# Homework Help: Electric Field Involving Gauss's Law Problem

1. Feb 5, 2010

### king_abu

1. The problem statement, all variables and given/known data
An atom that has Z protons, each with positive charge of magnitude e, reside in the nucleus at the center, surrounded by a uniformly distributed spherical cloud of radius R consisting of Z electrons, each with negative charge magnitude e, centered on the nucleus e. Use Gauss's Law to determine an expression for the electric field as a function of r, the distance from the center of the atom.
(1) Find the value of electric field for r<R (magnitude + direction)
(2) Find the value of electric field for R>r (mag + dir)
(3) If the atom were ionized, such that it was deficient by one electron, what would be the new expressions for the regions r<R and r>R

2. Relevant equations

Gauss's Law: Integration of E dA = Q/eo
E = kq1q2/r^2

3. The attempt at a solution
(1-2) I treated this problem as a hollow sphere
The electric field is zero inside a conducting sphere.
The electric field outside the sphere is given by: E = kQ/r2, just like a point charge.
(3) This one I have no idea......

2. Feb 5, 2010

### xcvxcvvc

A conducting sphere alone has an electric field of zero within it. However, do you think there would be an electric field in a conducting sphere if there were charged particles inside of it - say, a positive nucleus with Z protons?

Think of the nucleus as a point charge meaning regardless of how small your Gauss surface is, you still have the z*e Coulombs enclosed, so the E field is not zero within the electron cloud.

Your second answer seems wrong, also. Once you've drawn a Gauss surface that encloses both the electron cloud and the nucleus, what does Q enclosed become?

3. Feb 7, 2010

### sweetdion

So for part one (r<R) would u use E=Q/4PiE0r^2 to give you that the electric field=Ze/4PiE0r^2

4. Feb 7, 2010

### sweetdion

Or would it be Integration EdA=Qenc/E0 and dA=2Pi*rL and Qenc=Ze so E=Ze/2Pi*rLE0

5. Feb 7, 2010

### xcvxcvvc

yes that's right
I'm unsure what rL is and i'm unsure how you got the integral to equal that.

you can integrate to get the same answer as above. Your E field is constant at all points on your surface, so your integral becomes:
$$E\int_A dA=\frac{Q_{enc}}{\epsilon_o}$$

That integral, I hope you see, becomes the total area of your Gauss surface. What is your surface? It is a sphere, so it becomes the surface area of a sphere:
$$E(4\pi r^2)=\frac{Q_{enc}}{\epsilon_o}$$
$$E=\frac{Q_{enc}}{4\pi\epsilon_or^2}$$

6. Feb 7, 2010

### sweetdion

Thank you very much.

As for part B for r>R if it encloses both the electron cloud and the nucleus cloud then Qenc=0 therefore the Electric field=0?

And for Part C: If it were deficient by one electron, does that mean the overall charge would be positive since there will be more protons then electrons so

r<R (when deficient by one electron) I'm a little bit confused on this one.

Now there are Z-1 electrons. Would you just plug that into the equation that we got in part a to give E=(Z-1)e/4Pi*E0r^2

r>R (when deficient by one electron)

Qenc=+1?? so would i just plug that in for the electric field?

7. Feb 7, 2010

### xcvxcvvc

You are right about part B. For part C when r < R, it's the same problem as before! Does $$Q_{enc}$$ change? No, so the E-field doesn't either!

For when r > R, $$Q_{enc}$$ becomes +e like you say. Just solve the integral the same way I solved it for part A

8. Feb 7, 2010

### sweetdion

So let me get this right,

For when r>R (with one electron deficient), I solved the integral to get E=Z*+e/4Pi*E0r^2

correct? the only thing that changes is the Qenc. The Qenc becomes Z*+e

9. Feb 8, 2010

### xcvxcvvc

No, Qenc becomes +1e

If there are Z - 1 electrons and Z protons, you get:
Z(e) + (Z -1)(-e)
Ze + (e - Ze)
e

So the total charge enclosed is 1 electron's worth but positive.