Ask Questions, Get Answers

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

Potential due to a system at polar coordinates $\;(r , \theta)\;$ is given by $\;\large\frac{\alpha \;cos \theta}{r^2}\;$ Find the magnitude of electric field at that point if $\;\alpha=3 Vm^2\;,(r , \theta)=(\sqrt{3} ,tan^{-1} \large\frac{1}{\sqrt{2}}). $

$(a)\;\sqrt{3} \;Vm^{-1}\qquad(b)\;3\;Vm^{-1}\qquad(c)\;3 \sqrt{3} \;Vm^{-1}\qquad(d)\;1 \;Vm^{-1}$

Can you answer this question?

1 Answer

0 votes
Answer : (d) $\;1 \;Vm^{-1}$
Explanation :
$E_{r}=-\large\frac{\partial V }{\partial r} = - \large\frac{\partial}{\partial r}\;(\large\frac{\alpha\; cos \theta}{r^2})$
$E_{r}=\large\frac{2\;\alpha\;cos \theta}{r^3}$
$E_{\theta}=- \large\frac{\partial}{\partial \theta}\;(\large\frac{\alpha\;cos \theta}{r^2})\times \large\frac{1}{r}$
$E_{\theta} = \large\frac{\alpha\;sin \theta}{r^3}$
$|E|=\sqrt{E^{2}_{r}+E^{2}_{\theta}}=\large\frac{\alpha}{r^3}\;\sqrt{4 \cos^{2} \theta+sin^2 \theta}$
$cos \theta=cos\; tan^{-1} (\large\frac{1}{\sqrt{2}})$
$cos \theta =\large\frac{\sqrt2}{\sqrt{3}}$
$sin \theta =\large\frac{1}{\sqrt{3}}$
$|E|=\large\frac{3}{3 \sqrt{3}}\times\sqrt{4\times\large\frac{2}{3}+\large\frac{1}{3}}$
$|E|=\large\frac{\sqrt{3}}{\sqrt{3}} =1\;.$


answered Feb 7, 2014 by yamini.v
edited Aug 14, 2014 by thagee.vedartham

Related questions

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