Electricity and Magnetism: Verifying the Inverse Square Law

  • #1
giodude
30
1
Hello,

I'm currently working through Purcell and Morin, Electricity and Magnetism textbook and came across a problem in which the goal is to verify the inverse square law. I'm worked through and completed the problem. However, I'm confused how this verifies the inverse square law, I'm posting the images of the solution below.
 
Physics news on Phys.org
  • #2
The attached files are not in .pdf format.
How were they produced ?
 
  • #3
I took images of them on my phone and then airdropped them to my laptop, I'll fix them up when I get back to my laptop and edit the post. Thank you.
 
  • #4
I've here the 3rd edition, where it seems that the authors try to derive the Coulomb field of a static point charge. As to be expected from this book, it's all buried in some strange pedagogics, making the problem more complicated than it is.

The idea is simply to use the spherical symmetry of the problem. So let the point charge, ##Q##, sit at rest in the origin of a Cartesian coordinate system. We want to calculate ##\vec{E}(\vec{r})## at any position ##\vec{r} \neq \vec{0}##, because at the origin we have obviously a singularity, which is characteristic for the assumption of a "point charge" in classical field theory.

Mathematically the problem is simple because of spherical symmetry. There's no other vector in the problem than ##\vec{r}##, because no direction is in any way special except the direction of the position vector itself. Thus you can make the Ansatz
$$\vec{E} = E_r \vec{e}_r,$$
where ##\vec{e}_r=\vec{r}/r##. The "radial component" ##E_r## can only depend on ##r=|\vec{r}|##, again due to the spherical symmetry.

Now you simply use Gauss's Law in integral form
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=Q_V/\epsilon_0.$$
It's obvious, again because of the spherical symmetry, to choose a spherical shell of radius ##r## around the origin for ##\partial V##. The surface-normal vectors are ##\vec{e}_r## and thus with our ansatz for ##\vec{E}##
$$E_r (r) 4 \ pi r^2=Q/\epsilon_0 \; \Rightarrow \; E_r(r)=\frac{Q}{4 \pi \epsilon_0 r^2}.$$
That's it! It's simply spherical symmetry and Gauss's Law!
 

Similar threads

Replies
12
Views
443
Replies
15
Views
786
  • Electromagnetism
Replies
5
Views
1K
Replies
25
Views
1K
Replies
9
Views
2K
  • Mechanics
Replies
5
Views
1K
Replies
8
Views
1K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
1
Views
790
Back
Top