Finding the E field direction of integration

[Identity]

If you have a line of charge with charge density $$\lambda=\frac{dq}{dl}$$ and you want to find the electric field at a perpendicular distance z from the midpoint, you get

$$dE = \frac{1}{4\pi\epsilon_0}\frac{\lambda}{r^2}dl$$

Then you integrate $$dE$$ from one end of the line of charge to the other. (e.g. $$\int_{-L}^L ... dl$$)

Obviously if you reverse the integral terminals you get the negative of your original answer, but physically, why should reversing integral terminals even matter? (i.e. $$\int_L^{-L}...dl$$)

After all, the physical interpretation of the integral is just summing up the little $$dq$$s over the line, what does it matter which direction you do it in? And importantly, how do you know which is the correct direction to sum up the $$dq$$s?

Thanks

 

[Hurkyl]

what does it matter which direction you do it in?

Because one way has positive length, and the other way has negative length.

And importantly, how do you know which is the correct direction

The real line is oriented so that displacements towards $+\infty$ are positive.


Why do the signs creep in? Because you are using an oriented notion of integration -- a parametrized curve (from -L to L) with respect to a differential form ($d\ell$).

There are unsigned notions of integration. If you have a measure (say, $\mu$), you can define integrals over sets -- e.g.

$$\int_{S} \ldots d\mu$$​

Of course, it turns out that integrals of the standard length measure on R over intervals can be computed in the oriented way:

$$\int_{[-L,L]} f \, d\mu = \int_{-L}^L f(x) \, dx = -\int_L^{-L} f(x) \, dx$$​

(I'm assuming L>0 in the above)

 

[Identity]

Thanks hurkyl