# Query regarding solution to Griffith's "Introduction to electrodynamics", Fourth edition, Problem 4.13 [closed]

I was solving the following problem from Griffith's Introduction to Electrodynamics 4th edition:

Q.$$4.13$$ : A very long cylinder of radius $$a$$ carries a uniform polarisation $$P$$ perpendicular to the axis. Find the electric field inside and outside the cylinder.

My Attempt:

Since $$P$$ is uniform, so $$\rho_b=0$$ and $$\sigma_b=\vec P\cdot \hat n=\vec P\cdot \hat s=P \cos \theta$$

Now we consider 2 Gaussian cylinders of length $$a$$ and radius $$r$$, $$r in cylinder $$1$$ and $$r>a$$ in cylinder $$2$$.

Applying Gauss' Law, for $$r, \begin{align}\vec E\cdot {\mathrm d\vec A}=\oint \frac{1}{\epsilon_0}Q_\textrm{enc} &=\frac{1}{\epsilon_0} \int_V \rho_b \,\, ~\mathrm dV=0\\ \implies \vec E &=\vec 0\end{align}

Applying Gauss' Law, for $$r, \begin{align}\vec E\cdot {\mathrm d\vec A}=\oint \frac{1}{\epsilon_0}Q_\textrm{enc} &=\frac{1}{\epsilon_0} \left(\int_V\rho_b \,\,~\mathrm dV + \int_S \sigma_b ~\mathrm dA\right)\\ \implies E\cdot 2\pi rl &=\frac{1}{\epsilon_0} (0 + P \cos \theta \cdot 2\pi a l)\\ \implies~~~~~~~~~~ \vec E &=\frac{Pa\cos \theta}{\epsilon_0r} \hat s \end{align}

But Griffith's is providing a different solution which is entirely different from mine which I can't understand at all.

Can someone explain why I am wrong or if I am wrong at all?

First of all, I think you have a problem with your notation. Gauss' Law should read $$\int_{\partial V} \vec{E} \cdot d\vec{S} = \frac{q_{\text{enc}}}{\epsilon_{0}},$$ where the enclosed charge can be written as $$q_{\text{enc}} = \int_{V}\rho(\vec{r}^{\, \prime}) dV^{\prime}.$$ It is very important that you remember this.

On the other hand, once you set $\sigma_{b} = P \cos \theta$, you should be able to use separation of variables to solve Laplace's equation to obtain the potential in both regions, and then you can compute the electric field by simply taking its gradient.

I'm not totally sure about this, but I think you should not consider the enclosed charge as the volume integral of the volume density plus the surface integral of the induced surface density. Also, I don't think you can use Gauss' law directly, unless you proceed as Griffiths does.

I can help with But Griffith's is providing a different solution which is entirely different from mine and which I can't understand at all.
Usually one has a problem in understanding
"Think of it as two cylinders of opposite uniform charge density"
part in the image that you have attached.
It is similar to the uniformly polarised sphere's case; see the image.
What it means that when the cylinder is polarised the bound charges distribute themselves in such a fashion that the positive charge is accumulated in the direction in which the polarisation vector points (here, the upper side of the curved surface of the cylinder) while the negative charge gets accumulated in the opposite direction (lower side of the curved surface)

I have given a similar representation for a cylinder in the figures below. The charges on the cylinder in image (a) can be assumed to be due to two different cylinders with opposite bound charge densities, positive for red and negative for the green one. The charges for these two cancel out in the overlapping regions but produce approximately the same effect as the original cylinder due to the charges in the non-overlapping regions.

Using these two new assumed cylinder, the net electric field of the original cylinder can be calculated using the principle of superposition using the individual field due to the charge densities of the two cylinders.
Hope it will help.

Here $$\rho_b=0$$ but $$\rho_{free}$$ is not zero. In order to use gauss law, you have to consider total charge, i,e, free and bound charge.