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Often in an E&M problem, I'm having to "chop" an extended object into an infinite sum of smaller extended objects which I know more about to find a potential or electric field or whatever. The part I'm having trouble with is how to convert the surface density (line density) to volume density (surface density) and vice versa.

In one problem I did recently, I just thought maybe I should just collapse a dimension in volume density $\rho$ to get surface density $\sigma$ via: $\rho dr = \sigma$. It gave me the right answer, but I doubt that's the right way to think about it and that that formula will always work.

Right now I'm trying to "cut" a cylinder of uniform volume density $\rho$ into disks of uniform surface density $\sigma$. I thought maybe the right approach would be to relate the total charges. I've got $$Q_{\text{cylinder}}=\int \rho d\tau=\rho \pi r^2 h\\ \text{and}\\ Q_{\text{disk}}=\int \sigma dS=\sigma \pi r^2\; .$$ However then I'm at a loss as to how I should relate $Q_\text{cylinder}$ and $Q_\text{disk}$. What's the general process here?

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I have personally also contemplated this issue, and have come up with a simple solution that is satisfactory, to me at least. I'm sure this can also be found in many textbooks. In general, we have

$$\tag{$\star$} Q=\int \rho\ d\tau$$

because we are considering a three dimensional space. Intuitively, we feel it should be possible to talk about a three dimensional charge distribution in every case. The question is how to conceptualize this when discussing surface, line or point charges. The solution comes in the form of the Dirac delta distribution (or function, depending who you ask).

Let's take a look at an example: Consider a 2-sphere of radius $R$, with some charge distribution $\sigma(\theta,\phi)$ on it. What is the three dimensional charge distribution $\rho(r,\theta,\phi)$ corresponding to this situation? Like I said, we have to use the Dirac delta:

$$\rho(r,\theta,\phi)=\delta(r-R)\sigma(\theta,\phi)$$

Now, $(\star)$ gives us: $$ Q=\int\rho\ d\tau=\int_{0}^{2\pi}\int_{0}^\pi\int_0^\infty \rho\ r^2\sin\theta\ dr\ d\theta\ d\phi=R^2\int_{0}^{2\pi}\int_{0}^\pi\sigma\ \sin\theta\ d\theta\ d\phi$$

Similarly, when considering a line or point charge, one uses two or three Dirac delta's to describe the distribution in 3-space.

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Given a volume with finite charge density $\rho$, the surface charge density $\sigma$ of an embedded surface will be infinitesimal. The charge of the volume is the integral of the infinitesimal charges of the embedded surfaces. Conversely, a finite surface charge density would give you an infinite charge density there - specifically a delta function which, integrated over, would still be a finite total charge. In your example above, the cylinder and disc charges are related by:

$$Q_\text{cylinder} = \int_\text{all discs} dQ_\text{disc} = \int d\sigma_\text{disc} \ dS_\text{disc} = \int \rho\ dh\ dS_\text{disc} =\int \rho\ dh\ r\ dr\ d\theta$$

Here I've interpreted the surface charge density $d\sigma$ as the volume charge density times the thickness of the surface $\rho\ dh$. But integrating over the surface charge density is unintuitive and obscures how the surfaces add up to a volume.

A better way to look at it: You don't slice the integration volume into surfaces; you slice it into thin, infinitesimal volumes. Then you don't need to convert between surface and volume densities and it's clear how the chunks add up to the total volume. In your example, the volume of each disc is $dV_\text{disc} = \pi R^2 dh$ and you would integrate

$$Q_\text{cylinder} = \int_\text{all discs} \rho\ dV_\text{disc} = \int_0^H \rho \pi R^2 dh$$

Now, the "general process" to solve these problems is to directly evaluate the volume integral. Chopping up a volume and integrating the slices is just a way of skipping some of the steps you'd do evaluating the volume integral directly. If you don't know what to use as the volume element $dV$ - it depends on the coordinates and you'll almost always be using one of:

$$\\\text{Cartesian}: dV = dx\ dy\ dz \\ \text{Polar}: dV = r\ dr\ dh\ d\theta \\ \text{Spherical}: dV = r^2 \sin\theta\ dr\ d\theta\ d\phi$$

(You might want to look at the Wikipedia articles on volume elements and the Jacobian)

The following gets a little rambly but discusses why you should sometimes take the slice approach and illustrates where the volume element comes from:

Alternatively, it is sometimes faster to break the volume down into slices whose volume you can write down, and on which the charge density is constant. This allows you to do an integral over a single variable rather than the volume.

If you don't know the volume of the slice, or the function to integrate varies over the finite dimensions of the slice, you can slice the slice! This gives you another nested integral, and one of the finite dimensions of the slice becomes infinitesimal. You can repeat this until all the sides are infinitesimal, and you will be left with a parallelepiped, which is just the volume element $dV$ and you're back to doing the volume integral.

To illustrate this, consider a cylinder with charge density $\rho(r)$. The useful slicing scheme (to get surfaces of constant $\rho$) is to slice this cylinder into concentric tubes; each tube is of height $H,$ inner radius $r,$ and thickness $dr$ so the outer radius is $r+dr.$ I know the volume of this tube is

$$dV_\text{tube} = 2\pi rH\ dr$$

so I need to integrate

$$Q = \int \rho(r) dV_\text{tube} = \int_0^R \rho(r) 2\pi rH\ dr$$

If I didn't know the volume of a tube like this, or $\rho$ had a dependency on another variable, I'd slice it again into rings of height $dh,$ thickness $dr,$ and radius $r.$ Say I still don't know what the volume of these rings are, so I slice up my rings radially along the $\theta$ axis into rectangular parallelepipeds whose dimensions are $dV = dh \times dr \times r\ d\theta$. We're back to doing the volume integral in polar coordinates.

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I understood it this way.

Example:A uniformely charged sphere has linear charge density lambda.And sigma be surface charge density. lambda=sigma d(Surface area)/dx. for a cylinder Surface area=2 pi r x (where x is variable height)

Thus we get lambda=sigma(2 pi r.1)

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  • $\begingroup$ Please try to learn latex, it is very useful. $\endgroup$ – peterh Oct 8 '17 at 12:32
  • $\begingroup$ What is latex?. $\endgroup$ – Krishna Deshmukh Nov 15 '17 at 18:04
  • $\begingroup$ With it, you can insert beautiful formulas into your posts. Check this: physics.meta.stackexchange.com/a/9134/32426 $\endgroup$ – peterh Nov 16 '17 at 0:04

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