Question

-How does one derive Gauss' Law from Coulomb's?
-How do you arrive at the divergence law from the integral law (or vise versa, beginning on the one which is arrived at from Coulomb's)?
-It it intuitively obvious, I know, but throughout the proceeds do I take it as a given that [Flux=Surface integral (x-Field parallel to the normal vector of a point on the surface)]?

Answer 1

*

Answer 2

\[ F = \frac{q_1 q_2}{4 \pi \epsilon_0r^2} \\
E = \frac{q}{4 \pi \epsilon_0r^2} \\
\vec E \cdot {4 \pi r^2} = {q \over \epsilon_0} = {\rho V\over \epsilon_0}\]

\[ \oint_s\vec E \cdot ds = {q \over \epsilon_0} = \iiint _v{\rho dV\over \epsilon_0} \\
\text{Divergence theorem } \\
\iiint_v \nabla \cdot \vec E \cdot dV = \iiint _v{\rho dV\over \epsilon_0}\]

Answer 3

\[ \nabla \cdot E = {\rho \over \epsilon_0} \\
\oint_s E \cdot dS = \iiint_v \nabla \vec E dV = {\rho \over \epsilon_0}\]

Answer 4

The answer to the third question is yes?

I'm close to covering it (maybe I'll understand it in a month or so), but could you quickly summarise the divergence theorem?

Thanks for this and that

Answer 5

A closed surface

Answer 6

if you do summations all over this closed surface
\[ \oint_s F \cdot dS = F_1 .dS_1 + F_2 .dS_2+F_3 .dS_3 + .... \]

Answer 7

then this should be equal to the divergence of the field
\[\oint_s F \cdot dS = \iiint_v\nabla \cdot F dV \]

Answer 8

To ask before I get lost- in general terms, what does the right hand side mean (I am comfortable with divergence notation)?

Answer 9

Damn this draw app. my work is lost.

Answer 10

\[\int\limits \int\limits \int\limits \frac{dF}{dx} dxdydz+\int\limits \int\limits \int\limits \frac{dF}{dy} dxdydz+\int\limits \int\limits \int\limits \frac{dF}{dz} dxdydz\]
\[=Fyz+Fxz+Fxy+f(x,y,z)\]

Answer 11

Is it that?
And curse that draw app!