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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}}). $

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

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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
 

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