# Potential difference on a conical surface

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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tiny-tim
Homework Helper
hi issacnewton!
… 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.

thanks tiint tim.....