Higher dimensional FTC in electrostatics: Mathematically rigorous or not?

oliverkahn
Messages
27
Reaction score
2
I have two volumes ##V## and ##V'## in space such that:

1. ##∄## point ##P## ##\ni## ##[P \in V ∧ P\in V']##

2. ##V## is filled with electric charge ##q##

3. ##\rho = \dfrac{dq}{dV}## varies continuously in ##V##

4. ##V'## is filled with electric charge ##q'##

5. ##\rho' = \dfrac{dq'}{dV'}## varies continuously in ##V'##

Let ##r## be the distance between a point ##P_1 \in V## and a point ##P_2 \in V'##

In electrostatics, we use the implication:

##\displaystyle \dfrac{d^2\vec{F}}{dq\ dq'}=k\dfrac{\hat{r}}{r^2} \implies \vec{F}=k\int_q \int_{q'}\dfrac{\hat{r}}{r^2} dq'\ dq##

Does this implication has any mathematically rigor?

NOTE: I know:

##G(x)## is differentiable on interval ##[a,b]##

##\land## ##g(x)## is Riemann integrable function in interval ##[a,b]##

##\land## ##\dfrac{d\ G(x)}{dx} = g(x)## in interval ##(a,b)##

##\implies \displaystyle G(b)-G(a) = \int^b_a g(x) dx##

But I don't see why this works for integral over a volume, integral over a surface, etc...
 
Physics news on Phys.org
@oliverkahn - what does FTC stand for? Some of us mentors are confused.. I'm one of them.
 
oliverkahn said:
But I don't see why this works for integral over a volume, integral over a surface, etc...
In case the dimension is higher than one, it is called Stoke's theorem.
 
jim mcnamara said:
@oliverkahn - what does FTC stand for? Some of us mentors are confused.. I'm one of them.
Fundamental Theorem of Calculus
 
oliverkahn said:
Fundamental Theorem of Calculus
Rats. I had "Federal Trade Commission" in the Mentor pool. Oh well...

(It's always best to define your acronyms when you use them when they may be non-obvious...) :smile:
 
  • Like
Likes hutchphd
And assume they are never obvious.
 
  • Like
Likes berkeman