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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?

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

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  • $\begingroup$ Theoretically, shouldn't a uniformly charged body have a uniformly polarized charge density? The entire body lenght with $$ \sigma > 0 $$ or $$ \sigma < 0 $$? $\endgroup$ – July H. Sep 2 '20 at 0:45
  • $\begingroup$ I guess they mean that each of the two parts has a uniform charge density. Clearly if one part is positive and one part is negative then the whole thing cannot be uniform. The phrase “uniformly polarized” is inappropriate here. $\endgroup$ – G. Smith Sep 2 '20 at 0:48

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