Interpreting charge density with delta functions for surface charge

[homology]

Okay so say we have charge Q on a 2-sphere of radius R then the charge distribution will be rho=(Q/2piR^2)delta(r-R), which gives Q when integrated over space.

1) So my question is, what does this say about rho? To me, it says that rho is zero everywhere except on the surface of the sphere where it is infinite.

P.S. yes I know the delta function is really a distribution. What I'm asking for is a Physicist's interpretation of the charge density.

Thanks,

kevin

 

[vincentchan]

To me, it says that rho is zero everywhere except on the surface of the sphere where it is infinite.

you are right.. you can imagine there are total charge Q distribute on an infinite thin spherical metal sheet.. since the metal sheet is so thin, pho=Q/V goes to infinite on the surface of sphere

 

[Crosson]

Physically, there is a skin depth to the current of a few nanometers in a good conductor. But for things like surface integrals it is convenient to regard it as infinitesimally thin so that we can use the calculus.

Sometimes you just have to step outside the math (lthe way we all know there is no ifinite force do to a charged point particle just because of 1/r^2).