(a) Find the fields, and the charge and current distributions, corresponding to

$\mathit{v}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{t}}{{\mathbf{r}}^{\mathbf{2}}}\overline{\mathbf{r}}$

(b) Use the gauge function $\mathit{\lambda }\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{)}\mathbf{(}\mathbf{q}\mathbf{t}\mathbf{/}\mathbf{r}\mathbf{)}$to transform the potentials, and comment on the result.

The function is given as:

Here,${\epsilon }_0$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).

$$\begin{gathered} \frac{\partial A}{\partial t}\frac{\partial }{\partial t}\left(-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{qt}{r^2}\stackrel{⏜}{r}\right) \\ \frac{\partial A}{\partial t}=-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}\stackrel{⏜}{r} \end{gathered}$$

Write the expression for the electric field strength.

$$\begin{gathered} E=-\nabla V-\frac{\partial A}{\partial 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{\partial \mathrm{A}}{\partial \mathrm{t}}=-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\stackrel{⏜}{\mathrm{r}}\mathrm{in}\mathrm{equation}\left(2\right). \\ \mathrm{E}=0-\left(-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}}\right) \\ \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}} \\ \mathrm{Write}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{strength}. \\ \stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}\mathbf{=}}\mathbf{\nabla }\mathbf{\times }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}\mathbf{}\mathbf{}\mathbf{}\dots \dots \left(3\right) \\ \mathrm{Find}\mathrm{the}\mathrm{curl}\mathrm{of}\mathrm{A}. \\ \nabla \times \mathrm{A}=\left[\begin{array}{c}\frac{1}{\mathrm{r}\sin \theta }\left[\frac{\partial }{\partial \mathrm{\theta }}\left(\sin {\mathrm{A}}_{\mathrm{\varphi }}\right)-\frac{\partial {\mathrm{A}}_{\mathrm{\theta }}}{\partial \mathrm{\varphi }}\right]\stackrel{⏜}{\mathrm{r}}+\frac{1}{\mathrm{r}}\left[\frac{1}{\mathrm{sin\theta }}\frac{\partial {\mathrm{A}}_{\mathrm{r}}}{\partial \mathrm{\varphi }}-\frac{\partial }{\partial \mathrm{r}}\left({\mathrm{rA}}_{\mathrm{\varphi }}\right)\right]\stackrel{⏜}{\mathrm{\theta }}\\ \frac{1}{\mathrm{r}}\left[\frac{\partial }{\partial \mathrm{r}}\left({\mathrm{rA}}_{\mathrm{\varphi }}\right)-\frac{\partial {\mathrm{A}}_{\mathrm{r}}}{\partial \mathrm{\varphi }}\right]\stackrel{⏜}{\mathrm{\varphi }}\end{array}\right] \\ \mathrm{Substitute}{\mathrm{A}}_{\mathrm{\varphi }}={\mathrm{A}}_{\mathrm{\theta }}=0\mathrm{and}{\mathrm{A}}_{\mathrm{r}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \end{gathered}$$ …… (2)

$$\begin{gathered} ∆\times A=\left[\begin{array}{cccc}\frac{1}{r\sin \theta }\left[\frac{\partial }{\partial \theta }\left(\sin \theta \left(0\right)\right)-\frac{\partial \left(0\right)}{\partial \varphi }\right]& \stackrel{⏜}{r}+\frac{1}{r}\left[\frac{1}{\sin \theta }\frac{\partial \left({\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}}\frac{qt}{r^2}\right)}{\partial \varphi }\frac{\partial }{\partial \theta }\left(r\times 0\right)\right]\varphi & & \\ \frac{1}{r}\left[\frac{\partial }{\partial r}\left(r\times 0\right)-\frac{\partial \left(-{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}}{\displaystyle \frac{qt}{r^2}}\right)}{\partial \varphi }\right]\stackrel{⏜}{\varphi }& & & \end{array}\right] \\ ∆\times A=\left[\begin{array}{c}0+\frac{1}{2}\left[\frac{1}{\sin \theta }\frac{\partial \left(-{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}}\frac{qt}{r^2}\right)}{\partial \varphi }-0\right]\\ \frac{1}{2}\left[0-\frac{\partial \left(-{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}}\frac{qt}{r^2}\right)}{\partial \varphi }\right]\stackrel{⏜}{\varphi }+\end{array}\stackrel{⏜}{\varphi }+\right] \\ ∆\times A=0 \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}\nabla \times \stackrel{\rightharpoonup }{\mathrm{A}}=0\mathrm{in}\mathrm{equation}\left(3\right). \\ \stackrel{\rightharpoonup }{\mathrm{B}}=0 \\ \mathrm{Write}\mathrm{the}\mathrm{divergence}\mathrm{of}\mathrm{an}\mathrm{electric}\mathrm{field}. \\ \mathbf{\nabla }\mathbf{\times }\mathbf{E}\mathbf{=}\frac{\mathbf{p}}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathrm{Here},p\mathit{}\mathrm{is}\mathrm{the}\mathrm{charge}\mathrm{density} \\ \mathrm{Substitute}\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ \nabla .\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}}\right)=\frac{\mathrm{p}}{{\mathrm{\epsilon }}_0} \\ \mathit{}\mathit{}\mathit{}p={\mathrm{\epsilon }}_0\nabla .\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}}\right) \\ p\mathit{=}\frac{\mathit{1}}{\mathit{4}\pi }q{\delta }^{\mathit{3}}\mathit{\left(}\stackrel{\mathit{\rightharpoonup }}{r}\mathit{\right)}\mathit{4}\mathrm{\pi } \\ \mathrm{p}\mathit{=}q{\delta }^{\mathit{3}}\mathit{\left(}\stackrel{\mathit{\rightharpoonup }}{r}\mathit{\right)} \\ \mathrm{Write}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{current}\mathrm{density}. \\ \mathbf{\nabla }\mathbf{\times }\mathbf{B}\mathbf{=}\mathbf{-}{\mathbf{\mu }}_{\mathbf{0}}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{J}} \end{gathered}$$

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

$$\begin{gathered} 0=-{\mu }_0\stackrel{\rightharpoonup }{J} \\ \stackrel{\rightharpoonup }{J}=0 \\ \mathrm{Therefore},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{is}\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}},\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{is}\stackrel{\rightharpoonup }{\mathrm{B}},\mathrm{the} \\ \mathrm{charge}\mathrm{density}\mathrm{is}\mathrm{p}={\mathrm{q\delta }}^3\left(\stackrel{\rightharpoonup }{\mathrm{r}}\right),\mathrm{and}\mathrm{the}\mathrm{current}\mathrm{density}\mathrm{is}\stackrel{\rightharpoonup }{\mathrm{J}}=0. \end{gathered}$$

(b)

Write the gauge transformation of the vector potential.

$$\begin{gathered} A^{\text{'}}=A+\nabla \lambda \\ \mathrm{Substitute}\mathrm{\lambda }=\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\mathrm{and}\mathrm{A}=-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ {\mathrm{A}}^{\text{'}}=\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}}\right)+\nabla \left(\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\right) \\ {\mathrm{A}}^{\text{'}}=\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{⏜}{\mathrm{r}}\right)+\left(\frac{\mathrm{qt}}{4{\mathrm{\pi \epsilon }}_0}\right)\frac{\partial }{\partial \mathrm{r}}\left(\frac{1}{\mathrm{r}}\right)\stackrel{⏜}{\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{\partial \mathrm{\lambda }}{\partial \mathrm{t}} \\ \mathrm{Substitute}\mathrm{\lambda }=\left(-\frac{1}{4{\mathrm{\pi \epsilon }}_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{\partial \mathrm{\lambda }}{\partial \mathrm{t}}\left(\left(-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\right) \\ {\mathrm{V}}^{\text{'}}=\frac{1}{4{\mathrm{\pi \epsilon }}_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{\pi \epsilon }}_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