The potential on the rim of a uniformly charged disk

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SUMMARY

The discussion centers on solving problem 2.51 or 2.52 from Griffiths' Electrodynamics, specifically regarding the potential on the rim of a uniformly charged disk. The correct approach involves using the law of cosines, leading to the expression R² + r² - 2Rr cos(θ), rather than the incorrect R² + r² - 2Rr. The participants clarify the importance of the inner product, represented as \(\mathbf{R} \cdot \mathbf{r}\), in deriving the correct formula for the distance between vectors. This understanding resolves the confusion surrounding the initial vector formulation.

PREREQUISITES
  • Understanding of Griffiths' Electrodynamics, specifically problems 2.51 and 2.52
  • Familiarity with the law of cosines
  • Knowledge of vector operations, particularly inner products
  • Basic concepts of electric potential and surface charge density
NEXT STEPS
  • Study the derivation of the law of cosines in vector calculus
  • Explore Griffiths' Electrodynamics for deeper insights into electrostatics
  • Learn about vector fields and their applications in physics
  • Investigate the implications of surface charge density on electric potential
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as educators seeking to clarify concepts related to electric potential and vector analysis.

chaos333
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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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Here not Rr but ##\mathbf{R}\cdot\mathbf{r}##, an inner product, for |\mathbf{R}-\mathbf{r}|^2 =(\mathbf{R}-\mathbf{r})\cdot(\mathbf{R}-\mathbf{r}).
 
anuttarasammyak said:
View attachment 347821

Here not Rr but ##\mathbf{R}\cdot\mathbf{r}##, an inner product, for |\mathbf{R}-\mathbf{r}|^2 =(\mathbf{R}-\mathbf{r})\cdot(\mathbf{R}-\mathbf{r}).
This makes a lot of sense now, thanks.
 
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