# Electrostatics - spherically symmetric charge density

PhysicsUnderg

## Homework Statement

Imagine a spherically symmetric charge density

p(x) = Cr for r≤a and 0 for r>a

(a) determine the electric field E(x) and potential V(r). Notice that V(r) and E(x) are continuous at r=a.
(b) Now suppose additional charge is placed uniformly on the surface at r=a, with surface density σ0. Determine E(x) and V(r) for this case. E(x) is discontinuous at r=a, but V(r) is continuous. Explain why.

## Homework Equations

I am not sure where even to start... can someone point me in the right direction?

dacruick

If you have a sphere with a spherically symmetric charge density, that charge can be simplified as a point charge. So now you need to find the magnitude of that point charge.

Once you have that, it shouldn't be hard to find E(r) and V(r).

Chopin

If you have a sphere with a spherically symmetric charge density, that charge can be simplified as a point charge. So now you need to find the magnitude of that point charge.

Once you have that, it shouldn't be hard to find E(r) and V(r).

Only outside of the sphere, though, right? This problem is looking inside of the sphere too, so that's not going to work.

PhysicsUnderg: Start with trying to find the electric field. Do you know which equation connects the electric field with charge density?

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dacruick

Only outside of the sphere, though, right? This problem is looking inside of the sphere too, so that's not going to work.

If the sphere is radially charged doesn't that mean that it can also be so on the inside?

Its been a while since I've done this stuff.

guyguy

This seems odd to me for the charge density

$$\rho (r) = C r$$

I don't think you can simplify to a point charge, since the potential for a point charge drops as $$1/r^2$$ and this density says that there is more charge as you get further away from the origin. That is not the same field as a point charge.

dacruick

If your sphere is radially charged, that means there are an infinite amount of hollow spheres all with uniform charge densities with radii ranging from 0 to a.

Can't each of those hollow spheres be treated as a point charge?

You would just have to take the integral of the electric field * dq from 0 to r for r < a.

The only step in between would be to find the total charge at the centre due to each of the hollow spheres.

Chopin

You're correct that a spherical charge density can be considered as a series of spherical shells. The field generated by a spherical shell of charge falls off at $$1/r^2$$ outside of the sphere, meaning you can simplify it to a point charge. But inside the shell, the generated field is 0. So to calculate the field at any point, you can indeed model the charge distribution as a series of shells and integrate, but you only integrate up to the radius of the point of interest, not all the way out. That means the integral will be different at each point, and so the resulting field will not fall off at $$1/r^2$$ inside of the charge distribution.

Usually, the reason that you simplify a distribution to a point charge is because it allows you to calculate a single equivalent charge, and then easily determine the field at any point away from the distribution, because the field will just fall off at $$1/r^2$$. If you're still inside of the distribution, and the field isn't falling off at $$1/r^2$$, then this simplification doesn't do anything for you--you still have to do a separate integral for every point, so the math will be exactly the same as it was before.

I think the easiest way to solve this is just to ignore these kinds of simplifications for the moment, and go back to the relevant equations for electrostatics. Once they've been solved for this problem, the simplifications we've discussed here will emerge as a consequence of them (in fact I suspect that might have been the point of this problem in the first place.)

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dacruick

You're correct that a spherical charge density can be considered as a series of spherical shells. The field generated by a spherical shell of charge falls off at 1/r^2 outside of the sphere, meaning you can simplify it to a point charge. But inside the sphere, the generated field is 0. So to calculate the field at any point, you can indeed model the charge distribution as a series of shells and integrate, but you only integrate up to the radius of the point of interest, not all the way out. That means the integral will be different at each point, and so the resulting field will not fall off at 1/r^2 inside of the charge distribution.

Usually, the reason that you simplify a distribution to a point charge is because it allows you to calculate a single equivalent charge, and then easily determine the field at any point away from the distribution, because the field will just fall off at 1/r^2. If you're still inside of the distribution, and the field isn't falling off at 1/r^2, then this simplification doesn't do anything for you--you still have to do a separate integral for every point, so the math will be exactly the same as it was before.

I think the easiest way to solve this is just to ignore these kinds of simplifications for the moment, and go back to the relevant equations for electrostatics. Once they've been solved for this problem, the simplifications we've discussed here will emerge as a consequence of them (in fact I suspect that might have been the point of this problem in the first place.)

Ahh thats true, the "simplifcation" wouldn't be very much of one in this case. the understanding is the most important part of this. Although the person asking the question doesn't seem to be too interested in the answer.

And was that a pun on the "point of this problem"?

guyguy

I agree, because the main part of the problem is inside the sphere. You can use

$$\mathbf{E} = (1/4 \pi \epsilon_o) \int \frac{\rho}{r^3} \mathbf{r} dV$$

and then break down the differential volume element to integrate

PhysicsUnderg

I apologize for taking so long to reply... thank you to everyone who posted! :-) I am working out the problem now with your suggestions. If I find myself stuck again, I will be back... and I promise not to take a year to reply next time.