# E Field Hollow sphere

1. Sep 10, 2013

### bowlbase

1. The problem statement, all variables and given/known data
Standard E field problem where I'm to find the field at 3 positions of a hollow sphere that has a charge density k/r^2
r ≤ a
a ≤ r < b
b ≤ r

2. Relevant equations
∫Eda=Q/ε

3. The attempt at a solution

I guess the thing that is tripping me up are the limits. I know that before the radius gets to the inside surface of the sphere the E field is 0. But at a there must be something. But I've gone no measurable distance into a charge region, correct? So it must still be zero, I think.

The second part is just the volume integral of the density from a to r divided by the surface area.

And for three I want to say it will behave like a point charge but the hollowness of the sphere leads me to believe this isnt true. But rather, it should be the whole charge of the sphere from a to b.

I just want to understand conceptually what I'm doing. So no solutions really needed.

Thanks.

2. Sep 10, 2013

### Tsunoyukami

I've been reviewing electrostatics recently and actually just worked this problem this morning. This problem is a simple application of Gauss' Law, which is one of the most powerful tools for solving problems in E&M (at least when symmetry permits).

If you're having difficulty understanding the problem conceptually I suggest you reread the section on Gauss' Law. Presumably you're using Griffths' Introduction to Electrodynamics, and from hereon I will refer to the third edition (which is the edition I own). I recommend reviewing Example 2.2 on page 70, and completing Problems 2.11, 2.12 and 2.14 on page 75 in order to test your understanding of Gauss' Law so far.

Gauss' Law can be stated in the following manner:

$$\int_{S} \stackrel{\rightarrow}{E} \cdot d\stackrel{\rightarrow}{a} = \frac{Q_{in}}{\epsilon_{o}}$$

As you have stated, the electric field will be zero in the region where r < a (because the enclosed charge up to this point is 0). If you work the problem carefully you will notice that it never asks you to find the field when r = a or when r = b - this is something that will be discussed later in Griffths (see section 2.3.5, page 87 and Problem 2.30 on page 90) - and so your concern about what happens at r = a should not impede your progress with the problem.

In the second region recall that Gauss' Law is interested only in the enclosed charge which will change as a function of the radius (you can find the enclosed charge by completing the integral found on page 68:

$$Q_{in} = \int_{V} \rho d\tau$$

You can calculate the field in the third region in a similar manner.

3. Sep 10, 2013

### bowlbase

I am using Griffith's but the problem was adjusted by the teacher to be as I stated. I've performed this evaluation before but the limits were never as they are here. Well, the second and third are but the first is not.

4. Sep 10, 2013

### Tsunoyukami

Ah, my apologies. I thought you had simply typed the inequalities incorrectly. As I stated, I would refer to section 2.3.5 Summary: Electrostatic Boundary Conditions on page 87. The section essentially says that the electric field is discontinuous at boundaries.

I believe that for your purposes solving the problem under the conditions r < a, a < r < b, and r < b should suffice. I think its reasonable to assume that when r = a the "enclosed" charge is zero because when r = a there is no charge inside this surface (it is instead an infinitesimal amount of charge smeared over the surface of the sphere described by $V = \frac{4}{3} \pi r^{3}$ which is not contained within).

Furthermore, it only makes sense to me that the field in each of the regions should have a formula valid for the entire region - that is, for r < a we know E = 0; if the problems asks us to find the field for r <= a I would expect to have a single expression valid in that entire region, including when r = a, namely E = 0.

However, if in doubt I would ask your professor as to what he considers the proper interpretation to be.

5. Sep 10, 2013

### bowlbase

I'll be asking tomorrow during office hours. I've just gone over 2.3.5 and it says for boundaries that are curved I need to use a "very small A" which I don't entirely understand. I'll work the problem tonight as the book has it and adjust it later I guess. Like you said, I would expect to have an equation that explains from the origin to a to describe the internal field. However, he may expect a discussion of the field exactly at 'a' as an addition to E=0 before it. Though, still, it must be so small as to be basically 0 anyway.

The reason that you need to pick A very small for curved surfaces (or surfaces where $\sigma$ is non-constant) is to be able to draw a very small Gaussian pillbox that will permit the symmetry for which the following results are valid (ie. that the sides of the pillbox contribute nothing to the flux and that the component of the electric field parallel to the surface is continuous across the surface).