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

In summary, the issue with the original calculation was that the unit vector r depends on the angle of the charge element in the ring, so the horizontal component of the electric field cannot be evaluated by simply integrating over r. Instead, one must use ring-shaped charge elements with area 2πrdr and integrate over both r and the angle φ. This results in the cancellation of the r-dependent terms in the horizontal component of the electric field, as expected from the cylindrical symmetry of the problem.
  • #1
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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.
 
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  • #2
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
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
ergospherical said:
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
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
They seem to cancel in different ##\theta##.
1636761668941.jpeg
 
  • #7
zenterix said:
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
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


ergospherical said:
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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1. How do I calculate the electric field due to a charged disk?

To calculate the electric field due to a charged disk, you will need to use the formula E = k * Q * r / (r^2 + h^2)^(3/2), where k is the Coulomb's constant, Q is the charge of the disk, r is the distance from the center of the disk to the point where you want to calculate the electric field, and h is the height of the disk.

2. What is the Coulomb's constant?

The Coulomb's constant, denoted by k, is a proportionality constant that appears in the Coulomb's law, which describes the force between two charged particles. Its value is approximately 8.99 x 10^9 Nm^2/C^2.

3. How do I perform the integration for the electric field due to a charged disk?

To perform the integration for the electric field due to a charged disk, you will need to use the substitution method and integrate the formula E = k * Q * r / (r^2 + h^2)^(3/2) with respect to r. This will result in the final formula E = k * Q / (sqrt(r^2 + h^2)).

4. What are the units for the electric field?

The units for the electric field are Newtons per Coulomb (N/C) or Volts per meter (V/m).

5. Can I use the formula for the electric field due to a charged disk for any shape of charged object?

No, the formula for the electric field due to a charged disk is only applicable for a disk-shaped object. For other shapes, you will need to use different formulas or techniques to calculate the electric field.

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