# Finding the E field direction of integration

1. Jul 3, 2010

### 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

2. Jul 3, 2010

### Hurkyl

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

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)

3. Jul 4, 2010

### Identity

Thanks hurkyl