. \\ \mathbf{鈭�}\mathbf{脳}\mathbf{E}\mathbf{=}\frac{\mathbf{p}}{{\mathbf{蔚}}_{\mathbf{0}}} \\ \mathrm{Here},p\mathit{}\mathrm{is}\mathrm{the}\mathrm{charge}\mathrm{density} \\ \mathrm{Substitute}\mathrm{E}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ 鈭�.\left(\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}}\right)=\frac{\mathrm{p}}{{\mathrm{蔚}}_0} \\ \mathit{}\mathit{}\mathit{}p={\mathrm{蔚}}_0鈭�.\left(\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}}\right) \\ p\mathit{=}\frac{\mathit{1}}{\mathit{4}蟺}q{未}^{\mathit{3}}\mathit{\left(}\stackrel{\mathit{鈬赌}}{r}\mathit{\right)}\mathit{4}\mathrm{蟺} \\ \mathrm{p}\mathit{=}q{未}^{\mathit{3}}\mathit{\left(}\stackrel{\mathit{鈬赌}}{r}\mathit{\right)} \\ \mathrm{Write}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{current}\mathrm{density}. \\ \mathbf{鈭�}\mathbf{脳}\mathbf{B}\mathbf{=}\mathbf{-}{\mathbf{渭}}_{\mathbf{0}}\stackrel{\mathbf{鈬赌}}{\mathbf{J}} \end{gathered}$$

Here, B is the magnetic field, and J is the current density.

$$\begin{gathered} 0=-{渭}_0\stackrel{鈬赌}{J} \\ \stackrel{鈬赌}{J}=0 \\ \mathrm{Therefore},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{is}\mathrm{E}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}},\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{is}\stackrel{鈬赌}{\mathrm{B}},\mathrm{the} \\ \mathrm{charge}\mathrm{density}\mathrm{is}\mathrm{p}={\mathrm{辩未}}^3\left(\stackrel{鈬赌}{\mathrm{r}}\right),\mathrm{and}\mathrm{the}\mathrm{current}\mathrm{density}\mathrm{is}\stackrel{鈬赌}{\mathrm{J}}=0. \end{gathered}$$

(b)

Write the gauge transformation of the vector potential.

$$\begin{gathered} A^{\text{'}}=A+鈭�位 \\ \mathrm{Substitute}\mathrm{位}=\left(\frac{1}{4{\mathrm{蟺蔚}}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\mathrm{and}\mathrm{A}=-\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ {\mathrm{A}}^{\text{'}}=\left(\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}}\right)+鈭�\left(\left(\frac{1}{4{\mathrm{蟺蔚}}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\right) \\ {\mathrm{A}}^{\text{'}}=\left(\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{铃�}{\mathrm{r}}\right)+\left(\frac{\mathrm{qt}}{4{\mathrm{蟺蔚}}_0}\right)\frac{鈭�}{鈭�\mathrm{r}}\left(\frac{1}{\mathrm{r}}\right)\stackrel{铃�}{\mathrm{r}} \\ {\mathrm{A}}^{\text{'}}=0 \\ \mathrm{Write}\mathrm{the}\mathrm{gauge}\mathrm{transformation}\mathrm{of}\mathrm{the}\mathrm{scalar}\mathrm{potential}. \\ {\mathrm{V}}^{\text{'}}=\mathrm{V}-\frac{鈭�\mathrm{位}}{鈭�\mathrm{t}} \\ \mathrm{Substitute}\mathrm{位}=\left(-\frac{1}{4{\mathrm{蟺蔚}}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\mathrm{and}\mathrm{V}=0\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ {\mathrm{V}}^{\text{'}}=0-\frac{鈭�\mathrm{位}}{鈭�\mathrm{t}}\left(\left(-\frac{1}{4{\mathrm{蟺蔚}}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\right) \\ {\mathrm{V}}^{\text{'}}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{\mathrm{r}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{gauge}\mathrm{transform}\mathrm{of}\mathrm{the}\mathrm{vector}\mathrm{potential}{\mathrm{isA}}^{\text{'}}=0,\mathrm{and}\mathrm{the}\mathrm{gauge} \\ \mathrm{transform}\mathrm{of}\mathrm{the}\mathrm{scalar}\mathrm{potential}\mathrm{is}{\mathrm{V}}^{\text{'}}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\mathrm{q}}{\mathrm{r}}. \end{gathered}$$

Write the gauge transformation of the scalar potential.

Therefore, the gauge transform of the vector potential is, and the gauge transform of the scalar potential is