# Electric Field

1. Sep 26, 2008

### Shackleford

#29

(a) The charge is zero because it's within the sphere.

(b) I don't know how to do this one.

(c) = E(4 pi r^2) = Q/e0

E = kQ/r2

2. Sep 26, 2008

### Defennder

You know, it'll help a lot if you were to reduce the size of that image. I had to zoom out just to read it.

a) Just use Gauss law and symmetry here. What is the flux through a spherical closed surface situated within the cavity?

b) Again use Gauss law and symmetry. Your Gaussian surface should be spherical and this time it should terminate in the non-conducting spherical shell.

c) Yeah this appears correct.

3. Sep 26, 2008

### sha_virgo

hi,, i have a doubt.. we can generate magnetic field using electric field.. then why cant we generate electric field using magnetic field?

4. Sep 26, 2008

### sha_virgo

hi guys... i have a doubt.. we can generate magnetic field using electric field.. then why cant we generate electric field using magnetic field?

5. Sep 26, 2008

### Defennder

We can in fact induce an electric field with a time-varying magnetic field. This is captured mathematically by Faraday's law, where the magnitude of the induced emf is proportional to the rate of change of magnetic flux. Generators make use of this principle to 'create' electricity from motion magnets.

6. Sep 26, 2008

### Shackleford

Sorry about the size. I just got my new 19" LCD monitor yesterday. lol.

There's a similar example problem in the chapter. After finding EA with a solid sphere, they set that equal to Qe over e0. Since the Qe is partial not the total charge Q, it looks like they simply setup a ratio of the partial charge to the total charge.

Here's what they did. Let's make sure I followed correctly.

charge density = charge per unit volume = dQdV

r0 = radius of sphere, r < r0

Qe = ( (4/3) pi r^3 pe ) / ( (4/3) pi r0^3 pe ) = (r^3) Q / (0r^3)

They just setup a ratio using the volume and charge density to reflect the partial charge?

For (b) setting up the charge ratio, I get ( r^3 - r1^3 / r0^3 - r^3).

= EA = E(4pi r^2) = Qe / e0 = ( 1 / (4pi r^2 e0) ) ( r^3 - r1^3 / r0^3 - r^3)

7. Sep 27, 2008

### Defennder

In that case, the problem was concerning a solid sphere of charge. But here your problem is a shell of charge so you can't make use of any result from that example.

What is r1 here? And what is r0 supposed to be? The inner radius of the shell?

8. Sep 27, 2008

### Shackleford

Well, this is a sphere with a cavity in it.

According to the diagram, r1 is the inner radius and r0 is the outer radius.

I looked at the back of the book to see the answer they got. The example problem I cited gave me the idea to setup a ratio, so that's how I got the right answer finally.

9. Sep 27, 2008

### Defennder

Yeah your final answer looks ok. Except you forgot to include Q. And don't abuse the '=' sign. It gets confusing.