Can You Calculate Electric Fields Without Infinitesimals?

[jaydnul]

I'm trying to find a way to use calculus without infinitesimals and I'm stuck on this physics problem.

It's a uniform charge distribution question. Basically a half circle with radius r and you have to find the electric field at a point that is along its x-axis. The E_y component will be 0 because of symmetry. So all you need is is E_x. The equation ends up being:
E_x=∫dEcosθ
The only way I know how to solve this is using infinitesimals and finding one in terms of the other, like this:
dE=k\frac{dQ}{r^2}
dQ=\frac{Q}{πr}dy
dy=rdθ
Then substituting those in all the way up so I have my integral in terms of θ.

So how would I go about solving this without using infinitesimals? Thanks

 

[verty]

This is always difficult to explain.

$E_x = \int cos(\theta(E)) dE$.

On the one hand, $E_x$ is the area under $y = cos(\theta(E))$, a Riemann sum of rectangles having width $dE$. This seems to imply that $dE$ is a width, an infinitesimal width.

On the other hand, we have a rate of change $dE$ and a related rate of change $cos(\theta(E)) dE$ which when integrated, gives $E_x$. The rate of change of $E_x$ is $cos(\theta(E)) dE$, it is related to $dE$, the rate of change of $E$.

Which is it? Is $dE$ a width or is it a rate? This is a little like asking what is a number. One person will say, the thing you count with, another will say, an equivalence class of sets by cardinality, another will say, a Church numeral, etc, etc. There is no answer to that question. In the same way, it doesn't matter what $dE$ is, what matters is how differentials are used.

Of course, there is also the view that $dE$ is a finite width and $cos(\theta(E))$ is the gradient of the tangent to $E_x$ at $E$, so $cos(\theta(E)) dE$ is the $E_x$ offset of the tangent at distance $dE$ from $E$.

 

[gopher_p]

So how would I go about solving this without using infinitesimals? Thanks

I think the general answer to this question (as it pertains to setting up integrals) is "Set up an approximating Riemann Sum and take a limit." Most of the time it's just a matter of changing all of the ds to \Deltas. In my experience, the physicists aren't using infinitesimals so much as they are being a bit lazy with limits and abusing notation. It gets the job done, though. So more power to 'em.