Electric Field Calculation for a Rotating Thread with Charge Density

[LostInSpace]

Hi

I have been assigned a problem that I can't solve.

I have a rotating axis to which a thread is connected in one end. The thread is perpendicular to the rotating axis. The thread has a charge density \lambda and a length L.

First of all, I need a mean value of the charge density of the circular disc described by the rotating thread with respect to time. I have interpreted this as the charge density:
\sigma(r) = \frac{\lambda \mathrm{d}r}{2\pi r\mathrm{d}r} = \frac{\lambda}{2\pi r}<br />

The charge density is supposed to be a function of r; the distance to the center of the disc. However, in the density above, the charge density is infinite close to the center. I can't interpret this conceptually.

Second, I am supposed to determine the electrical field a distance r_0 from the center of the disc along the rotational axis. Coulombs law yields:

E = \frac{1}{4\pi\epsilon_0}\iint_\Omega \frac{\mathrm{d}q}{R^2} = \frac{1}{4\pi\epsilon_0}\iint_\Omega\frac{\sigma\mathrm{d}x\mathrm{d}y}{r_0^2 + x^2 + y^2} = [\mathrm{Polar\ coordinates}] = \frac{\lambda}{4\pi\epsilon_0}\int_0^L\frac{\mathrm{d}r}{r_0^2 + r^2} = \ldots = \frac{\lambda\theta}{4\pi\epsilon_0r_0}

This result is a bit strange, if you consider the extreme values. For instance:
<br /> \begin{array}{ll}<br /> \lim_{\theta\rightarrow 0}E = 0 &amp; \mathrm{Ok!} \\<br /> \lim_{\theta\rightarrow \frac{\pi}{2}}E = k &amp; \mathrm{Ok?} \\<br /> \lim_{r_0\rightarrow 0}E = \infty &amp; \mathrm{Not\ Ok??} \\<br /> \lim_{r_0\rightarrow\infty}E = 0 &amp; \mathrm{Ok}<br /> \end{array}<br />

From the third extreme value, I must conclude that the result is wrong, as it should be 0 in the disc (the forces cancel each other)

What is wrong? Is it the mathematics or the physics that fail?

Please Help!

Nille

 

[clive]

1) First of all, it seems that the analogy of the rotating thread with a charged disc is not very appropiate (in my opinion).

In any point of the rotating axis you will have an electric field with two vectorial components:
\vec{E}=\vec{E_{axial}+\vec{E_{norm}}}
The normal component will rotate syncronic with the thread whereas the axial component will be constant. Because of sinusoidal variation of the normal component, its average is zero. Then you have to evaluate only the axial component as the wire is at rest.

2) When you evaluate the electric field, you add vectors, then I think you need a cos(\alpha). The expression for \vec{E} which you use seems to be wrong even if the analogy with the charged disc is correct.

 

[wais]

Hi Nille,

Thank you for sharing your problem with us. It seems like you are on the right track with your calculations, but there are a few conceptual and mathematical errors that are leading to the strange results you are getting. Let's go through them one by one.

Firstly, the mean charge density of the rotating thread should not be infinite at the center. This is because although the thread is infinitely thin, it still has a finite length L. Therefore, the charge density should be a function of both r and L. The correct expression for the mean charge density should be:

\sigma(r) = \frac{\lambda}{2\pi rL}

This takes into account the finite length of the thread. Now, let's move on to your calculation of the electric field. You are correct in using Coulomb's law to calculate the electric field, but there is a slight error in your integration limits. In polar coordinates, the limits for the integral should be from 0 to 2π instead of 0 to L. This is because you are integrating over the entire circular disc, not just along the length of the thread. Therefore, the correct expression for the electric field should be:

E = \frac{\lambda}{4\pi\epsilon_0}\int_0^{2\pi}\int_0^L\frac{r\mathrm{d}r\mathrm{d}\theta}{(r_0^2 + r^2)^{3/2}} = \frac{\lambda L}{4\pi\epsilon_0r_0}\left(\frac{1}{\sqrt{1 + (\frac{L}{r_0})^2}} - 1\right)

This expression takes into account the finite length of the thread and the correct integration limits. Now, let's look at the extreme values you mentioned. As you correctly noted, the electric field should be 0 at the center of the disc and as the distance r_0 increases to infinity. However, at the edge of the disc, the electric field should not be infinite. This is because the charge on the thread is distributed over a finite length, so the electric field will also decrease as you approach the edge of the disc. Therefore, the correct result should be:

\lim_{r_0\rightarrow 0}E = \frac{\lambda}{4\pi\epsilon_0} \frac{L}{r_0