# Surface charge and volume charge density mathematical confusion

1. Apr 11, 2014

### user3

If you have a charged solid sphere with uniform volume charge density ρ, then the total charge on the sphere is

Q = ρ*4/3*∏*R^3 , where R is the radius of the sphere.

Now, if you also know that any single spherical shell from within the sphere has uniform surface charge density σ.

Then the total charge on the sphere could also be written as ∫σ 4 ∏ r^2 dr with the limits from 0 to R. That would give

Q = σ*4/3*∏*R^3

but how is σ = ρ ?

2. Apr 11, 2014

### suvendu

How can you write Q= ∫σ 4 ∏ r^2 dr?
It is Q= ∫σda
where da=rsinθdθdø

That will come out as σ.4πr^2.

3. Apr 11, 2014

### user3

Q= ∫σda is the charge on a spherical shell, a layer from the infinitely many that form the solid sphere. The total charge of the sphere is what I need.

4. Apr 11, 2014

### CAF123

Yes, you are right to question this. One is a volume charge density and the other is a surface charge density and so at the very least they should be related by a single power of R.

To obtain the area of a spherical shell of thickness dr, find the area of a sphere of radius r+dr and subtract this from the area of a sphere of radius r, supposing that dr is very small.

Alternatively, the integral ∫4πr2dr from 0 to R gives the volume enclosed by a sphere of radius R. Find this and multiply by ρ (since ρ is not a function of r) and you get the same answer.

5. Apr 11, 2014

### TSny

It might help to consider a simpler analog. Suppose you have charge uniformly spread along the x axis with a linear charge density λ. So, if you picked an infinitesimal interval dx, the charge in dx would be λdx. If you picked a mathematical point on the x-axis, how much charge would that point have?

You have a similar situation with the sphere. If you pick a mathematical surface in the sphere, how much charge would that surface have?

In order to capture some charge, you need to have a "surface" with some thickness, dr.

6. Apr 11, 2014

### Tanya Sharma

This isn't correct .4πr2dr isn't the surface area of the spherical shell .Rather 4πr2dr is the volume of the spherical shell at a distance r from the center having thickness dr .This volume needs to be multiplied by the volume charge density in order to get the charge within the spherical shell . Further this charge needs to be summed up covering the entire volume of the sphere.

You are calculating charge within the sphere i.e charge within the volume of the sphere ,so volume charge density is required .

Moreover , Q = σ*4/3*∏*R^3 is dimensionally incorrect .

Last edited: Apr 11, 2014
7. Apr 11, 2014

### user3

I get it. Thank you all.