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.

  • 1
    $\begingroup$ What exactly do you need to find out? the potential on the $z$ axis, at large $z$? $\endgroup$ – Emilio Pisanty Sep 5 '13 at 22:20
  • $\begingroup$ Some dimensional analysis auditing would definitely help. Since $\rho\propto \epsilon_0 az$, the units of $\epsilon_0 a$ must be $\text{charge}/\text{length}^4$, which makes $q\propto \epsilon_0 a R^5$ dimensionally incorrect, and similarly for your equation for $p$. $\endgroup$ – Emilio Pisanty Sep 5 '13 at 22:25
  • $\begingroup$ @EmilioPisanty I need to find the potential at very big distance to the origin (not necessary on the z axis). but the result of p is wrong, so also the result for V in wrong $\endgroup$ – sunrise Sep 5 '13 at 22:41

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

| cite | improve this answer | |
  • $\begingroup$ thanks for your answer. Can't I find q and then multiply it for a? $\endgroup$ – sunrise Sep 5 '13 at 22:48
  • $\begingroup$ @sunrise Well, for one, $q$ is zero, isn't it? $\endgroup$ – Emilio Pisanty Sep 5 '13 at 22:49
  • $\begingroup$ sure! the total charge is zero. I'm sorry, I meant $Q_+$.. $\endgroup$ – sunrise Sep 5 '13 at 22:50
  • $\begingroup$ is my reasoning about the centers of positive and negative charges wrong? $\endgroup$ – sunrise Sep 5 '13 at 22:51
  • $\begingroup$ Yes. You are changing your problem for a different one, and you can't be sure that the exact details will carry over. While you can indeed model your cylinder as two charges of $\pm Q$ at $\pm L$, you have two free parameters and you can't be sure that the optimal position for your charges is at $L=2R$. (In fact, it won't be: there's plenty of charge at smaller $z$'s, so the effective position will be closer to the origin.) $\endgroup$ – Emilio Pisanty Sep 5 '13 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.