# Gauss' law and conducting spheres

1. May 31, 2012

### srn

1. The problem statement, all variables and given/known data
A conducting sphere of radius R2 has a central cavity of radius R1 that holds a charge q in its centre. Determine the electrical field for r > R2, r < R1 and R1 < r < R2 and determine the charge density induced by q.

I'm not allowed to include a link to my figure, but I'm sure you understand. In 2D it's 2 circles with a common centre, one has radius R2 and the other R1 where R1 < R2.

2. Relevant equations (not sure why it's not displaying LaTeX from this point on)
$∫e.dA = \frac{q}{\epsilon_0}$

3. The attempt at a solution
Q induces a charge on the sphere.

r < R2
The electrical field due to the sphere is zero (E-field inside hollow conductors in electrostatic equillibrium is zero). That means the E-field is only due the q.
For a random r < R2:
$E = \frac{q}{\epsilon_0}\cdot 4\pi r^2$

R1 < r < R2
$E \cdot 4\pi r^2 = \frac{q + (3/4 \pi r^3 - R_1^3)\rho}{\epsilon_0}$

r > R2
$E \cdot 4\pi r^2 = \frac{q + (3/4 \pi R_2^3 - R_1^3)\rho}{\epsilon_0}$

I think this is correct, however, I still need the charge that is incuded on the sphere. Intuitively I would say it is equal to q itself, because the electrical field inside (the solid sections) of the sphere should be zero. Though I'm confused by the 1/r^2 decay; for a uniform field it's easy...

Come to think of it, suppose you start off with the sphere and somehow turn the charge off. Suppose you then turn it on, the electrons would then migrate torwards the cavity. While that happens there is an electrical field inside the sphere, but after a while these electrons should settle, more specifically when the e-field generated by the seperation of e- and positive molecules counteracts the e-field generated by the charge. Right? So that would mean the e-field for r1 < r < r2 is zero?

second edit: then again, the field in r1 < r < r2 is opposite to that of the charge, so my solution for r > r2 cannot be correct as I add both. Do I need to substract the "weakening" of q's field as it passes through the sphere?

Last edited: May 31, 2012
2. May 31, 2012

### Sleepy_time

Hi srn. For your r<R1 (think you might have a typo and called it R2). It would be $\frac{1}{4{\pi}r^2}$. For R1<r<R2, electric fields inside a perfect conductor is always zero. You can also use that fact to determine the charge induced on the inside surface and thus the charge induced on the outer surface and so the E-field for r>R2.

3. May 31, 2012

### SammyS

Staff Emeritus
I have corrected some typos above, and made some of the LaTeX more readable.

I take it that the spheres are concentric and that the conducting sphere is electrically neutral.

Your solution for r < R1, should have the 4πr2 in the denominator.

$\displaystyle E = \frac{q}{4\pi r^2\epsilon_0}$

What is the electric field within conducting material itself under static conditions? Is it not zero?
The answer to this should tell you the electric field for R1 < r < R2. It plus Gauss's Law should also tell you how much charge resides on the inner surface of the cavity.​

The total charge on the conducting sphere should give you how much charge resides on the outer surface of the sphere.

4. Jun 1, 2012

### srn

Thank you both for replying and thank you for fixing my Latex, I'll be more accurate next time.

For r < R1: that should have been in the denominator indeed, sorry, clumsy, I did know that.

For R1 < r < R2: you're right, it should be zero. I was thorougly confused by an exercise in my book; calculating the electrical field in a solid charged sphere. They had a non-zero value, but upon reading it again it said insulating sphere. Given that it is conducting for this exercise, E = 0 for R1 < r < R2 since it is in electrostatic equillibrium. (I assume so, doesn't state otherwise.)

Now, constructing a Gaussian surface for R1 < r < R2; Gauss' law says $q_{in}$ must be zero since $E = 0$. There's still the charge $q$ however, so: (inner is the charge on the inside wall of the surrounding sphere) $q_{in} = q + q_{inner} = 0$ so that $q_{inner} = -q$.

Given that the outer shell has no charge, its outer surface must have charge $+q$.

For r > R2: the included charge is q + 0 = q so that
$E \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$
which is identical to the r < R1 field.

Which I do find a bit peculiar, I had expected it to be weakened by the opposing field in the conducting sphere but I'm probably thinking of it too concretely.

5. Jun 1, 2012

### cupid.callin

Well your solution appears to be correct.

Remember that for outside sphere r > R2 and for inside r < R1. And also R2 > R1
So any field you find at any point inside will be greater than field outside (by the sphere alone).

6. Jun 1, 2012

### srn

Yep, I realize that; though what I meant was that I would have suspected that the field for r > R2 would be weaker in comparison to a situation where there is no conducting sphere. I.e. it is weaked more than just from the distance from q. But I suppose from these results that the field for r > R2 is in no way affected by the counteracting field in the sphere due to the seperation of particles.

I was incorrectly comparing it to a Newtonian force, I think. E.g. suppose you have a rope that is attached to a wall and that experiences a tensile force at the free end. Suppose you put a stick through and perpendicular to the rope, and that you apply a force opposite the tensile force. The section between the stick and the wall would then experience less force.

But I guess I cannot compare them because the Newtonian force exists throughout the rope because it has been "transmitted" by action-reaction pairs on every small piece of rope, whereas the electrical field "exists" in the space and is not influenced by properties of the field nearby. Something like that?

7. Jun 1, 2012

### cupid.callin

No the charges inside the shell cannot influence the field outside. You can understand this inside shell, both +q and -q exist. So field due to these 2 charges at any point gets cancelled. Also, any new charge when introduced outside the shell also cannot change the field inside the shell. This is because on introducing a new charge, the surface charge will rearrange so as to cancel any effect of outside charge inside the body. This property of conductors is also used in electrostatic shielding.

PS: My phone is running low on battery so I will post rest of my reply later ...