# Homework Help: Gauss's Law

1. Aug 18, 2008

### yti1211

y is this statement wrong?

The electric field on a gaussian surface is generally not influenced by charge that is not enclosed by the surface.

ps: isn't the electric field proportional only to the ENCLOSED CHARGE, but not the outside charge??

This is getting me so fRustrated

2. Aug 18, 2008

### Defennder

Gauss law is a law of electric flux, not E-field. If the situation is symmetric enough, you may extract the value of the E-field from it. Consider a Gaussian surface enclosing an electric dipole. Are you able to determine the E-field at any point on the surface? Now imagine the distance between the positive and negative charges increasing along with the size of the Gaussian surface. What is the E-field at any point on the surface?

3. Aug 18, 2008

### yti1211

hmm, first, thx Defender! I no that u need a constant E throughout the Gaussian surface in order to extract E from the integral. but how do you explain why the E- field on Gaussian surface is affected by a charge outside of the surface?? cuz the ENCLOSED Q is the key right? or do u mean the size of the Gaussian surface is changing when a charge is placed outside of it?

THXXX :)

4. Aug 18, 2008

### yti1211

ohh, I think I got it now! So the distribution of E field is changing on the Gaussian surface, but the total E dA is the same. yAy