1. Feb 16, 2008

### cryptoguy

1. The problem statement, all variables and given/known data

I know that the electric flux through some solid is $$E\pi\\r^2$$. E is an outside electric field directed towards the solid.
Does that mean that there is a net charge of magnitude $$\epsilon$$ $$E\pi\\r^2$$ inside the solid?

My idea is that the net charge is opposite of the entry point of the field so the solid is in electrostatic equilibrium (Electric field inside the solid is 0). Is that correct?

2. Feb 16, 2008

### Mindscrape

The electric flux actually depends on what kind of symmetry you are exploiting. For example, the one you are talking about would be a Gaussian surface for a radially symmetric field, and would mean that the electric flux, the density of the field lines, is uniform across the whole surface of the sphere E*4πr^2.

As far as directions and signs are concerned, physics has established the normal for a closed surface is always outward, so if the field lines are also outward then the flux is positive. If the field moves opposite to the normal, such as a negative charge, then the flux will be negative.

3. Feb 16, 2008

### cryptoguy

Right so if the electric field E is hitting 1/2 of the surface area (like a projector illuminating a ball), would the positive charge move to the other hemisphere's surface?

4. Feb 16, 2008

### Mindscrape

I'm not quite sure if I understand your question. Gaussian surfaces are figurative so charges don't build on them, they are basically like creating an unalterable, imaginary soap bubble for fields to pass through. What surface are you referring to?

5. Feb 17, 2008

### physixguru

6. Feb 17, 2008

### cryptoguy

My question is if E*A = Qinside/epsilon, and E is directed towards one face of the "soap bubble" where is the charge Q located?

7. Feb 17, 2008

### Staff: Mentor

All Gauss's law tells you is the net charge within the Gaussian surface, not necessarily where that charge is.