Constant Electric Field in Sphere-ception Cavity?

[johnqwertyful]

Homework Statement

There is a sphere of uniform charge, with a hollow sphere inside it, centered on the x-axis. Show that the electric field in the cavity is constant, and find its value.

Homework Equations

The Attempt at a Solution

The problem suggested to use superposition. So I thought of breaking it into two parts. The first one being the biggest sphere around the cavity, and the second being everything else. The first part by symmetry cancels out, then we have the second part. There I am stuck. Any help please?

 

[johnqwertyful]

I mean I could just use Gauss' law. Flux would be the charge, 0. So constant field. Right? But it says to use superposition.

 

[TSny]

I mean I could just use Gauss' law. Flux would be the charge, 0. So constant field. Right? But it says to use superposition.

Zero charge inside a Gaussian surface only means that the total flux through the surface is zero. It does not imply that the field is uniform within the Gaussian surface.

Think about creating the cavity by superimposing charge densities of opposite sign.

 

[johnqwertyful]

So it's the same as if you were to have the sphere with a charge density, σ, but superimpose a charge density -σ in the one part.

Since both charge densities are constant, the resulting field should be constant as well, right?

 

[TSny]

So it's the same as if you were to have the sphere with a charge density, σ, but superimpose a charge density -σ in the one part.

Yes. (Note, I think most people use σ for surface charge densities and ρ for volume charge densities. But, that's not too important.)

Since both charge densities are constant, the resulting field should be constant as well, right?

Unfortunately, it's not that simple. Uniform charge densities do not imply uniform electric field.

Since you are going to add together the fields of two spheres of uniform density, you are going to need an expression for E inside a sphere of uniform density.

 

[johnqwertyful]

Yes. (Note, I think most people use σ for surface charge densities and ρ for volume charge densities. But, that's not too important.)



Unfortunately, it's not that simple. Uniform charge densities do not imply uniform electric field.

Since you are going to add together the fields of two spheres of uniform density, you are going to need an expression for E inside a sphere of uniform density.

Yes, it's ρ in the problem. The rest makes sense, thank you!

 

[sudhirking]

Dear Johnqwertyful,

Before we move on to the problem let us consider the following logic on natural numbers. It is agreed that the identity number 0 does nothing when added to some other number x. That is x +0 is of course x. And another way of writing 0 is by a number subtracted from its very self like y-y! So given any number, say 5, I can add to it by a special form of 0 without changing the value. That is 5 = 5 + (y-y)!

We will be doing the E-M equivalent of this mathematical trickery. Consider an entire sphere uniformly charged with some voluem density ρ and radius R. The actual sphere is just this with a spherical cavity palced about the center with a radius r (and hopefully it is understaood that r ≤ R for obvious reasons).

The electric field ordinarily inside an entirely charged of uniform density is proportional to the radial distance measured from the centerd pointing outwards. The exact relation may be pbtained by abusing Guass's Law (due to my inablity to type equations I will hide the details). Make sure though you express it in vector form.

You might charge me with wasting your time as we have calculated the electric field for an entireyl different problem! But consider this: the electric field of the entire spehere charge MINUS the electric field of a smaller sphere with the same center and smaller radii r. Isn't this just the electric field of the sphere with the cavity! If this is still not convincing, consider a the electric field of the sphere with the cavity PLUS the electric field of the cavity endowed with charge. Isn't that just the electric field of the entire sphere charged!

Another way of seeing this in terms of the original mathematics analogy:
(the electric field of sphere with cavity) = (the electric field of sphere with cavity) + 0
where I choose 0 cleverly enough so that we get this equation:

E(sphere with cavity) = E(sphere with cavity) + {E(cavity charged) - E(cavity charged)}

Invoking "associativity" we will shift out brakets from the form of 0 to:

E(sphere with cavity) = {E(sphere with cavity) + E(cavity charged)} - E(cavity charged)

Recognizing the sum in brackets is nothing bu tthe electric field inside of sphere totally charged,

E(sphere with cavity) = E(sphere entirely) - E(cavity charged)

We may even move the negative "inside", much like x - y = x + (-1)y:

E(sphere with cavity) = E(sphere entirely) + E(cavity charged with reverse polarity)

I will hide the rest of the details but they require just the use of electric fields inside unifmorly charged spheres.

 

[vanhees71]

I guess, it's electrostatics. Then, I'd solve the Poisson equation
$$\Delta \Phi=-\rho, \quad \vec{E}=-\vec{\nabla} \Phi$$
for the given problem. It's spherically symmetric, so you can make the ansatz $\phi(\vec{x})=\phi(r)$ with $r=|\vec{x}|$ and use the Laplace operator in spherical coordinates. Then you can solve this equation for the given charge distribution.