# Doubt about uniformly charged bodies

I was studying the electric field behavior at one point with respect to a uniformly charged disk, and while analyzing examples, this specific one caused me doubts.

The example talks about a disk fixed at the origin of the xy plane, and its charge density being $$> 0$$ for $$0$$ $$\leq$$ r $$\leq$$ $$B/2$$ and $$<0$$ for $$B/2 \leq r \leq B$$, where $$B$$ symbolizes the radius and $$r$$ symbolizes the distance to the disk's center.

I would like to generalize a formula to obtain $$\vec E$$ at any point in $$z$$. So, considering that a disk is formed by several rings, I integrated the $$\vec E$$ formula of the ring from $$0$$ to $$B$$:

$$\vec E =$$$$\int d\vec E =$$ $$\int_{0}^{B} \frac{k*2\pi \sigma R *z\hat k}{(z^2 + R^2)*\sqrt {(z^2+R^2)}} dR =$$ $$-2k\hat k\sigma \pi z*(\frac{1}{\sqrt {z^2+B^2}} - \frac{1}{\sqrt{z^2}})$$

I am having difficulties in applying the formula obtained in the example situation, because it has different $$\sigma$$ depending on $$r$$.

Any thoughts?

• What is $dx$ supposed to mean? Why is $R$ in the integrand? – G. Smith Sep 2 '20 at 0:40
• Thanks, that's a typo. Already fixed it. – July H. Sep 2 '20 at 0:47
• $R$ is a constant: the radius of the disk and the limit of integration. The variable of integration should be the radial coordinate $r$. – G. Smith Sep 2 '20 at 0:51
• I forgot to explain that $$dq = 2\pi R dR$$, I'll fix – July H. Sep 2 '20 at 0:52
• In fact, I forgot to define the disk radius as another variable, I will use $B$. I believe that the integral its right, now. – July H. Sep 2 '20 at 0:58

I am having difficulties in applying the formula obtained in the example situation, because it has different $$\sigma$$ depending on $$r$$.
Integrate from $$0$$ to $$B/2$$ using the positive charge density. Integrate from $$B/2$$ to $$B$$ using the negative charge density. Add the two contributions to get the total field.
• Theoretically, shouldn't a uniformly charged body have a uniformly polarized charge density? The entire body lenght with $$\sigma > 0$$ or $$\sigma < 0$$? – July H. Sep 2 '20 at 0:45