# Point charge at the center of a spherical conductor

1. Jan 23, 2008

### snoweangel27

I have the problem

A positive point charge of magnitude 2.4 micro Coulombs is at the center of an uncharged spherical conducting shell of inner radius 65 cm and outer radius 85 cm.

I have been trying to calculate the total charge for the inner sphere. I solved for the Electric Field by using E=kq/r^2, then using Coulombs law to obtain Q in micro Coulombs, but I end up with a wrong answer. Could someone please explain what I am doing wrong.

2. Jan 23, 2008

### Staff: Mentor

Not exactly sure what you did. Why not use Gauss's law?

3. Jan 23, 2008

### snoweangel27

Sorry, I feel really dumb, I was forgetting that the Total Charge on the smaller sphere would be negative. It's always nice to find out when you've wasted an hour trying to figure out what was wrong and it was just missing a negative sign

4. Feb 5, 2010

### ll1111

i am doing this problem now. i am just learning gauss's law and am very confused.

i did end up getting the problem right, but i do not understand how you know whether or not it will be positive or negative. can someone help?

5. Feb 5, 2010

### ideasrule

The whole thing's uncharged (according to the problem) and there's a positive charge in the middle. Of course the shell has to be negative.

6. Feb 5, 2010

### ll1111

i'm sorry.. i'll reiterate, i really don't understand physics very much, even though i honestly do try. i took physics 1 in high school my junior year and somehow passed the AP test. i'm now in college physics 2, we just started electricity a week ago. i'll admit that i don't understand the basic concepts at all, but i really don't feel like the teacher explains anything much. i'm just trying my best to follow formulas and get by. could someone please explain a bit why the inner one must be negative but the outer one is positive? (i needed to find the densities & electric fields for both the inner and outer surfaces)

Last edited: Feb 5, 2010
7. Feb 5, 2010

### ideasrule

Oh, I see the problem: you misunderstood the question. The question talks about a shell of finite thickness (like a soccer ball, except with a thicker shell). There's a small, positive object inside the shell, but the system (small object + shell) is neutral. So the shell must be negative to cancel out the positive charge of the small object.

8. Feb 5, 2010

### ll1111

what about the charge on the surface of the outer shell? is that positive to cancel the negative of the... inner shell? i'm sorry to be a pain.

9. Feb 5, 2010

### ideasrule

No, the outer skin of the shell is neutral. You can prove this using Gauss' law. Take a Gaussian surface surrounding the inner skin of the shell. The total electric flux through that surface will be 0, since the surface is totally inside a conductor. That means the enclosed charge must be 0. The rest of the shell must be neutral, or else the system won't be neutral.

10. Feb 5, 2010

### fluidistic

I'm also having problems to understand the question. So it's like the surface of a soccer ball, but with an object in the middle. Then what does the inner and outer radius represent? It's 20 cm thin. Does that mean that the conductor fills within both radii?

By the way, what is the question? Calculate the total charge of the shell? But as you said, the charge of the small object+shell=0 and we know the charge of the small object... hence there's almost nothing to do.

I'm completely misunderstanding the problem/situation.

11. Feb 5, 2010

### ll1111

A positive point charge of magnitude 1.6 µC is at the center of an uncharged spherical conducting shell of inner radius 65 cm and outer radius 110 cm.

(a) Find the charge densities on the inner and outer surfaces of the shell.
inner -.301358 µC/m2
outer .10522 µC/m2
Find the total charge on each surface.
inner -1.6 µC
outer 1.6 µC

maybe my question is slightly different? i think it's the same though. these answers are correct, it's online homework and it tells you if you're right or wrong.
what i'm getting confused about is where to put negative signs.
to solve the problem, i used Qenclosed = charge density * A

12. Feb 5, 2010

### ideasrule

That's the way I interpreted it, but it is ambiguous.

The question was to calculate the charge of the "inner sphere". I interpreted that to mean the inner surface of the shell, but again, the question's ambiguous.

13. Feb 5, 2010

### fluidistic

Ok thanks ideasrule.
Edit: So it's like a 3 dimensional hollow sphere. In its center there's a small charge. So if we set the place where the charge is as to be the origin, in order to reach the surface of the sphere I would have to pass through 65 cm of vacuum and then 20 cm of metal (conductor). Am I right?
And they ask what is the charge over the inner surface of the sphere. Wow, I've no idea how to do this. I don't really see how Gauss can help me. Ok so the total charge is 0. Hence the whole sphere (without the charge in its middle) must have a charge of $$-1.16 \mu C$$. I don't really know how to find the charge distribution.

Last edited: Feb 5, 2010
14. Feb 5, 2010

### ideasrule

I don't know, but that's how I interpreted the question.
I talked about how to find it in post 9.

15. Feb 6, 2010

### ehild

As for the signs: Remember, the charges can move freely in a conductor. There is a positive charge in the middle of a metal shell: it attracts the electrons of the metal, and they go as close to the positive charge as they can, but they can not leave the metal. So the electrons accumulate at the inner surface, making it negative.
Meanwhile the bulk of the metal shell becomes positive for a very short time: some electrons migrated to the inner surface and left the positive metal ions behind. These positive charges attract electrons from the region further away from the centre and at the end the bulk of the metal becomes neutral, leaving the outer surface positively charged.

ehild

16. Feb 6, 2010

### ehild

In the original question, the metal shell was neutral.

"A positive point charge of magnitude 2.4 micro Coulombs is at the center of an uncharged spherical conducting shell of inner radius 65 cm and outer radius 85 cm."

The geometry has spherical symmetry, so Gauss Law is easy to apply. The electric field lines are radial and the electric field intensity depends only on r, distance from the centre.
Inside the empty sphere, from r=0 to r=Rin, a sphere of radius r encloses the charge at the centre. So the surface integral of E for the sphere is

$$E*4r^2\pi=\frac{Q}{\epsilon_0}\rightarrow E=\frac{Q}{4\pi\epsilon_0 r^2}$$

The electric field lines always start out at positive charges and end at negative ones, except those lines which go to infinity. There are Q/epsilon0 field lines emerging from the central charge. The same number of field lines hit the surface of the inner sphere: so there must be -Q charge there. This means -Q/(4piRin2 surface charge density. But it is true for any charge Q that Q/epsilon0 field lines emerge from it or end in it. The electric field at the surface of the inner sphere is

$$E=\frac{Q}{4\pi\epsilon_0 R_in^2}$$.

There are no field lines inside the metal. All the field lines ending in a surface charge come from the empty sphere. So the electric field at the inner surface corresponds to surface charge density of magnitude

$$\sigma=E*\epsilon_0= \frac{Q}{4\pi R_{in}^2}$$

For Rin<r<Rout, the enclosed charge is 0, and so is the electric field inside the metal shell.

Outside the shell and on the outer surface, the enclosed charge is Q as the shell itself is neutral. The electric field is

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

and the corresponding surface charge density is

$$\sigma=E*\epsilon_0= \frac{Q}{4\pi R_{out}^2}$$ .

If the total arrangement is neutral (it is so when the metal shell is grounded) the field in the empty sphere is the same as before, the surface charge density at the inner surface of the shell is the same again, but the field outside the shell is zero, as the enclosed charge is zero.

ehild

17. Feb 6, 2010

### fluidistic

Thanks to both. I've understood.

18. Feb 6, 2010

### ll1111

edit - nevermind i think i misread the question.