Potential difference on a conical surface

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SUMMARY

The discussion centers on calculating the potential difference on a conical surface as presented in problem 2.26 of Griffiths' "Introduction to Electrodynamics" (3rd Edition). A participant questions the use of a rectangular patch as a differential area element for double integration, while another confirms that although this method is valid, Griffiths' approach using a ring is more efficient. The consensus is that while multiple methods can yield correct results, the simplest approach is preferred for efficiency and exam scoring.

PREREQUISITES
  • Understanding of electrostatics and potential difference concepts
  • Familiarity with Griffiths' "Introduction to Electrodynamics" (3rd Edition)
  • Knowledge of differential calculus and integration techniques
  • Basic comprehension of conical geometry in physics
NEXT STEPS
  • Study the method of calculating potential using differential area elements in electrostatics
  • Review Griffiths' problem-solving techniques in "Introduction to Electrodynamics"
  • Explore the implications of using different differential elements in integration
  • Practice similar problems involving potential differences on various geometries
USEFUL FOR

Students of electromagnetism, physics educators, and anyone preparing for exams in electrodynamics who seeks to understand efficient problem-solving methods in electrostatics.

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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hi IssacNewton! :smile:
IssacNewton said:
… 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.

yes, you certainly can do it that way, but griffiths' way is easier and quicker.

(your way gives you extra work, and one integral ends up being 2πr anyway, which you can work out just by looking at the diagram)

as a general rule, every method works, but you get more marks in the exam (and you save time) if you use the simplest or quickest method. :biggrin:
 
thanks tiint tim...
 

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