# Two questions on electrostatics

1. Jun 19, 2006

### r4nd0m

Hi,
I'm a bit stuck with some things in electrostatics.

My first problem:

in my textbook, when they try to derivate the formula for the potential of a point charge: $$V(b) = - \int E.d\mathbf{l} = -\frac{q}{4 \pi \varepsilon_0} \int_\infty^b \frac{1}{r^3} \mathbf{r}.d \mathbf{l}$$

they say that $$\mathbf{r}.d \mathbf{l} = r.dr$$
There's also a picture which looks like this:

http://img232.imageshack.us/img232/6044/charge3zj.jpg [Broken]

My question is: Why isn't $$d \mathbf{r} = d \mathbf{l}$$ ? Why does $$d \mathbf{r}$$ have the same direction as $$\mathbf{r}$$ ?

My second problem:

There is some law or theorem which says that there is no electric field inside a conductor. Can this be proved, or is it just an empirical law?

Last edited by a moderator: May 2, 2017
2. Jun 19, 2006

### Galileo

The dl is a line element of the path you are taking to evaluate the integral (it doesn't matter what path you take), just some path from infinity to b.
$$\vec r$$ is a vector pointing in the direction of the electric field (if the charge is positive), thus in the radial direction.
The line element in spherical coordinates is something like $$d\vec l = dr\hat r+rd\theta \hat \theta+r\sin \theta d\phi \hat \phi[/itex], so [tex]\vec r \cdot d\vec l=rdr$$.

3. Jun 19, 2006

### siddharth

Well, you need to be a bit more precise. The electric field is zero in a region completely surrounded by a conductor, provided there is no charge enclosed in that region.

It can be proved easily using the first uniqueness theorem.

4. Jun 19, 2006

### r4nd0m

great, thank you very much for your help.