- 9

- 0

## Main Question or Discussion Point

Can anyone explain this to me?

- Thread starter skyliner34
- Start date

- 9

- 0

Can anyone explain this to me?

- 2,221

- 7

if the enclosing surface is a sphere surrounding the point charge, with the point at center, it's pretty easy to see how Gauss's law is directly compatible with the inverse-square relationship.

- 249

- 0

All information you have is the location of 2 points: the source and the point where you want to evaluate the vector. The problem has cylindric symmetry around the axis connecting the two points, so vector MUST be directed parallel to this axis: either straigth away or towards the source.

The magnitude of the vector must be independent of the choise of coordinate system, so it can only depend on the distance between the two points. Under this circumstances the integral of the electric field over a sphere is not difficult to evaluate.

- 2,221

- 7

there are a lot of different scaler functions of that distance.First you must think what are the possible vector fields caused by a single point source (charge).

All information you have is the location of 2 points: the source and the point where you want to evaluate the vector. The problem has cylindric symmetry around the axis connecting the two points, so vector MUST be directed parallel to this axis: either straigth away or towards the source.

The magnitude of the vector must be independent of the choise of coordinate system, so it can only depend on the distance between the two points..

the reason why that surface integral evaluates to a constant that is proportional to the enclosed point charge is because the magnitude of that field vector is inversly proportional to the square of the distance and proportional to the charge. why it's proportional to the amount of charge is because if you had two different charges sitting virtually next to each other, the total force you get from both should come from adding the force you get due to each charge. why it's inverse-square with distance is because of this concept ofUnder this circumstances the integral of the electric field over a sphere is not difficult to evaluate.

think of a 100 watt light bulb surrounded by a bunch of concentric spheres all centered on the 100 watt light bulb. intensity of radiant energy is how much radiating power crosses a unit area that is held perpendicular to the flow of this radiant power. so if you were 10 meters out, that 100 watts is distributed over 4[itex]\pi[/itex]10

Gauss's law does not work unless it's an inverse-square relationship with distance and proportional to the amount of "stuff" at the point source that we're measuring distance from.

Last edited:

- 249

- 0

I only solved the first part of the problem: we are not proving Gauss law from Coulomb's. We are proving the opposite, which is more difficult. For this proof we must first use symmetry to show that the electric field vector is directed away from the source and it's magnitude depends only on the distance (when I was taught about Coulomb's law I felt cheated because this first part was omitted and the proof Coulomb->Gauss was sold as Gauss->Coulomb proof).

nicksauce

Science Advisor

Homework Helper

- 1,272

- 5

[tex]\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}[/tex]

One can verify that given

[tex]\rho = q\delta^3(\vec{r})[/tex]

A solution for E is

[tex]\vec{E} = \frac{q}{4\pi\epsilon_0 r^2}[/tex]

- Last Post

- Replies
- 6

- Views
- 5K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 9

- Views
- 5K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 528