A very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis. Find the electric field inside the cylinder. Show that the field outside the cylinder can be expressed in the form

$\mathit{E}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{s}}^{\mathbf{2}}}\left[2\left(P-\hat{s}\right)\hat{s}-P\right]$

[Careful: I said "uniform," not "radial"!]

Consider a very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis/

Assume the center of the positively charged one be at, and negatively at $-\overrightarrow{d}/2$.

Write the formula ofelectric field inside the cylinder.

${\overrightarrow{\mathbf{E}}}_{\mathbf{i}\mathbf{n}}\mathbf{=}{\overrightarrow{\mathbf{E}}}_{\mathbf{+}\mathbf{,}\mathbf{i}\mathbf{n}}\mathbf{+}{\overrightarrow{\mathbf{E}}}_{\mathbf{-}\mathbf{,}\mathbf{i}\mathbf{n}}$ …… (1)

Here, ${\overrightarrow{\mathbf{E}}}_{\mathbf{+}\mathbf{,}\mathbf{in}}$is positive field inside the cylinder and ${\overrightarrow{\mathbf{E}}}_{\mathbf{-}\mathbf{,}\mathbf{in}}$is outside the cylinder.

Write the formula ofelectric field outside the cylinder.

${\overrightarrow{\mathbf{E}}}_{out}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\pi }}\mathbf{+}\mathbf{\left(}\mathbf{2}\hat{\mathbf{r}}\mathit{d}\mathit{c}\mathit{o}\overrightarrow{\mathbf{E}}\mathbf{\right)}$ …… (2)

Here,$\mathit{\lambda }$ are linear charge densities,$\overrightarrow{\mathbf{d}}$is offset, $\overrightarrow{\mathbf{r}}$is radius of cylinder and${\mathit{\epsilon }}_{\mathbf{0}}$ is relative pemitivity.

The field within the cylinder is homogeneous, oppositely charged, and slightly displaced, and it is identical to one of the two stacked cylinders. Let the centres of the two charge d objects be at $\overrightarrow{d}/2$and $-\overrightarrow{d}/2$, respectively.

Figure 1

First we determine the field of a homogenously charged by Gauss’ law:

$\begin{array}{rcl}2rL{E}_{in}& =& \frac{Q_{in}}{{\epsilon }_0}\\ & =& \frac{1}{{\epsilon }_0}{r}^2\mathrm{\pi L\rho }\\ \overrightarrow{\mathrm{E}}& =& \frac{\mathrm{\rho }}{2{\mathrm{\epsilon }}_0}\overrightarrow{\mathrm{r}}\end{array}$

But, the centers of the cylinders are offset, so determine fields inside are:

Determine the positive field inside the cylinder.

${\overrightarrow{E}}_{+,in}=\frac{\rho }{2{\epsilon }_0}\left(\overrightarrow{r}-\frac{\overrightarrow{d}}{2}\right)$

Determine the negative field inside the cylinder.

${\overrightarrow{E}}_{-,in}=\frac{\rho }{2{\epsilon }_0}\left(\overrightarrow{r}+\frac{\overrightarrow{d}}{2}\right)$

Determine the total inside field is the sum of these two:

Substitute $\frac{\rho }{2{\epsilon }_0}\left(\overrightarrow{r}-\frac{\overrightarrow{d}}{2}\right)$for ${\overrightarrow{E}}_{+,in}$and $-\frac{\rho }{2{\epsilon }_0}\left(\overrightarrow{r}+\frac{\overrightarrow{d}}{2}\right)$for ${\overrightarrow{E}}_{-,in}$into equation (1).

$\begin{array}{rcl}{\overrightarrow{E}}_{in}& =& \frac{\rho }{2{\epsilon }_0}\left(\overrightarrow{r}-\frac{\overrightarrow{d}}{2}\right)-\frac{\rho }{2{\epsilon }_0}\left(\overrightarrow{r}+\frac{\overrightarrow{d}}{2}\right)\\ & =& -\frac{\rho }{2{\epsilon }_0}\overrightarrow{d}\\ & =& -\frac{\rho }{2{\epsilon }_0}\end{array}$

Since, $\rho \overrightarrow{d}=\overrightarrow{P}$.

Therefore, thevalue of electric field inside the cylinder is ${\overrightarrow{E}}_{in}=-\frac{\overrightarrow{P}}{2{\epsilon }_0}$.

The field outside is one of two lines of charge that are offset by $\overrightarrow{d}$, and have linear charge densities of $\lambda$.

localid="1658389270826" $\lambda =\rho R^2\mathrm{\pi }$

Figure 2

The field outside is therefore:

$\begin{array}{rcl}{\overrightarrow{E}}_{out}& =& {\overrightarrow{E}}_{+,out}+{\overrightarrow{E}}_{-,out}\\ & =& \frac{\lambda }{2\pi r+{\epsilon }_0}{\hat{r}}_{+}-\frac{\lambda }{2\pi r-{\epsilon }_0}{\hat{r}}_{-}\\ & =& \frac{\lambda }{2\pi r{\epsilon }_0}\left(\frac{{\overrightarrow{r}}_{+}}{{r^2}_{+}}-\frac{{\overrightarrow{r}}_{-}}{{r^2}_{-}}\right)\end{array}$

Then use the following:

The positive radius of the cylinder.

localid="1657629883067" $$\begin{gathered} {\overrightarrow{r}}_{+}=\overrightarrow{r}-\frac{\overrightarrow{d}}{2} \\ Takingsquareboththesidesasfollows: \\ {r}_{+}^2=r^2+{\left(\frac{d}{2}\right)}^{*}-rd\cos \theta \end{gathered}$$

The negative radius of the cylinder.

${\overrightarrow{r}}_{-}=\overrightarrow{r}+\frac{\overrightarrow{d}}{2}$

Taking square both the sides as follows:

$r_{-}^2=r^2+{\left(\frac{d}{2}\right)}^{*}+rd\cos \theta$

Next expand the terms and approximate:

$\begin{array}{rcl}\frac{1}{r_{+}^2}& =& \frac{1}{r^2+{\left({\displaystyle \frac{d}{2}}\right)}^2\mp rd\cos \theta }\\ & =& \frac{1}{r^2}\frac{1}{1+{\left({\displaystyle \frac{d}{2r}}\right)}^2\mp rd\cos \theta }\\ & \approx & \frac{1}{r^2}\left(1-{\left(\frac{d}{2r}\right)}^2\pm \frac{d}{r}\cos \theta \right)\end{array}$

Using these identities we get:

$\begin{array}{rcl}{\overrightarrow{E}}_{out}& =& \frac{\lambda }{2{\epsilon }_0\pi }\left[\left(\overrightarrow{r}-\frac{\overrightarrow{d}}{2}\right)\frac{1}{r^2}\left(1-{\left(\frac{d}{2r}\right)}^2+\frac{d}{r}\cos \theta \right)-\left(\overrightarrow{r}+\frac{\overrightarrow{d}}{2}\right)\frac{1}{r^2}\left(1-{\left(\frac{d}{2r}\right)}^2-\frac{d}{r}\cos \theta \right)\right]\\ & =& \frac{\lambda }{2{\epsilon }_0\pi }\left[-\overrightarrow{d}+\overrightarrow{d}{\left(\frac{d}{2r}\right)}^2+2\overrightarrow{r}\frac{d}{r}\cos \theta \right]\end{array}$

Now ignore the middle term, on basis that it is much smaller than the others.

The solution is thus:

Determine theelectric field outside the cylinder.

Substitute $\rho R^2\mathrm{\pi }$for $\lambda$and $\hat{r}$for $\cos \theta$into equation (2).

$\begin{array}{rcl}{\overrightarrow{E}}_{out}& =& \frac{\lambda }{2{\epsilon }_0{\mathrm{\pi r}}^2}\left(2\overrightarrow{r}d\cos \theta -\overrightarrow{d}\right)\\ & =& \frac{\rho R^2\mathrm{\pi }}{2{\epsilon }_0{\mathrm{\pi r}}^2}\left(2(\overrightarrow{P}·\hat{r})\hat{r}-\overrightarrow{P}\right)\\ & =& \frac{1}{2{\epsilon }_0}{\left(\frac{R}{r}\right)}^2\left(2\left(\overrightarrow{P}·\hat{r}\right)\hat{r}-\overrightarrow{P}\right)\end{array}$

Where, again, $\rho \overrightarrow{d}=\overrightarrow{P}$.

Therefore, the value of electric field outside the cylinder is ${\overrightarrow{E}}_{out}=\frac{1}{2{\epsilon }_0}{\left(\frac{R}{r}\right)}^2\left(2\left(\overrightarrow{P}·\hat{r}\right)\hat{r}-\overrightarrow{P}\right)$.