Electric field inside a uniformly polarized cylinder

[Potatochip911]

Homework Statement

This is problem 4.13 from Griffiths. A long cylinder of radius a carries a uniform polarization P perpendicular to its axis. Find the electric field inside the cylinder.

Homework Equations

$\int \vec{E}\cdot dA = q_{encl}/\varepsilon_0$

The Attempt at a Solution

[/B]
We solve the problem by using a positive and negatively polarized cylinder:
Using Gauss' law we find

$$ \vec{E}(2\pi \vec{s} l) = \frac{\rho \pi \vec{s}^2 l}{\varepsilon_0} \\ \Rightarrow \vec{E} = \frac{\rho}{2\varepsilon_0}\vec{s}$$

then
$$\vec{E_+} = \frac{\rho}{2\varepsilon_0}\vec{s_+} \\ \vec{E_-} = -\frac{\rho}{2\varepsilon_0}\vec{s_-}$$

Summing together gives

$$\vec{E} = \frac{\rho}{2\varepsilon_0}\left(\vec{s_+} - \vec{s_-} \right)$$

and since for dipoles we define the vector $\vec{d}$ going from the negative to the positive charge $\vec{d} = s_+ - s_-$

$$\vec{E} = \frac{\rho \vec{d}}{2\varepsilon_0} = \frac{\vec{P}}{2\varepsilon_0}$$

However, in the solutions manual they claim $\vec{d}$ is going from the negative to the positive axis and that $\vec{s_+}-\vec{s_-} = -\vec{d}$ which I am having a hard time believing as from a simple diagram this seems incorrect.

 

[Charles Link]

I can't draw a diagram, but I think I can describe the picture: Assume the + cylinder is translated to the right by a vector $\vec{d}$. (The cross-sectional view involves two circles of radius $a$ in the x-y plane, with the + circle a distance $d$ to the right of the - circle). Then, at a given location $\vec{r}$ where the electric field is being determined, e.g. upward and to the right of the origin, $\vec{s_-}$ will be the longer vector in reaching the location $\vec{r}$. We can write $\vec{s_-}=\vec{s_+}+\vec{d}$.

 

[Potatochip911]

I can't draw a diagram, but I think I can describe the picture: Assume the + cylinder is translated to the right by a vector $\vec{d}$. (The cross-sectional view involves two circles of radius $a$ in the x-y plane, with the + circle a distance $d$ to the right of the - circle). Then, at a given location $\vec{r}$ where the electric field is being determined, e.g. upward and to the right of the origin, $\vec{s_-}$ will be the longer vector in reaching the location $\vec{r}$. We can write $\vec{s_-}=\vec{s_+}+\vec{d}$.

Am I drawing it incorrectly? I still don't find the same result as you.

 

[Charles Link]

Am I drawing it incorrectly? I still don't find the same result as you.
View attachment 221841

Your drawing is incorrect. $d$ is a shift of the origins of two circles, each of radius $a$, along the x-axis. A point $\vec{r}$ is common to both of them. $\vec{s}_+$ is a vector from the rightmost origin to $\vec{r}$, and $\vec{s}_-$ is a vector from the leftmost origin to $\vec{r}$.