Potential difference on a conical surface

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SUMMARY

The discussion centers on Problem 2.26 from Griffiths' "Introduction to Electrodynamics" (3rd Edition), specifically regarding the calculation of electric potential on a conical surface. The original solution utilizes a ring as the differential area element, while a participant questions the feasibility of using a small rectangular patch instead. The consensus suggests that using a rectangular patch is valid, and participants are encouraged to explore the outcomes of this alternative approach through double integration.

PREREQUISITES
  • Understanding of Griffiths' "Introduction to Electrodynamics" concepts
  • Familiarity with electric potential and integration techniques
  • Knowledge of differential area elements in calculus
  • Basic principles of electromagnetism
NEXT STEPS
  • Explore the application of double integration in electromagnetism
  • Study the use of differential area elements in electric field calculations
  • Review Problem 2.26 in Griffiths' EM book for deeper insights
  • Investigate alternative methods for calculating electric potential on curved surfaces
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as educators looking for alternative teaching methods for potential calculations on conical surfaces.

issacnewton
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hi

i am posting a problem 2.26 from griffiths EM book third edition.i am also attaching the solution from the book's solution manual. in the solution, griffiths has taken the ring as the differential element. but i want to know if we can take the small rectangular patch on the conical surface as the differential area element so that we can do double integration for calculating the potential at two points.

Issac n
 
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IssacNewton said:
hi

i am posting a problem 2.26 from griffiths EM book third edition.i am also attaching the solution from the book's solution manual. in the solution, griffiths has taken the ring as the differential element. but i want to know if we can take the small rectangular patch on the conical surface as the differential area element so that we can do double integration for calculating the potential at two points.

Issac n

i forgot to attach. let me see how to do it
 
I don't see why not. What do you get when you try it?
 

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