The potential on the rim of a uniformly charged disk

  • #1
chaos333
11
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
 
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  • #3
anuttarasammyak said:
The law of cosines https://en.m.wikipedia.org/wiki/Law_of_cosines might be helpful.
I understand the answer involves the law of cosine formula, but how was that derived from the vector?
 
  • #4
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].
 
  • #5
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.
 
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