Electric field a distance z above flat circular disk.

[-Dragoon-]

Homework Statement

Find the electric field a distance above the center of a flat circular disk of radius R, which carries a uniform surface charge σ.

Homework Equations

The Attempt at a Solution

Basically, I want to solve this usually trivial problem without using symmetry arguments and the superposition principle but rather the more laborious method introduced in Griffith's section 2.1. Hence:

[itex]\vec{E(r)} = \frac{1}{4\pi\epsilon_{0}}\int\frac{\sigma(r)}{r'^2}\hat{r'}da[/itex]

Where [itex]\hat{r'} = \frac{z\hat{z} - r\hat{r}}{\sqrt{z^2+r^2}}[/itex] and [itex]r' = \sqrt{r^2 + z^2}[/itex], thus:

[itex]\vec{E(r)} = \frac{\sigma}{4\pi\epsilon_{0}}\int\frac{z\hat{z} - r\hat{r}}{(z^2 + r^2)^{3/2}}da[/itex]

Converting to polar coordinates we have:
[itex]\vec{E(r)} = \frac{\sigma}{4\pi\epsilon_{0}}\iint\frac{z\hat{z} - r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta[/itex]

=> [itex]\frac{\sigma}{4\pi\epsilon_{0}}\int_0^{2\pi}\int_0^r\frac{z\hat{z} - r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta[/itex]
Which gives:
[itex]\frac{\sigma2\pi}{4\pi\epsilon_{0}}[z\hat{z}\int_0^r\frac{r}{(z^2 + r^2)^{3/2}}dr - \int_0^r\frac{r^2\hat{r}}{(z^2 + r^2)^{3/2}}dr][/itex]

Evaluating the first integral gives me the correct answer for the electric field:
[itex]\frac{\sigma2\pi z}{4\pi\epsilon_{0}}[- \frac{1}{\sqrt{z^2+r^2}}\Big|_0^r]\hat{z}[/itex]

Which means, the second integral must be zero: [itex]\int_0^r\frac{r^2\hat{r}}{(z^2 + r^2)^{3/2}}dr = 0[/itex]

Except I am not exactly sure on how to show that is the case. The radial unit vector seems to be the problematic term, and it cannot be simply taken outside the integral. Would like to have some help and insights on this, thanks.

 

[TSny]

Go back and think about the ##\theta## integration in the expression $$\iint\frac{ r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta$$ Note that ##\hat{r}## is not a constant when integrating over ##\theta##.

 

[-Dragoon-]

Go back and think about the ##\theta## integration in the expression $$\iint\frac{ r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta$$ Note that ##\hat{r}## is not a constant when integrating over ##\theta##.

Ah, I see. Thanks. Is then ##\hat{r}## a constant when integrating with respect to ##r## then?

 

[TSny]

Yes. If you integrate over ##r## first, then you are keeping ##\theta## constant. So, the unit vector ##\hat{r}## has a fixed direction.

 

[Nickie]

First of all, great job on using the more laborious method to solve this problem! It's always good to practice different techniques and gain a deeper understanding of the concepts.

To show that the second integral is indeed equal to zero, we can use the fact that the electric field is a conservative vector field. This means that the line integral of the electric field along any closed path is always equal to zero. In other words, the work done by the electric field in moving a charge from one point to another is independent of the path taken.

In this case, we can choose a circular path of radius r centered at the origin. This means that the electric field will be tangential to the path at all points. Therefore, the work done by the electric field along this path will be zero. Mathematically, this can be written as:

\oint_C\vec{E}\cdot d\vec{l} = 0

where C is the chosen circular path and d\vec{l} is an infinitesimal displacement along the path.

Now, let's consider the integral in question:

\int_0^r\frac{r^2\hat{r}}{(z^2 + r^2)^{3/2}}dr

We can rewrite this integral as:

\int_0^r\frac{r\hat{r}}{(z^2 + r^2)^{3/2}}rdr

Notice that r\hat{r} is the infinitesimal displacement along the circular path, which means that the integral can be written as:

\int_C\frac{\vec{r}}{(z^2 + r^2)^{3/2}}\cdot d\vec{l}

This is essentially the line integral of the electric field along the circular path, which we know is equal to zero. Therefore, the integral in question is also equal to zero. This shows that the second integral in your solution is indeed equal to zero, and you have correctly solved the problem.

I hope this helps! Keep up the good work.