Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
37.6k views
in Physics by (75.3k points)

Find the potential φ at the edge of a thin disc of radius R carrying the uniformly distributed charge with surface density σ.

1 Answer

+1 vote
by (69.1k points)
selected by
 
Best answer

By definition, the potential in the case of a surface charge distribution is defined by integral  In order to simplify integration, we shall choose the area element dS in the form of a part of the ring of radius r and width dr in (Fig.). Then dS - 2θ r dr, r = 2R cos θ and dr = - 2R sin θ d θ. After substituting these expressions into integral 

we obtain the expression for φ at the point O:

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...