# Delta function charge densities

1. Feb 2, 2010

### kreil

1. The problem statement, all variables and given/known data
Using Dirac delta function in the appropriate coordinates, express the following charge distributions as three-dimensional charge densities p(x).

(a) In spherical coordinates, a charge Q uniformly distributed over a spherical shell of radius a.

(b) In cylindrical coordinates, a charge $\lambda$ per unit length uniformly distributed over a cylindrical surface of radius b.

(c) In cylindrical coordinates, a charge Q spread uniformly over a flat circular disc of negligible thickness and radius R.

(d) The same as in (c), but using spherical coordinates.

Note that
$$\delta(x-x')= \frac{1}{|J(x_i, \zeta_i)|} \delta( \zeta_1 - \zeta_1')\delta( \zeta_2 - \zeta_2')\delta( \zeta_3 - \zeta_3')$$

where $J(x_i, \zeta_i)$ is the Jacobian relating cartesian coordinates $(x_1,x_2,x_3)$ to new coordinates $(\zeta_1,\zeta_2,\zeta_3)$.

2. Relevant equations

$$\int_{- \infty}^{+ \infty} f(x) \delta(x-a) dx = f(a)$$

3. The attempt at a solution

For some reason I can't wrap my head around how to methodically do this problem. Let's start by discussing part (a). Here is what I have:

(a)

The definition of charge density is: $$\rho = \frac{TotCharge}{TotArea}$$

In this case then, we get: $$\rho = \frac{Q}{4 \pi a^2}$$.

However, this charge is localized to the surface of the sphere where r=a, so:

$$\rho(r)=\frac{Q}{4 \pi a^2} \delta(r-a)$$

is this a correct answer? it doesnt seem to be three dimensional, but then again $\theta$ and $\phi$ seem like irrelevant values since the charge is localized to the surface where r=a. Doing a check integration of the charge density in 3d space produces the total charge Q, so I have reason to believe it is correct..

As far as the rest of them go, I could use some help as to how to get started/use that jacobian equation.

2. Feb 2, 2010

Oh, really?

3. Feb 2, 2010

### kreil

lol just trying to be clear.

4. Feb 2, 2010

### gabbagabbahey

The point of my question was to intimate that the definition you gave is incorrect, and is leading to your confusions.....I suggest you look up the definition of "volume charge density".

5. Feb 2, 2010

### kreil

im aware of that, but in part a it is a surface charge density isn't it?

6. Feb 2, 2010

### gabbagabbahey

No, in all 4 parts of the question your are asked to find the volume charge density....in part (a), the surface charge density is finite everywhere, so no delta function would be needed....however, the volume charge density is not finite everywhere, and yet the total charge is finite and so a delta function is needed.
So, what is the general definition of volume charge density?

7. Feb 2, 2010

### kreil

charge/volume, so for a it would be $$\rho = \frac{3Q}{4 \pi a^3}$$

8. Feb 2, 2010

### gabbagabbahey

Not quite, and no.

The definition of volume charge density, in a region $\mathcal{V}$, is given mathematically by the integral

$$Q=\int_{\mathcal{V}}\rho(\textbf{x})d^3\textbf{x}$$

(using the notation of Jackson)

It is only when the charge density is uniform over the entire region, that this definition is equivalent to $\rho=\text{charge}/\text{volume}$.

Is the charge density of a spherical shell really uniform over the entire volume of the sphere?

9. Feb 2, 2010

### kreil

ohh ok, i didnt think to use that definition.

I suppose the charge density would not be uniform since the charge is localized to the surface and is not present inside/outside the sphere.

how do i proceed using that definition?

10. Feb 2, 2010

### gabbagabbahey

Okay, so you know the following things:

(1)The charge density is zero everywhere except on the surface ($r=a$) where it is uniformly distributed
(2)The total charge is Q

Now, what else can we say?

Let's look at te above definition:

$$Q=\int_{\mathcal{V}}\rho(\textbf{x})d^3\textbf{x}\implies \rho(\textbf{x})=\frac{dQ}{d^3\textbf{x}}$$

What is the volume element $d^3\textbf{x}$ in spherical coordinates?

11. Feb 2, 2010

### kreil

$$d^3 \textbf{x}=r^2 sin \theta dr d\theta d\phi$$

Last edited: Feb 2, 2010
12. Feb 2, 2010

### gabbagabbahey

Right, so

$$\rho(\textbf{x})=\frac{dQ}{r^2\sin\theta dr d\theta d\phi}$$

Now, clearly dQ will be zero everywhere except on the surface, and hence so will $\rho$.....so let's look at the surface...It has finite extent in both the azimuthal and polar dir ections ($\phi$ varies from zero to $2\pi$ and $\theta$ varies from zero to $\pi$ over the surface)....But what about the radial direction...for an infinitesimally thin surface, doesn't it have zero extent (meaning that $dr=0$)?....What does that tell you about $\rho$ at the surface?

13. Feb 2, 2010

### kreil

so rho is zero everywhere except at the surface (r=a), where it blows up to infinity (since if dr=0 p=dQ/0) so it can be identified with a delta function of the form $\delta(r-a)$?

14. Feb 2, 2010

### gabbagabbahey

You've listed two of the three qualities of that delta function (zero everywhere except at $r=a$, where it is infinite)....what about the 3rd property of a delta function....does the charge density satisfy that property as well?

If so, then surely you can say that it must be of the form $\rho(\textbf{r})=f(\textbf{r})\delta(r-a)$ where $f$ is some well behaved function....right?

15. Feb 2, 2010

### kreil

$$\int_{-\infty}^{+\infty} \delta(r-a)dr = 1$$

and we know $$\frac{1}{Q} \int_{\mathcal{V}}\rho(\textbf{x})d^3\textbf{x}=1$$

so it satisfies the property.

Therefore, $$\rho(\textbf{r})=f(\textbf{r})\delta(r-a)$$.

Now, how do we find f(r)? Would we plug this into the above integral and solve for f(a)?

16. Feb 2, 2010

### gabbagabbahey

The more general property I was looking for from you, is

$$\int_{-\infty}^{+\infty} f(r)\delta(r-a)dr = f(a)$$

which means that the integral of the charge density over any volume enclosing the surface must be finite (despite the density itself being infinite)

Well, plug this form for $\rho[/tex] into this integral you have and what do you get?...What is the simplest [itex]f(\textbf{r})$ that satisfies this relation?

17. Feb 2, 2010

### kreil

$$\int_{all}f(r)r^2\delta(r-a)dr \int_0^{\pi}sin\theta d \theta \int_0^{2 \pi}d \phi=Q$$

$$Q= 4 \pi a^2 f(a) \implies f(a)=\frac{Q}{4 \pi a^2}$$

So $$\rho(r)= \frac{Q}{4 \pi r^2}\delta(r-a)$$....?

18. Feb 2, 2010

### gabbagabbahey

Yup, $\rho(r)= \frac{Q}{4 \pi a^2}\delta(r-a)[/tex] would also work...in fact, there are several [itex]f(\textbf{r})$ that would work, but all are mathematically equivalent.

Now, apply similar reasoning to the rest f the charge distributions....se what you can come up with.

19. Feb 2, 2010

### kreil

thanks a lot for all your help and patience

20. Aug 25, 2011

### jfy4

Hi,

Using what was talked about in this thread I have been able to do (a) and (b), but I am still stuck on (c) and (d). Its the fact that there is continuous charge distribution from the center of the disc out to a finite radius. I'm not sure how to deal with that part. From what I gather, I need a delta function for the fact that the disc is flat in the z-direction, $\delta(z)$, but also there is an infinity at $r=0$. but I'm not sure how to add a delta function such that I still get a continuous range of values out to a finite radius.

Any help would be great, thanks,