# Gauss's Law question

1. Mar 3, 2013

### ehabmozart

Consider a point charge surrounded by a gaussian sphere. In the derivation of the formula, we said that E is constant everywhere on the sphere. Hence, we used Flux= EA.. However, isn't the E created on the surface is due to one field line out of the pint charge. I mean the formula kQ/r^2 is the field of the charge in a particular direction? I need clarification in this. Here, i just want to know kq/r^2.. What does it respond to?

2. Mar 3, 2013

### Staff: Mentor

Field lines are just an aid to visualizing the field. The field exists everywhere.
That tells you the field surrounding a point charge, which depends only on the distance r from the charge. The direction of the field is radially outward (for a positive charge).

3. Mar 4, 2013

### cwilkins

The field due to the point charge is constant on the surrounding sphere because every point on that sphere is at the same distance away from the charge. You can think of $r^2$ as the (squared) distance from the point where the charge is to wherever you're measuring the field. Every point on the sphere has the same value of $r^2$ so $E$ has the same magnitude at every point on that sphere.

4. Mar 4, 2013

### BruceW

yeah, I agree with these dudes. Another point: the kq/r^2 you can think of as a consequence of how the field lines are spreading out. The 'closeness' of the field lines is proportional to the strength of the electric field. And since the field lines are spreading out as 1/r^2 , then the electric field also weakens as 1/r^2. So in this way, you can think of the electric field as due to all the field lines coming out of the point charge, not due to any single field line.