Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Find the potential on the edge of a disc of charge density $\;\sigma$ and radius $R$


Can you answer this question?

1 Answer

0 votes
Answer : (c) $\;4\;\sigma\;k\;R$
Explanation : $\;dV=\large\frac{k\;d\;q}{r}$
$dr=-2\;R\;sin\;\theta\;d \theta$
$dr=-\large\frac{\sigma}{2\;\pi\;\in_{0}}\;2\;R\;\theta\;sin\;\theta\;d \theta$
$V=\int_{\large\frac{\pi}{2}}^{0}\;dV=\large\frac{\sigma\;R}{\pi\;\in_{0}}\;\int_{0}^{\large\frac{\pi}{2}}\;\theta\;sin\;\theta\;d \theta$
answered Feb 4, 2014 by yamini.v
edited Feb 4, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App