The potential on the rim of a uniformly charged disk

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
chaos333
Messages
11
Reaction score
1
This comes from Griffiths' Electrodynamics and is problem 2.51 or 2.52, the disk has a surface charge density and my usual approach to solving these problems is to pick an area element and find a way to create a vector to the point(s) at which the potential is evaluated at. I sent a picture of my thought process and attempt at the problem. The solution involves a R^2+r^2-2Rrcos(theta) instead of R^2+r^2-2Rr that I have and I don't know how they arrived to that. Is my vector wrong or something else?
1720118052126.png
 
Physics news on Phys.org
1720144524123.png


Here not Rr but ##\mathbf{R}\cdot\mathbf{r}##, an inner product, for [tex]|\mathbf{R}-\mathbf{r}|^2 =(\mathbf{R}-\mathbf{r})\cdot(\mathbf{R}-\mathbf{r})[/tex].
 
anuttarasammyak said:
View attachment 347821

Here not Rr but ##\mathbf{R}\cdot\mathbf{r}##, an inner product, for [tex]|\mathbf{R}-\mathbf{r}|^2 =(\mathbf{R}-\mathbf{r})\cdot(\mathbf{R}-\mathbf{r})[/tex].
This makes a lot of sense now, thanks.
 
  • Like
Likes   Reactions: anuttarasammyak