# Electromagnetism: Charge Density of a Shell

1. Sep 6, 2009

### fusionshrimp

1. The problem statement, all variables and given/known data
We learned in E&M about the Dirac Delta function and its applications to representing the densities of point charges in space. For example, the charge density of space with a point charge q at the origin is given by q*delta(r). How does this density representation work with continuous distributions, such as a charged spherical shell?

2. Relevant equations
None really

3. The attempt at a solution
Haven't got a clue, just need a hint to point me in the right direction. New to all this dirac delta nonsense

2. Sep 7, 2009

### latentcorpse

do you havea copy of Riley, Hobson, Bense? it was good when i did this.

basically the Dirac delta function "is" the continuous analogue of the Kronecker delta.

one definition is:

$\int_{-\infty}^{\infty} f(x) \delta(x-x_0) dx=f(x_0)$ for 1D or
$\int \int \int f(r) \delta(r-r_0) dV=f(r_0)$ for 3D

now let's say you have a sphere containing a uniformly distributed charge, the total charge of said sphere is $Q=\int \int \int \rho(r) dV$ by definition where $\rho(r)$ is the charge density inside the sphere.

you're told that instead of having a sphere of uniformly distributed charge, you have instead just a point charge at the centre of the sphere (i.e. located at $r=0$ in that particular coordinate system). In this situation we can describe the charge density as $\rho(r)=q \delta(r)$ where q is the charge of the point charge.

now we want to find the total charge:
$Q=\int \int \int q \delta(r) dV=\int \int \int q \delta(r-0) dV = q$ using the definition of the Dirac delta function.

hopefully that helps as to why you can write $\rho(r)=q \delta(r)$

3. Sep 7, 2009

### kuruman

Think of it this way: For a charged spherical shell of radius R, the volume charge distribution is zero everywhere in space except at r = R where the volume charge distribution is infinite (because the thickness of the shell is zero.) Therefore, we write the volume charge density as

$$\rho(r)=A\delta(r - R)$$

If you know that the total charge on the shell is Q, you find constant A by integrating the volume charge distribution over all space (triple integral) and setting it equal to Q.

4. Sep 7, 2009

### fusionshrimp

Ok, thanks. That makes sense now; so now I should say, for a volume containing a spherical shell of radius R centered at the origin:

$$\int_V \rho\cdot\delta (r-R)\cdot r^2\,dr\,sin\,\theta\, d\theta\, d\varphi=Q$$

$$\rho\int_0^\infty r^2 \delta (r-R)\, dr \int_0^\pi sin\,\theta\, d\theta \int_0^{2\pi} \, d\varphi$$

$$\rho\cdot R^2\cdot 4\pi = Q$$

Therefore the volume charge density should be $$\rho = {Q\over 4\pi R^2}\cdot\delta(r-R)$$ for the total charge to come out to Q, right? (Sorry, I'm not very experienced with TeX)

5. Sep 7, 2009

### fusionshrimp

One more thing, does this mean that the one-dimensional delta function is endowed with the units of 1/length?

6. Sep 7, 2009

### kuruman

Yes, if the independent variable has dimensions of length.