# Gauss' Law for Electrostatics

• May 10th 2011, 09:00 AM
topsquark
Gauss' Law for Electrostatics
I admit that I have something of a block when it comes to Electromagnetism, but this question is so basic it's really bugging me. So time to throw my pride aside.

I have a member at PHF that asked me if an exterior non-uniform charge distribution can create an electric field inside a closed surface. My initial answer was "no" based on Gauss' Law because I could swear I've done a number of them. However it is true that a single point charge outside the surface generates an electric field inside the surface. My problem is that the net flux through the surface is 0, but I thought that implied that E was also 0? There's a big flaw in my reasoning but I can't seem to find it.

-Dan

Addendum: As I've been typing there's something that's trying to worm its way into my head. I've leave the question as posted. If I figure it out I'll let you all know the hows and whys.
• May 10th 2011, 09:05 AM
Ackbeet
Quote:

Originally Posted by topsquark
I admit that I have something of a block when it comes to Electromagnetism, but this question is so basic it's really bugging me. So time to throw my pride aside.

I have a member at PHF that asked me if an exterior non-uniform charge distribution can create an electric field inside a closed surface. My initial answer was "no" based on Gauss' Law because I could swear I've done a number of them. However it is true that a single point charge outside the surface generates an electric field inside the surface. My problem is that the net flux through the surface is 0, but I thought that implied that E was also 0?

No, it just means that the "number" of field lines going into the surface is the same as the "number" of field lines going out. You will have an electric field in this case! An exterior non-uniform charge distribution cannot create any net flux through a closed surface. But in order not to do that, the field it does generate will be just right so that the new field lines coming in one portion of the surface cancel out the field exiting in another portion of the surface.

Quote:

There's a big flaw in my reasoning but I can't seem to find it.

-Dan

Addendum: As I've been typing there's something that's trying to worm its way into my head. I've leave the question as posted. If I figure it out I'll let you all know the hows and whys.
• May 10th 2011, 12:07 PM
topsquark
Quote:

Originally Posted by Ackbeet
No, it just means that the "number" of field lines going into the surface is the same as the "number" of field lines going out. You will have an electric field in this case! An exterior non-uniform charge distribution cannot create any net flux through a closed surface. But in order not to do that, the field it does generate will be just right so that the new field lines coming in one portion of the surface cancel out the field exiting in another portion of the surface.