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VortexLattice
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VOLUME charge density from SURFACE/LINE charge density??
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!
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!