91Ó°ÊÓ

The function is given as:

Here,${\mathrm{Î\mu }}_0$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).

$$\begin{gathered} \frac{∂A}{∂t}\frac{∂}{∂t}\left(-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{qt}{r^2}\stackrel{Áåœ}{r}\right) \\ \frac{∂A}{∂t}=-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q}{r^2}\stackrel{Áåœ}{r} \end{gathered}$$

Write the expression for the electric field strength.

$$\begin{gathered} E=-∇V-\frac{∂A}{∂t}......\left(2\right) \\ \mathrm{Here},\mathrm{V}\mathrm{and}\mathrm{A}\mathrm{are}\mathrm{the}\mathrm{scalar}\mathrm{and}\mathrm{vector}\mathrm{potential}. \\ \mathrm{Substitute}\mathrm{v}=0\mathrm{and}\frac{∂\mathrm{A}}{∂\mathrm{t}}=-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\stackrel{Áåœ}{\mathrm{r}}\mathrm{in}\mathrm{equation}\left(2\right). \\ \mathrm{E}=0-\left(-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}}\right) \\ \mathrm{E}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}} \\ \mathrm{Write}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{strength}. \\ \stackrel{\mathbf{â‡\P Ä}}{\mathbf{B}\mathbf{=}}\mathbf{∇}\mathbf{×}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{A}}\mathbf{}\mathbf{}\mathbf{}……\left(3\right) \\ \mathrm{Find}\mathrm{the}\mathrm{curl}\mathrm{of}\mathrm{A}. \\ ∇×\mathrm{A}=\left[\begin{array}{c}\frac{1}{\mathrm{r}\sin \mathrm{θ}}\left[\frac{∂}{∂\mathrm{θ}}\left(\sin {\mathrm{A}}_{\mathrm{Ï•}}\right)-\frac{∂{\mathrm{A}}_{\mathrm{θ}}}{∂\mathrm{Ï•}}\right]\stackrel{Áåœ}{\mathrm{r}}+\frac{1}{\mathrm{r}}\left[\frac{1}{\mathrm{²õ¾\pm ²Ôθ}}\frac{∂{\mathrm{A}}_{\mathrm{r}}}{∂\mathrm{Ï•}}-\frac{∂}{∂\mathrm{r}}\left({\mathrm{rA}}_{\mathrm{Ï•}}\right)\right]\stackrel{Áåœ}{\mathrm{θ}}\\ \frac{1}{\mathrm{r}}\left[\frac{∂}{∂\mathrm{r}}\left({\mathrm{rA}}_{\mathrm{Ï•}}\right)-\frac{∂{\mathrm{A}}_{\mathrm{r}}}{∂\mathrm{Ï•}}\right]\stackrel{Áåœ}{\mathrm{Ï•}}\end{array}\right] \\ \mathrm{Substitute}{\mathrm{A}}_{\mathrm{Ï•}}={\mathrm{A}}_{\mathrm{θ}}=0\mathrm{and}{\mathrm{A}}_{\mathrm{r}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \end{gathered}$$ …… (2)

$$\begin{gathered} ∆×A=\left[\begin{array}{cccc}\frac{1}{r\sin \mathrm{θ}}\left[\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}\left(0\right)\right)-\frac{∂\left(0\right)}{∂\mathrm{Ï•}}\right]& \stackrel{Áåœ}{r}+\frac{1}{r}\left[\frac{1}{\sin \mathrm{θ}}\frac{∂\left({\displaystyle \frac{1}{4{\mathrm{Ï€Î\mu }}_0}}\frac{qt}{r^2}\right)}{∂\mathrm{Ï•}}\frac{∂}{∂\mathrm{θ}}\left(r×0\right)\right]\mathrm{Ï•}& & \\ \frac{1}{r}\left[\frac{∂}{∂r}\left(r×0\right)-\frac{∂\left(-{\displaystyle \frac{1}{4{\mathrm{Ï€Î\mu }}_0}}{\displaystyle \frac{qt}{r^2}}\right)}{∂\mathrm{Ï•}}\right]\stackrel{Áåœ}{\mathrm{Ï•}}& & & \end{array}\right] \\ ∆×A=\left[\begin{array}{c}0+\frac{1}{2}\left[\frac{1}{\sin \mathrm{θ}}\frac{∂\left(-{\displaystyle \frac{1}{4{\mathrm{Ï€Î\mu }}_0}}\frac{qt}{r^2}\right)}{∂\mathrm{Ï•}}-0\right]\\ \frac{1}{2}\left[0-\frac{∂\left(-{\displaystyle \frac{1}{4{\mathrm{Ï€Î\mu }}_0}}\frac{qt}{r^2}\right)}{∂\mathrm{Ï•}}\right]\stackrel{Áåœ}{\mathrm{Ï•}}+\end{array}\stackrel{Áåœ}{\mathrm{Ï•}}+\right] \\ ∆×A=0 \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}∇×\stackrel{â‡\P Ä}{\mathrm{A}}=0\mathrm{in}\mathrm{equation}\left(3\right). \\ \stackrel{â‡\P Ä}{\mathrm{B}}=0 \\ \mathrm{Write}\mathrm{the}\mathrm{divergence}\mathrm{of}\mathrm{an}\mathrm{electric}\mathrm{field}. \\ \mathbf{∇}\mathbf{×}\mathbf{E}\mathbf{=}\frac{\mathbf{p}}{{\mathbf{Î\mu }}_{\mathbf{0}}} \\ \mathrm{Here},p\mathit{}\mathrm{is}\mathrm{the}\mathrm{charge}\mathrm{density} \\ \mathrm{Substitute}\mathrm{E}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ ∇.\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}}\right)=\frac{\mathrm{p}}{{\mathrm{Î\mu }}_0} \\ \mathit{}\mathit{}\mathit{}p={\mathrm{Î\mu }}_0∇.\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}}\right) \\ p\mathit{=}\frac{\mathit{1}}{\mathit{4}\mathrm{Ï€}}q{\mathrm{δ}}^{\mathit{3}}\mathit{\left(}\stackrel{\mathit{â‡\P Ä}}{r}\mathit{\right)}\mathit{4}\mathrm{Ï€} \\ \mathrm{p}\mathit{=}q{\mathrm{δ}}^{\mathit{3}}\mathit{\left(}\stackrel{\mathit{â‡\P Ä}}{r}\mathit{\right)} \\ \mathrm{Write}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{current}\mathrm{density}. \\ \mathbf{∇}\mathbf{×}\mathbf{B}\mathbf{=}\mathbf{-}{\mathbf{μ}}_{\mathbf{0}}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{J}} \end{gathered}$$

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

$$\begin{gathered} 0=-{\mathrm{μ}}_0\stackrel{â‡\P Ä}{J} \\ \stackrel{â‡\P Ä}{J}=0 \\ \mathrm{Therefore},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{is}\mathrm{E}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}},\mathrm{the}\mathrm{magnetic}\mathrm{field}\mathrm{is}\stackrel{â‡\P Ä}{\mathrm{B}},\mathrm{the} \\ \mathrm{charge}\mathrm{density}\mathrm{is}\mathrm{p}={\mathrm{\pm çδ}}^3\left(\stackrel{â‡\P Ä}{\mathrm{r}}\right),\mathrm{and}\mathrm{the}\mathrm{current}\mathrm{density}\mathrm{is}\stackrel{â‡\P Ä}{\mathrm{J}}=0. \end{gathered}$$

(b)

Write the gauge transformation of the vector potential.

$$\begin{gathered} A^{\text{'}}=A+∇\mathrm{λ} \\ \mathrm{Substitute}\mathrm{λ}=\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\mathrm{and}\mathrm{A}=-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ {\mathrm{A}}^{\text{'}}=\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}}\right)+∇\left(\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\right) \\ {\mathrm{A}}^{\text{'}}=\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{qt}}{{\mathrm{r}}^2}\stackrel{Áåœ}{\mathrm{r}}\right)+\left(\frac{\mathrm{qt}}{4{\mathrm{Ï€Î\mu }}_0}\right)\frac{∂}{∂\mathrm{r}}\left(\frac{1}{\mathrm{r}}\right)\stackrel{Áåœ}{\mathrm{r}} \\ {\mathrm{A}}^{\text{'}}=0 \\ \mathrm{Write}\mathrm{the}\mathrm{gauge}\mathrm{transformation}\mathrm{of}\mathrm{the}\mathrm{scalar}\mathrm{potential}. \\ {\mathrm{V}}^{\text{'}}=\mathrm{V}-\frac{∂\mathrm{λ}}{∂\mathrm{t}} \\ \mathrm{Substitute}\mathrm{λ}=\left(-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\mathrm{and}\mathrm{V}=0\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression}. \\ {\mathrm{V}}^{\text{'}}=0-\frac{∂\mathrm{λ}}{∂\mathrm{t}}\left(\left(-\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\left(\frac{\mathrm{qt}}{\mathrm{r}}\right)\right) \\ {\mathrm{V}}^{\text{'}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{\mathrm{r}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{gauge}\mathrm{transform}\mathrm{of}\mathrm{the}\mathrm{vector}\mathrm{potential}{\mathrm{isA}}^{\text{'}}=0,\mathrm{and}\mathrm{the}\mathrm{gauge} \\ \mathrm{transform}\mathrm{of}\mathrm{the}\mathrm{scalar}\mathrm{potential}\mathrm{is}{\mathrm{V}}^{\text{'}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{q}}{\mathrm{r}}. \end{gathered}$$

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