91Ó°ÊÓ

At any boundary, the normal component of the electric field is discontinuous by a certain $\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}$amount. Write the expression for it.

${\mathit{E}}_{\mathbf{a}\mathbf{b}\mathbf{o}\mathbf{v}\mathbf{e}}\mathbf{-}{\mathit{E}}_{\mathbf{b}\mathbf{e}\mathbf{l}\mathbf{o}\mathbf{w}}\mathbf{=}\frac{\mathbf{σ}}{{\mathbf{Î\mu }}_{\mathbf{0}}}\hat{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}......\left(1\right)$

Here, $\mathrm{σ}$surface charge density and ${\mathrm{Î\mu }}_0$permittivity for free space.

a)

Refer the figure of Ex 2.5,

Write the expression for the electric field above the surface.

$E_{above}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}$

Write the expression for the electric field below the surface.

$E_{below}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}$

The value of $E_{above}-E_{below}$is calculate as

$$\begin{gathered} E_{above}-E_{below}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}\left(-\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}\right) \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}+\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n} \\ =\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{n} \end{gathered}$$

Hence, the result is consistent with the equation boundary condition of the electric field at any surface.

Consider the direction of the electric field pointing right as positive and the direction pointing left as negative.

Consider the figure 2.33's left plane.

Write the expression for the change in the electric field for the first plane.

$$\begin{gathered} E_{right}-E_{left}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}+\left(-\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}\right) \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n}+\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\hat{n} \\ =\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{n} \end{gathered}$$

Similarly, for the other plane, the change in electric field is $\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{n}$. As a result, the results are consistent with the electric field boundary condition equation at any surface.

Take the surface of the spherical shell as the boundary in problem 2.11. Inside the spherical shell, the electric field is zero.

Write the expression for the electric field outside the sphere.

$E_{out}=\frac{\mathrm{σ}{R}^2}{{\mathrm{Î\mu }}_0{r}^2}\hat{r}$ …… (3)

Substitute for in equation (3)

role="math" localid="1657600643193" $$\begin{gathered} E_{out}=\frac{\mathrm{σ}{R}^2}{{\mathrm{Î\mu }}_0{R}^2}\hat{r} \\ =\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r} \end{gathered}$$

Hence, the outside electric field at the surface of the spherical shell is $E_{out}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r}$.

Now, write the expression for the change in the electric field at the surface of the shell.

$$\begin{gathered} E_{out}-E_{in}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r}-0\hat{r} \\ =\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r} \end{gathered}$$

Thus, the result is consistent with the equation boundary condition of the electric field at any surface.

b)

Consider a cylindrical tube with a length and radius, and the cylindrical tube's surface as the boundary. Because the charge inside the cylindrical tube is zero, the electric field inside it is also zero.

Write the expression for the electric field outside the cylindrical tube from Gauss law.

$$\begin{gathered} E_{out}\left(2\mathrm{Ï€}rl\right)=\frac{q_{enc}}{{\mathrm{Î\mu }}_0} \\ {E}_{out}\left(2\mathrm{Ï€}rl\right)=\frac{\mathrm{σ}\left(2\mathrm{Ï€}Rl\right)}{{\mathrm{Î\mu }}_0} \\ {E}_{out}=\frac{\mathrm{σ}R}{{\mathrm{Î\mu }}_{0}r} \end{gathered}$$

Write the expression for the outside electric field of the cylindrical tube.

$E_{out}=\frac{\mathrm{σ}{R}^2}{{\mathrm{Î\mu }}_0{r}^2}\hat{r}$ …… (4)

Substitute R for r in equation (4)

$$\begin{gathered} E_{out}=\frac{\mathrm{σ}{R}^2}{{\mathrm{Î\mu }}_0{r}^2}\hat{r} \\ =\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r} \end{gathered}$$

Hence, the outside electric field at the surface of the tube shell is $E_{out}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r}$.

Now, write the expression for the change in the electric field at the surface of tube shell.

$$\begin{gathered} E_{out}-E_{in}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r}-0\hat{r} \\ =\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\hat{r} \end{gathered}$$

Thus, the result is consistent with the equation boundary condition of the electric field at any surface.

c)

Refer Ex2.7;

Write the expression for the outside potential.

$V_{out}=\frac{R^2\mathrm{σ}}{{\mathrm{Î\mu }}_{0}z}$ …… (5)

Substitute R for z in equation (5)

$$\begin{gathered} V_{out}=\frac{R^2\mathrm{σ}}{{\mathrm{Î\mu }}_{0}R} \\ =\frac{R\mathrm{σ}}{{\mathrm{Î\mu }}_0}.........\left(6\right) \end{gathered}$$

Thus, the outside potential at the surface of the cylinder is $\frac{R\mathrm{σ}}{{\mathrm{Î\mu }}_0}$. The inside potential is. Thus, the inside and outside potential at the surface of the cylinder are equal. Thus, the result is consistent with the boundary condition of the potential at the surface.

Differentiate equation (6) with respect to z.

$\frac{∂V_{out}}{∂z}=-\frac{R^2\mathrm{σ}}{{\mathrm{Î\mu }}_0{z}^2}$ …… (7)

Substitute R for z in equation (7)

$$\begin{gathered} \frac{\mathrm{σ}∂V_{out}}{∂Z}=\frac{R^2}{{\mathrm{Î\mu }}_0{R}^2} \\ =-\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0} \end{gathered}$$

Differentiate $V_{in}=\frac{R\mathrm{σ}}{{\mathrm{Î\mu }}_0}$with respect to z.

$\frac{∂V_{in}}{∂z}=0$

Now, find the value of$\frac{\mathrm{σ}∂V_{out}}{∂Z}=\frac{∂V_{in}}{∂Z}$

$\frac{\mathrm{σ}∂V_{out}}{∂Z}-\frac{∂V_{in}}{∂Z}=\frac{-\mathrm{σ}}{{\mathrm{Î\mu }}_0}$

Thus, the result is verified.