(Exercise review) Find the moment of the electric dipole

Question

I have solved an exercise, but the result I have obtined is wrong and I can't understand why. If you can help me, I'll be so grateful.

Let's consider a charged cylinder, its radius is equal to R and its height=4R. Its center of is in the origin of the cartesian axes and its axis is parallel to z axis; we know the volumic density of charge: $\rho= 2az \epsilon$ where $a, \epsilon$ constant $>0$.

There is an electric field inner to the cylinder, described by: $E_{ox}=0, E_{0y}=0, E_{0z}=az^2$

We want to find the electric potential to a great distance from the origin of the axes.

I have thought as follow.

Using Maxwell's first equation, I have found the volumic density of charge: $\rho=2\epsilon_0 a z$.

At a very big distance to the origin, I can consider the cylinder like an electric dipole, with the positive charge concetrated in the center uf the upper part of the cylinder and the negative one in the center of the lower part:

So $Q_+$ will be in (0,0,R) and $Q_-$ in (0,0,-R).

Now, I need the moment of dipole, ${\bf p}=q{\bf a}$ where ${\bf a}$ is the vector from the negative charge to the positive charge.

I have integrate $\rho$ and I have obtained $q=\int_0^{2R}\int_0^R\int_0^{2\pi}2az\epsilon r d\theta dr dz$ and I have obtained $q=4\pi a \epsilon_0 R^5$

Then I have posed a=2R, so I have obtained $p=8\pi a \epsilon_0 R^4$.The correct result is $p=32/3 \pi a \epsilon_0 R^5$

I can't understand where I was wrong and I wouldn't like to have made an error of proceeding.

Answer 1

You should calculate the dipole moment directly: $$ p=\int\rho(\mathbf r)z\,\text d z=\int_0^{2R}\int_0^R\int_0^{2\pi}2a\epsilon_0 {z^2}\, r \,\text d\theta \text dr \text dz =2\epsilon_0a\int_0^{2R}z^2 \text dz\int_0^R r \text dr\int_0^{2\pi} \text d\theta =2\epsilon_0a \left.\frac{z^3}{3}\right|_{-2R}^{2R} \left.\frac{r^2}{2}\right|_{0}^R 2\pi =2\epsilon_0 a\cdot \frac{2\cdot 8R^3}{3} \cdot \frac{R^2}{2}\cdot2\pi=\frac{32\pi}{3}\epsilon_0 aR^5. $$