# How to solve this without infinitesimals

• jaydnul

#### 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

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##.

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 $d$s to $\Delta$s. 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.