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

[zenterix]

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:

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.

 

[ergospherical]

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.

 

[anuttarasammyak]

I would get potential which is scalar, easier to handle than vector
\phi(x)=k_e \frac{Q}{\pi R^2}\int_0^R \frac{2\pi r dr}{\sqrt{x^2+r^2}}
Then
E=-\nabla \phi

 

[zenterix]

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.

 

[ergospherical]

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, $rdrd\varphi$.

 

[anuttarasammyak]

They seem to cancel in different $\theta$.

 

[vela]

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.

 

[zenterix]

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.