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VOLUME charge density from SURFACE/LINE charge density?

by VortexLattice
Tags: charge, density, surface or line, volume
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VortexLattice
#1
Jan5-12, 02:18 PM
P: 146
So I'm doing a problem from Jackson, but this is a more general question anyway. He says that there is a charge Q uniformly spread over a spherical shell of radius R, and asks for the three dimensional charge density ρ(x).

Obviously, there has to be a delta function δ(r - R), because charge can only be found at the distance R from the center of the sphere. And I think I got the answer right, because if I integrate ρ(x) over all space, I get Q (the right answer). But something is going wrong in my math -- units aren't matching up. I basically have to kind of write down the answer, and tack on units at the end (which isn't rigorous, but is this because it's not a very physical problem (i.e., always a finite width to things?) ?).

Here's what I tried:

I said the shell must have an area of [itex]4πR^2[/itex] and therefore a surface charge density of [itex]σ = \frac{Q}{4πR^2}[/itex].

So at the angle [itex]θ,\phi[/itex], there is the small bit of charge [itex] δq = σ r^2 sin(θ) δθ δ\phi[/itex]. Likewise, at this location, if we knew the volume charge density, it would be [itex] δq = ρ(x) r^2 sin(θ) δr δθ δ\phi[/itex].

So when I set these two equal, I get [itex] ρ(x) = \frac{1}{δr} \frac{Q}{4πR^2} [/itex]...which is a weird/useless answer to me.

I thought about it for a minute, and just intuitively got that the answer must be [itex] ρ(x) = δ(r - R) \frac{Q}{4πR^2} \frac{1}{m^3}[/itex]. This works if you integrate it over all space, but the units business is sketchy... R should have units of m, so this should really only look like a surface charge density. If I just treat R as a unitless variable and then tack on the [itex]\frac{1}{m^3}[/itex] then it works out, but there's definitely some stuff being skipped there...

Can someone explain how this can be done rigorously?

Thanks!!
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Antiphon
#2
Jan6-12, 12:18 AM
P: 1,781
The volume expression is supposed to have the delta function of (r-r0) in it. When you integrate on dr you should recover the surface charge expression. Instead of relating them through the integration over r, you merely equated them- hence the nonsensical delta function in the denominator.
VortexLattice
#3
Jan14-12, 11:00 PM
P: 146
Quote Quote by Antiphon View Post
The volume expression is supposed to have the delta function of (r-r0) in it. When you integrate on dr you should recover the surface charge expression. Instead of relating them through the integration over r, you merely equated them- hence the nonsensical delta function in the denominator.
Thanks, but this doesn't really answer my question... I know the answer I posted is correct (because I get the right answer if I integrate it) but the units business is funny, and that's what I'm not comfortable with.

Pengwuino
#4
Jan14-12, 11:10 PM
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VOLUME charge density from SURFACE/LINE charge density?

I'm confused, what is this [itex]1/m^3[/itex]??
VortexLattice
#5
Jan14-12, 11:24 PM
P: 146
Quote Quote by Pengwuino View Post
I'm confused, what is this [itex]1/m^3[/itex]??
Inverse meters cubed. My main problem with this problem is a kind of nitpicky one, but I'm still not sure how to solve it rigorously. Normally, if you have a charge density, it's in coulombs/m^3, right? But you don't write that out, it's included in the variables. Like, if we had a conducting sphere of radius R with charge Q on it, we would have ρ = Q/((4/3)πR^3), right? So the units are all in Q and R^3.

But if you look at the answer I got (that works), the units seem to be coulombs/m^2, unless you just explicitly declare it to have the right units (which is ghetto). Somehow, the delta function must have units of 1/m I guess?
Pengwuino
#6
Jan14-12, 11:30 PM
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Quote Quote by VortexLattice View Post
I thought about it for a minute, and just intuitively got that the answer must be [itex] ρ(x) = δ(r - R) \frac{Q}{4πR^2} \frac{1}{m^3}[/itex].
Sorry, this is the part I don't get. Why are you multiplying units with a formula?
VortexLattice
#7
Jan14-12, 11:41 PM
P: 146
Quote Quote by Pengwuino View Post
Sorry, this is the part I don't get. Why are you multiplying units with a formula?
Yeah, basically. That's the only way I can get units to work out. I know it's basically a wrong way of doing it, but I can't find a right one. Can you help me out?
Pengwuino
#8
Jan15-12, 12:12 AM
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Quote Quote by VortexLattice View Post
Yeah, basically. That's the only way I can get units to work out. I know it's basically a wrong way of doing it, but I can't find a right one. Can you help me out?
No, I mean, you're multiplying by dimensions, not by variables like you should. Do you mean the only way you can get it to work is if you say [itex]\rho \propto 1/R^5[/itex]?
Born2bwire
#9
Jan15-12, 01:07 AM
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I think mathematically the dirac delta function carries its own scaling of units here. Let's take the obvious problem of the charge distribution of a single point charge. That's simply,
[tex] \rho (\mathbf{r}) = q \delta (\mathbf{r}-\mathbf{r'}) [/tex]
What happens I believe is from the scaling property of the delta function. That is,
[tex] \int_{-\infty}^\infty \delta(\alpha x)\,dx
=\int_{-\infty}^\infty \delta(u)\,\frac{du}{|\alpha|}
=\frac{1}{|\alpha|} [/tex]
If the delta function is n-dimensional, then,
[tex]\delta(\alpha\mathbf{x}) = |\alpha|^{-n}\delta(\mathbf{x})[/tex]
In the end, we see that the dirac distribution will inherently scale by the dimensions of its argument when we take the integration. So your dirac distribution is only one dimensional and so it would reduce the dimensions of your integrand by 1/m when you perform the integration over the distribution.

EDIT: When you think about it, this inherently has to be true. The dirac delta distribution takes an n-dimensional integration and maps it onto a scalar. There has to be an associated reduction of n-dimensions for this to keep the bookkeeping straight.
vanhees71
#10
Jan15-12, 04:26 AM
Sci Advisor
Thanks
P: 2,333
A distribution is a distribution, not a function! A distribution is defined as a linear functional on an appropriate space of test functions. A convenient space of test functions is the space of fast falling smooth functions, i.e., functions which are differentiable at any order and which go to 0 at inifinity faster than any power of the arguments. This is a subspace of many convenient function spaces, particularly the Hilbert space of square integrable functions.

Take the here discussed case of the various singular charge distributions in electrodynamics. A usual (nonsingular) charge density gives the electric charge per volume at any point in space time in the sense of a macroscopic distribution (i.e., coarse grained over microscopically large but macroscopically small volume elements). The total charge in a region [itex]V[/itex] is at time [itex]t[/itex] thus given by

[tex]Q_V(t)=\int_{V} \mathrm{d}^3 \vec{x} \rho(t,\vec{x}).[/tex]

Now let' discuss the various cases of singular charge distributions usually discussed in classical electrodynamics courses.

First, there's the case of a single point charge [itex]q[/itex], sitting at the origin of the coordinate system. To discribe it by a charge distribution in the above sense of "charge per volume", we must define a generalized function or distribution in the above given sense of functional analysis with the property that [itex]Q_V=q[/itex] if [itex]V[/itex] contains the origin and [itex]Q_V=0[/itex] if it does not (for any finite region [itex]V[/itex] in space). This is the Dirac [itex]\delta[/itex] distribution:

[tex]Q_V=q \int_{V} \mathrm{d}^3 \vec{x} \delta^{(3)}(\vec{x}).[/tex]

This means that, as any charge distribution, this singular one for a point charge must have the dimension of "charge per volume", i.e., [itex]\delta^{(3)}(\vec{x})[/itex] must have the dimension of "1/volume=1/length^3".

Now let's see, how to describe surface charges. The most simple way to do this is to define the surface implicitly by

[tex]F(\vec{x})=0.[/tex]

E.g., for a sphere around the origin of radius [itex]R[/itex], we have

[tex]F(\vec{x})=\vec{x}^2-R.[/tex]

Now let [itex]\sigma(\vec{x})[/itex] be the charge per surface element at the point [itex]\vec{x}[/itex]. The charge on an area [itex]A[/itex] on the surface is given by

[tex]Q_A=\int_{A} \mathrm{d}A \sigma(\vec{x}).[/tex]

Now, to describe this distribution as a (singular) charge distribution per volume, take an infinitesimal cylindrical volume element with base area [itex]\mathrm{d} A[/itex] and height [itex]\mathrm{d} h[/itex]. This cylinder is taken perpendicular to the surface element. The charge in this little volume is

[tex]Q_{\mathrm{d} A}=\sigma(\vec{x}) \mathrm{d} A \stackrel{!}{=} \rho(\vec{x}) \mathrm{d} A \mathrm{d} h.[/tex]

Thus we have

[tex]\rho(\vec{x})=\frac{1}{\mathrm{d} h} \sigma(\vec{x}) .[/tex]

Now we express the deviation [itex]\mathrm{d} h[/itex] from the surface with help of the function [itex]F[/itex]: If you change [itex]F[/itex] by [itex]\mathrm{d} F[/itex] the perpendicular distance from the surface is given by

[tex]\mathrm{d} F=|\vec{\nabla} F| \mathrm{d} h,[/tex]

because [itex]\vec{\nabla} F[/itex] is perpendicular to the surface. Thus we have

[tex]\rho(\vec{x})=\frac{1}{\mathrm{d} h} \sigma = \sigma \frac{|\vec{\nabla} F|}{\mathrm{d} F}. [/tex]

From this we find, taking the limit [itex]\mathrm{d} h \rightarrow 0[/itex] of this procedure

[tex]\rho(\vec{x})=\sigma(\vec{x}) |\vec{\nabla} F(\vec{x})| \delta[F(\vec{x})].[/tex]

Indeed, then we have, if [itex]A[/itex] is the part of the surface inside the volume [itex]V[/itex]

[tex]\int_V \mathrm{d}^3 \vec{x} \sigma(\vec{x}) \delta[F(\vec{x})] = \int_{A} \mathrm{d} A \int \mathrm{d} h \delta(h) \sigma(\vec{x}) = \int_{A} \mathrm{d} A \sigma(\vec{x})=Q_{A}.[/tex]

As it should be.

For our example of the sphere we have in our parametrization

[tex]|\vec{\nabla} F|=2|\vec{x}|=:2r,[/tex]

and thus the charge density is

[tex]\rho(\vec{x})=2r \delta(\vec{x}^2-R^2) \sigma(\vec{x}) =2r \delta(r^2-R^2) \sigma(\vec{x})=\delta(r-R) \sigma(\vec{x}).[/tex]


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