Calculate the electric field due to a charged disk (how to do the integration?)

  • #1
zenterix
61
20
Homework Statement:
A disc of radius ##R## is uniformly charged with total charge ##Q>0##. Determine the direction and magnitude of the electric field at a point P lying a distance ##x>0## from the center of the disc along the axis of symmetry of the disc.
Relevant Equations:
I know how to solve this problem. The electric field vector points in only one direction, so there is really only one integral to do. However, I attempted to do the full integral, expecting the other two directions to cancel out, but my integrals don't seem to work out to the correct values. This is thus a question about integration.

Here I set up the integral for the electric field due to the disk at a point ##P##:

$$d\vec{E}_p=\frac{k_edq}{r_{dq,p}^2}\hat{r}_{dq,p}$$

$$dq=2\pi r dr \frac{Q}{\pi R^2}=\frac{2Qrdr}{R^2}$$

$$\vec{r}_{dq,p}=\vec{r}_{o,p}-\vec{r}_{o,dq}=x\hat{k}-r\hat{r}$$

$$r_{dq,p}=\sqrt{x^2+r^2}$$

$$\implies \vec{E}_p=\frac{2k_eQrdr}{R^2(x^2+r^2)^{3/2})}(x\hat{k}-r\hat{r})$$

$$\vec{E}_p=\frac{2k_eQ}{R^2}\left[\int_0^R \frac{rx}{(x^2+r^2)^{3/2}}dr\hat{k}-\int_0^R \frac{r^2}{(x^2+r^2)^{3/2}}dr\hat{r}\right]$$
I am interested in particular in the second integral, in the ##\hat{r}## direction.

Here is my depiction of the problem:
1636756487584.jpeg



As far as I can tell, due to the symmetry of the problem, this integral should be zero.

$$\int_0^R \frac{r^2}{(x^2+r^2)^{3/2}}dr\hat{r}$$

I don't believe I need to integrate anything inside ##\hat{r}##; it depends on an angle relative to some axis but not on ##r##.

$$r=x\tan{\alpha}$$
$$dr=x \sec^2{\alpha}d\alpha$$

$$(x^2+r^2)^{3/2}=(x^2(1+\frac{r^2}{x^2}))^{3/2}=x^3(1+\tan^2{\alpha})^{3/2}=x^3\sec^3{\alpha}$$

So the integral becomes

$$\int \frac{x^2\tan^2{\alpha} x \sec^2{\alpha}}{x^3\sec^3{\alpha}}d\alpha$$
$$=\int \frac{\tan^2{\alpha}}{\sec{\alpha}}d\alpha$$

$$=\int \frac{\sin^2{\alpha}}{\cos{\alpha}}d\alpha$$

$$=\int \frac{1}{\cos{\alpha}}d\alpha-=\int_0^R \cos{\alpha}d\alpha$$

$$=(\ln|\sec{\alpha}+\tan{\alpha}| - \sin{\alpha})$$

Note that

$$r=x\tan{\alpha}$$
$$\implies \sin{\alpha}=\frac{r}{(x^2+r^2)^{1/2}}$$
$$\implies \cos{\alpha}=\frac{x}{(x^2+r^2)^{1/2}}$$

The integral result in terms of ##r## is then

$$(\ln|\frac{(x^2+r^2)^{1/2}}{x}+\frac{r}{x}|-\frac{r}{(x^2+r^2)^{1/2}}) \big|_0^R$$

Which is not zero.

I don't know if I messed up setting up the integral, or in the integration itself.
 

Answers and Replies

  • #2
ergospherical
877
1,203
The issue with your work is that the unit vector ##\hat{\boldsymbol{r}}## depends on the angle ##\varphi## of the charge element in the ring around the ##x## axis. If you want to evaluate the horizontal component of the electric field explicitly, you'll need to e.g. write it out in Cartesians,\begin{align*}
dE_y = \dfrac{Q}{4\pi^2 \epsilon_0 R^2} \dfrac{r^2}{(x^2 + r^2)^{3/2}} \cos{\varphi} dr d\varphi \\
dE_z = \dfrac{Q}{4\pi^2 \epsilon_0 R^2} \dfrac{r^2}{(x^2 + r^2)^{3/2}} \sin{\varphi} dr d\varphi
\end{align*}and do a double integral over ##r## and ##\varphi##. Since they're separable, and the angle ranges over ##[0,2\pi]##, you can see immediately that both will be zero as expected from the cylindrical symmetry.
 
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  • #3
anuttarasammyak
Gold Member
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I would get potential which is scalar, easier to handle than vector
[tex]\phi(x)=k_e \frac{Q}{\pi R^2}\int_0^R \frac{2\pi r dr}{\sqrt{x^2+r^2}}[/tex]
Then
[tex]E=-\nabla \phi[/tex]
 
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  • #4
zenterix
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The issue with your work is that the unit vector ##\hat{\boldsymbol{r}}## depends on the angle ##\varphi## of the charge element in the ring around the ##x## axis. If you want to evaluate the horizontal component of the electric field explicitly, you'll need to e.g. write it out in Cartesians,\begin{align*}
dE_y = \dfrac{Q}{4\pi^2 \epsilon_0 R^2} \dfrac{r^2}{(x^2 + r^2)^{3/2}} \cos{\varphi} dr d\varphi \\
dE_z = \dfrac{Q}{4\pi^2 \epsilon_0 R^2} \dfrac{r^2}{(x^2 + r^2)^{3/2}} \sin{\varphi} dr d\varphi
\end{align*}and do a double integral over ##r## and ##\varphi##. Since they're separable, and the angle ranges over ##[0,2\pi]##, you can see immediately that both will be zero as expected from the cylindrical symmetry.
I thought I didn't have to worry about ##\varphi## because I wrote the integral in terms of infinitesimal rings. ##dq## is the charge on an infinitesimal ring When I integrate only over ##r##, aren't I summing up all the rings already?

I can see the intuition that I do need to somehow integrate over something that will cancel out, and the ##\hat{r}## vectors all cancel each other out when we consider ##\varphi## from 0 to ##2\pi##, but I can't understand the intuition based on the sum of the rings.
 
  • #5
ergospherical
877
1,203
If you want to find the vertical component of the electric field, then you can exploit the cylindrical symmetry by using ring-shaped charge elements of area ##2\pi r dr##. (The factor of ##2\pi## is equivalent to what you'd get if you wrote the double integral over the measure ##rdr d\varphi##, separated the ##r## and ##\varphi## integrals and put ##\displaystyle{\int_{0}^{2\pi}} d\varphi = 2\pi##.)

If you're considering the horizontal field, however, points at different angles around the ring produce contributions to the field at ##P## with different projections onto the ##y## and ##z## axes (depending on the angle). Different points on the ring are no longer equivalent, so you have no choice but to break it up into even smaller bits of area, ##rdrd\varphi##.
 
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  • #6
anuttarasammyak
Gold Member
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They seem to cancel in different ##\theta##.


1636761668941.jpeg
 
  • #7
vela
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I thought I didn't have to worry about ##\varphi## because I wrote the integral in terms of infinitesimal rings. ##dq## is the charge on an infinitesimal ring When I integrate only over ##r##, aren't I summing up all the rings already?
The field due to the ring points along the z-axis, so you've already integrated out x- and y- components. Try doing the full integral for the field of a ring of charge. That's where you'll see the cancellation occurs.
 
  • #8
zenterix
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20
They definitely do cancel, that is clear by simple inspection.

You are totally right about adding ##\varphi##, my question is more about where in my original calculations I missed adding that in.

And here's what I came up with.

There was a step where I wrote

$$\vec{r}_{dq,p}=\vec{r}_{o,p}-\vec{r}_{o,dq}$$

The issue is I wrote an expression for ##\vec{r}_{o,dq}##, the vector from the origin to ##dq##, but ##dq## is actually a ring. There isn't a single vector that points to it.

I need to actually consider a small ##dq## on the ring.

Previously I had

##dq_{ring}=\frac{2rQdr}{R^2}##

But now, I calculate a small charge on the ring

##dq=\frac{dq_{ring}}{2\pi r}=\frac{rQdrd\varphi}{\pi}##

Now the vector ##\vec{r}_{o,dq}## actually points to this small charge on the ring, which has a ##d\varphi## term in it.

$$d\vec{E}_p=\frac{k_erQdrd\varphi}{\pi(x^2+r^2)^{3/2}}(x\hat{k}-r\hat{r})$$

Now when we sub in ##\hat{r}=\cos{\varphi}\hat{i}+\sin{\varphi}\hat{j}##. The resulting integrals in ##\hat{i}## and ##\hat{j}## are 0 as expected.

Also, in the ##\hat{k}## direction we get the same result as before as well:

$$\frac{k_eQ}{R^2\pi}\int_0^{2\pi} \int_0^R \frac{rx}{(x^2+r^2)^{3/2}}drd\varphi$$
$$=\frac{2k_eQ}{R^2}\left [ \frac{x}{|x|} - \frac{x}{(x^2+r^2)^{3/2}} \right ]$$

Now, when this problem was done in a textbook, the symmetry argument was used to not do the integral in the ##\hat{r}## direction. However, the integral in the ##\hat{k}## direction (if my calculations above are correct) is also a double integral. Yet when using the symmetry argument we use a single integral to calculate the ##\hat{k}## direction electric field.

I just read this


If you want to find the vertical component of the electric field, then you can exploit the cylindrical symmetry by using ring-shaped charge elements of area ##2\pi r dr##. (The factor of ##2\pi## is equivalent to what you'd get if you wrote the double integral over the measure ##rdr d\varphi##, separated the ##r## and ##\varphi## integrals and put ##\displaystyle{\int_{0}^{2\pi}} d\varphi = 2\pi##.)

If you're considering the horizontal field, however, points at different angles around the ring produce contributions to the field at ##P## with different projections onto the ##y## and ##z## axes (depending on the angle). Different points on the ring are no longer equivalent, so you have no choice but to break it up into even smaller bits of area, ##rdrd\varphi##.
and I think it answers my doubts about why it is that when we consider just the ##\hat{k}## direction we can just integrate over rings, but the full problem considering all directions means we can't use rings.
 
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