Q50P - Introduction to Electrodynamics

Question

The electric potential of some configuration is given by the expression

                                       $\mathbf{V}\left(r\right)\mathbf{=}\mathbf{A}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda r}}}{\mathbf{r}}$

Where $\mathit{A}$ and $\mathit{\lambda }$are constants. Find the electric field$\mathit{E}\left(r\right)$ , the charge density$\mathit{\rho }\left(r\right)$, and the total charge$\left(Q\right)$ .

Step-by-Step Solution

Answer

The electric field is$E=A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{\stackrel{⏜}{r}}{r^2}$ .

The charge density is$A{\epsilon }_0=\left[4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)-\frac{{\mathrm{\lambda }}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{\lambda r}}\right]$ .

 

The total charge $Q$ is 0.

Step 1: Define functions

Consider the given expression.

 

$\mathbf{V}\left(r\right)\mathbf{=}\mathbf{A}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{\lambda r}}}{\mathbf{r}}$                                             …… (1)

 Here, $A$and $\lambda$are constant.

Step 2: Determine electric filed

Write the representation of electric filed is gradient of a scalar potential.

 

$E=-\nabla V$                                                   ……(2)

 

Hence, the electric filed is calculated as follows:

 $$\begin{gathered} E=-\nabla V \\ =-\frac{\partial }{\partial r}\left(A\frac{e^{-\lambda r}}{r}\right) \\ =-A\left[\frac{-r{e}^{-\lambda r}\left(-\lambda \right)-e^{-\lambda r}\left(1\right)}{r^2}\right]\stackrel{⏜}{r} \\ =-A\left[\frac{-r{e}^{-\lambda r}-e^{-\lambda r}}{r^2}\right]\stackrel{⏜}{r} \end{gathered}$$

Solve further.

 

$E=A\left(r\lambda e^{-\lambda r}+e^{-\lambda r}\right)\frac{\stackrel{⏜}{r}}{r^2}$                                                  …… (3)

 $=A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{\stackrel{⏜}{r}}{r^2}$

Thus, the electric field is $E=A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{\stackrel{⏜}{r}}{r^2}$ .

Step 3: Determine charge density

The Gauss’s law in different way[S1] 

 

$$\begin{gathered} \nabla .E=\frac{1}{{\epsilon }_0}\rho \\ \rho ={\epsilon }_0\nabla .E \end{gathered}$$                                                            …… (4)

 

Substitute $E=A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{\stackrel{⏜}{r}}{r^2}$ in equation (4).

 $$\begin{gathered} \rho ={\epsilon }_0\nabla .\left(A{e}^{-\lambda r}\left(1+r\lambda \right)\frac{r}{r^2}\right) \\ ={\epsilon }_0\left[\left(A{e}^{-\lambda r}\left(1+\lambda \right)\right)\left(\nabla .\frac{\stackrel{⏜}{r}}{r^2}\right)+\left(\frac{\stackrel{⏜}{r}}{r^2}\right).\nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)\right] \end{gathered}$$

But $\nabla .\frac{\stackrel{⏜}{r}}{r^2}=4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)$,

Therefore,

 $$\begin{gathered} \rho ={\epsilon }_0\left[\left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)\left(4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)\right)+\left(\frac{\stackrel{⏜}{r}}{r^2}\right)\nabla .\left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)\right] \\ ={\epsilon }_0\left[A4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)+\left(\frac{\stackrel{⏜}{\mathrm{r}}}{{\mathrm{r}}^2}\right)\nabla .\left({\mathrm{Ae}}^{-\mathrm{\lambda r}}\left(1+\mathrm{r\lambda }\right)\right)\right] \end{gathered}$$

 

Since$\left(e^{-\lambda r}\left(1+r\lambda \right)\right)\left(4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)\right)=4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)$ ,

$$\begin{gathered} \nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)=\stackrel{⏜}{r}\frac{\partial }{\partial r}\left[A{e}^{-\lambda r}\left(1+r\lambda \right)\right] \\ =\stackrel{⏜}{r}A\frac{\partial }{\partial r}\left[e^{-\lambda r}\left(1+r\lambda \right)\right] \\ =\stackrel{⏜}{r}A\left[\left(1+r\lambda \right)\frac{\partial }{\partial r}\left(e^{-\lambda r}\right)+e^{-\lambda r}\left(1+r\lambda \right)\right] \\ =\stackrel{⏜}{r}A\left[\left(1+r\lambda \right)\left(-\lambda \right)e^{-\lambda r}+e^{-\lambda r}\left(\lambda \right)\right] \end{gathered}$$

Solve further.

$$\begin{gathered} \nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)=\stackrel{⏜}{r}A\left[\lambda e^{-\lambda r}-\lambda \left(1+\lambda r\right)e^{-\lambda r}\right] \\ =\stackrel{⏜}{r}A\left[\lambda e^{-\lambda r}-\lambda e^{-\lambda r}\left(1+\lambda r\right)\right] \\ =\stackrel{⏜}{r}A\left[\lambda e^{-\lambda r}\left(1-\left(1+\lambda r\right)\right)\right] \\ =\stackrel{⏜}{r}A\left[\lambda e^{-\lambda r}-\left(1-1-\lambda r\right)\right] \end{gathered}$$

Solve further.

$$\begin{gathered} \nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)=\stackrel{⏜}{r}A\left[\lambda e^{-\lambda r}\left(-\lambda r\right)\right] \\ =\stackrel{⏜}{r}A\left[-{\lambda }^2{e}^{-\lambda r}\right] \end{gathered}$$

Multiply by$\left(\frac{\stackrel{⏜}{r}}{r^2}\right)$ on both sides.

 $$\begin{gathered} \left(\frac{\stackrel{⏜}{r}}{r^2}\right).\nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)=\left(\frac{\stackrel{⏜}{r}}{r^2}\right).\nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right) \\ =\left(\frac{\stackrel{⏜}{r}}{r^2}\right).A\left(\lambda e^{-\lambda r}+\left(1+r\lambda \right)e^{-\lambda r}\left(-\lambda \right)\right) \\ =\left(\frac{\stackrel{⏜}{r}}{r^2}\right).\left[A\left(\lambda e^{-\lambda r}-\lambda e^{-\lambda r}-r{\lambda }^2{e}^{-\lambda r}\right)\stackrel{⏜}{r}\right] \\ =\frac{1}{r^2}A\left[-{\lambda }^{2}r{e}^{-\lambda r}\right] \end{gathered}$$

Thus, 

 $\left(\frac{\stackrel{⏜}{r}}{r^2}\right).\nabla \left(A{e}^{-\lambda r}\left(1+r\lambda \right)\right)=-\frac{A{\lambda }^2}{r}{e}^{-\lambda r}$

Hence, 

$$\begin{gathered} \rho ={\epsilon }_0\left[A4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)+\left(\frac{\stackrel{⏜}{\mathrm{r}}}{{\mathrm{r}}^2}\right)\nabla .\left({\mathrm{Ae}}^{-\mathrm{\lambda r}}\left(1+\mathrm{r\lambda }\right)\right)\right] \\ ={\epsilon }_0\left[A4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)-\frac{A{\lambda }^2}{r}{e}^{-\lambda r}\right] \\ =A{\epsilon }_0\left[4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)-\frac{{\mathrm{\lambda }}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{\lambda r}}\right] \end{gathered}$$

Thus, the charge density is $A{\epsilon }_0\left[4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)-\frac{{\mathrm{\lambda }}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{\lambda r}}\right]$.

Step 4: Determine the total charge

Write the expression for the total charge.

 $$\begin{gathered} Q=\int \rho d\tau \\ =\int A{\epsilon }_0\left[4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)-\frac{{\mathrm{\lambda }}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{\lambda r}}\right]d\tau \\ =\int A{\epsilon }_{0}4{\mathrm{\pi \delta }}^3\left(\mathrm{r}\right)d\tau -\int A{\epsilon }_0\frac{{\mathrm{\lambda }}^2}{\mathrm{r}}{\mathrm{e}}^{-\mathrm{\lambda r}}\mathrm{d\tau } \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}\int {\mathrm{\delta }}^3\left(\mathrm{r}\right)\mathrm{d\tau }-{\mathrm{A\epsilon }}_0{\mathrm{\lambda }}^2\int \frac{{\mathrm{e}}^{-\mathrm{\lambda r}}}{\mathrm{r}}\mathrm{d\tau } \end{gathered}$$

 

Simplify further,

$$\begin{gathered} Q=4{\mathrm{\pi \epsilon }}_0\mathrm{A}\left(1\right)-{\mathrm{A\epsilon }}_0{\mathrm{\lambda }}^2\int \frac{{\mathrm{e}}^{-\mathrm{\lambda r}}}{\mathrm{r}}\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right) \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2\int {\mathrm{re}}^{-\mathrm{\lambda r}}\mathrm{dr} \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2{\int }_0^{\infty }{\mathrm{re}}^{-\mathrm{\lambda r}}\mathrm{dr} \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2{\left[-\frac{{\mathrm{re}}^{-\mathrm{\lambda r}}}{\mathrm{\lambda }}-\frac{{\mathrm{e}}^{-\mathrm{\lambda r}}}{{\mathrm{\lambda }}^2}\right]}_0^{\infty } \end{gathered}$$

 Also,

$$\begin{gathered} Q=4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0{\mathrm{\lambda }}^2\left(\frac{1}{{\mathrm{\lambda }}^2}\right) \\ =4{\mathrm{\pi \epsilon }}_0\mathrm{A}-4{\mathrm{\pi A\epsilon }}_0 \\ =0 \end{gathered}$$

Thus, the total charge Q is 0.