1
$\begingroup$

We consider a thin disk whose center is $O$ and radius is $R$, which has an electric surface charge density of $\sigma$. We want to find the expression of the electric field $\vec{E}$ at a point $M$ of the $(O_z)$ axis using the definition of an electric field at a point, where $(O_z)$ is perpendicular to the disk. The altitude of $M$ from the disk is $z$.

Let $\vec{k}$ be the unit vector in $(O_z)$.

I first calculated the potential $V$ at $M$ and then used the relation $\vec{E}=-\vec{grad}V$ and found that: $$\vec{E}=\dfrac{\sigma}{2\varepsilon _0}\left( 1-\dfrac{z}{\sqrt{R^2+z^2}}\right)\vec{k}$$

I checked this answer on the internet. It's correct.

Now I'll calculate it using the definition. Here I find a problem: the result is different. Using symeterie it's physically clear that $\vec{E}$ is a vector in $(O_z)$ and that's confirmed by the formula $\vec{E}=\dfrac{\sigma}{2\varepsilon _0}\left( 1-\dfrac{z}{\sqrt{R^2+z^2}}\right)\vec{k}$. However using this different reasoning I get a totally different result.


The alternative reasoning

Schema

According to the definition we have $d\vec{E}=\dfrac{dq}{4\pi\varepsilon _0 MP^2}\vec{u}=\dfrac{\sigma\,dS}{4\pi\varepsilon _0 MP^2}\vec{u}$ where $||\vec{u}||=1$

We have $\cos\theta =\dfrac{z}{MP}$ so $\dfrac{1}{MP^2}=\dfrac{\cos^2\theta}{z^2}$ and $dS=r\,dr\,d\phi$ in the polar coordinate system.

So $\vec{E}=\dfrac{\sigma\cos^2\theta\,r\,dr\,d\phi}{4\pi\varepsilon _0z^2}\vec{u}$

And $\tan\theta=\dfrac{r}{z}$ so $r=z\tan\theta$ and $dr=\dfrac{z}{\cos^2\theta}d\theta$

So $d\vec{E}=\dfrac{\sigma\tan\theta\,d\theta\,d\phi}{4\pi\varepsilon _0}\vec{u}$

Consider $\vec{j}$ the unit vector from $O$ to the right. We have: $$\vec{u}=\cos\theta\vec{j}+\sin\theta\vec{k}\tag{1}$$

So:

$$\vec{E}=\dfrac{\sigma}{4\pi\varepsilon _0}\left( \left(\iint\sin\theta\,d\theta\,d\phi\right)\vec{j}+\left(\iint\tan\theta\cos\theta\,d\theta\,d\phi\right)\vec{k}\right)$$

The first formula I found contains only $\vec{k}$, but in this last formula I have:

$$\left(\iint\sin\theta\,d\theta\,d\phi\right)\vec{j}=\left( \int\limits_0^{2\pi}d\phi\right)\left(\int\limits_0^{\theta_{max}}\sin\theta\,d\theta\right)\vec{j}\neq \vec{0}$$


I'm totally confused. This reasoning isn't false, so was my use of the polar coordinate systeme valid? And the choosen integrating domains is also valid?


Edit:

I found that I've made a mistake: $\vec{u}=\sin\theta\vec{j}+\cos\theta\vec{k}$. The relation $(1)$ is false.

But I'm still confused. We still find that $\vec{E}$ isn't on $(O_z)$:


Rectifying:

We have then:$$\vec{E}=\dfrac{\sigma}{4\pi\varepsilon _0}\left(\left(\iint\tan\theta\sin\theta\,d\theta\,d\phi\right)\vec{j}+ \left(\iint\sin\theta\,d\theta\,d\phi\right)\vec{k}\right)$$

Notice that:

$$\dfrac{\sigma}{4\pi\varepsilon _0}\left(\iint\sin\theta\,d\theta\,d\phi\right)\vec{k}=\dfrac{\sigma}{4\pi\varepsilon _0}\left( \int\limits_0^{2\pi}d\phi\right)\left(\int\limits_0^{\theta_{max}}\sin\theta\,d\theta\right)\vec{k}=\dfrac{\sigma}{2\varepsilon _0}\left( 1-\dfrac{z}{\sqrt{R^2+z^2}}\right)\vec{k}$$

because $\cos\theta_{max}=\dfrac{z}{\sqrt{R^2+z^2}}$. That's exactly the first formula we got.

But the problem is that we should then get:$$\left(\iint\tan\theta\cos\theta\,d\theta\,d\phi\right)\vec{j}=\vec{0}$$

which means that:$$\int\limits_0^{\theta_{max}}\tan\theta\sin\theta\,d\theta=0\tag{2}$$

While $\theta_{max}\in \left]0,\dfrac{\pi}{2}\right[$ it's mathematically clear that $(2)$ isn't correct because:$$\int\limits_0^{\theta_{max}}\tan\theta\sin\theta\,d\theta>0$$


After thinking, I found that I must introduce another vector essential to find exactly $\vec{u}$ which is $\vec{i}$ such as $\vec{i}\wedge\vec{j}=\vec{k}$ because if we look at the schema without rotating if we take the point where is wroten $\sigma$ then $\vec{u}$ isn't a vector of $(O,\vec{j},\vec{k})$. The coordinate related to $\vec{k}$ will not change but the one related to $\vec{j}$ will change.

What do you think about that?


Edit2:

Going from spheric to cartesian coordonate system we get $\vec{u}=\vec{e_r}=\cos\theta\vec{k}+\sin\theta\cos\phi\vec{i}+\sin\theta\sin\phi\vec{j}$. Here $\theta$ and $\phi are exactly the same we used previously.

We have:$$\int\limits_0^{2\pi}\sin\phi\,d\phi=\int\limits_0^{2\pi}\cos\phi\,d\phi=0$$

So finally by subtitution we get:$$\vec{E}=\dfrac{\sigma}{2\varepsilon _0}\left( 1-\dfrac{z}{\sqrt{R^2+z^2}}\right)\vec{k}$$ which is the same formula we got using the first reasoning.

$\endgroup$

closed as off-topic by ACuriousMind, BMS, Danu, user10851, Alfred Centauri Oct 28 '14 at 20:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, BMS, Danu, Community, Alfred Centauri
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Hi Scientifica, and welcome to Physics.SE! Please take a moment to read our homework policy. $\endgroup$ – ACuriousMind Oct 28 '14 at 18:57
  • $\begingroup$ Ok I'm reading it right now. Thanks for the welcome. $\endgroup$ – Scientifica Oct 28 '14 at 19:00
  • $\begingroup$ Please note that check my work-type questions are off-topic on this site, falling under the homework policy (even if they arise from something else than an academic assignment). $\endgroup$ – Danu Oct 28 '14 at 19:47
  • $\begingroup$ Thanks for your answer but that's not a check my work question because I know that it's false and I'm asking why it's false. $\endgroup$ – Scientifica Oct 28 '14 at 20:37
  • $\begingroup$ No really, this is a "check my work" question as in "check my work and show me my mistake". However, if you're convinced your math is correct and can phrase the question as something like is my reasoning in step c valid? then I think your question will be received better. $\endgroup$ – Alfred Centauri Oct 28 '14 at 20:53