Electric Field in a Non-symmetric Sphere (Purcell 1.16)

1. Jan 15, 2009

r16

1. The problem statement, all variables and given/known data
In the Berkley physics course E&M book (by Purcell) problem 1.16 is giving me some issues.

A sphere of radius a was filled with positive charge at uniform density $$\rho$$. Then a smaller sphere of radius a/2 was carved out, as shown in the figure

What are the direction and magnitude of the electric field at A? at B?

2. Relevant equations
Gauss's Law
$$\int_S \vec{E} \cdot d\vec{A} = 4 \pi q$$

3. The attempt at a solution
I am having difficulty figuring out the strength of the electric field at point A. I know that E inside of a hollow sphere with a constant surface charge $$\sigma$$is 0, so I imagined a sphere of radius $$a/2$$ cut out from the center of a sphere of radius $$a$$ of a constant charge density. Due to superposition, there is no electric field inside because everything cancels out. Then I imagined moving the hollow sphere up a distance $$dr$$ along the z axis (the x and y axis is symmetrical). There is now $$2\rho \pi (r (1-b/a))^2 dr$$ -where $$a$$ is the radius of the big sphere and b is the distance the center of the hollow sphere is from the center of the big sphere-difference in charge between the two hemispheres-assuming a right circular cylinder for the differential volume. Now there is not an equal distribution of charge outside the sphere and there should be a resultant electric field inside, pointing in the +z direction because there is more charge in the bottom hemisphere vs the top hemisphere. I have a hard time figuring out how to calculate the magnitude of E at the center of the radius-$$a$$ sphere over the non-symmetric resultant charge distribution of the sphere.

Per gauss's law, there is no net electric flux through the hollow sphere and thus no charge enclosed (which makes sense). This was my original answer but it doesn't seem right.

For part B, I just assumed that superposition holds even though the geometry of the charge configuration is not completely spherical. The resultant charge $$Q = \rho (4/3 \pi r^3 - 4/3 \pi (r/2)^3)$$ acts as a point charge and then I applied gauss's law for a point charge. Is that correct as well?

Last edited: Jan 16, 2009
2. Jan 16, 2009

Redbelly98

Staff Emeritus
It seems like you're overcomplicating things somewhat. The principle of superposition you mentioned is the key here.

You'll need to calculate:

The field of a sphere of radius a, charge density ρ.
The field of a sphere of radius a/2, charge density -ρ.

3. Jan 16, 2009

r16

I like thinking things through quite thoroughly and i often over complicate, ive been dealing with this issue for as long as i can remember

I'm confused, isn't a charge density of -ρ different than there just being no charge in there at all? Wouldn't this just make a dipole-moment-ish field? How would the superposition work moving the smaller sphere around inside of the larger sphere?

I did have a flash of inspiration and figured out how to solve the problem. If the hollow sphere is centered at +a/2, I'll draw my Gaussian surface (a sphere) at -a/2 with the center of the large sphere being the origin. This way I can get the flux at the center of the large sphere and the odd shaped sides are symmetric and will cancel out.

This raises another problem, however for point B. I think the way I solved it before was incorrect, because I was assuming all the charge inside of a gaussian sphere enclosing the large sphere with the whole acts at the center. With my new gaussian sphere I am drawing at -a/2, point B is on the other side, so it will just be the opposite direction of the field at A. Both of these the approaches cannot be correct. I think the latter is correct.

4. Jan 16, 2009

Redbelly98

Staff Emeritus
Inside the smaller-sphere's volume, you can think of the charge density as a combination of +ρ (from the larger sphere) and -ρ (from a hypothetical -ρ, radius a/2, sphere), which combine to give us the situation shown in your figure.