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The intensity of an electric field depends on the coordinates x and y as follows $\;\overrightarrow{E}=\large\frac{a\;(x \hat{i} + y \hat{j})}{x^2+y^2}\;$ where a is constants. Find the charge within a sphere of radius R with the centre at the origin

$(a)\;8 \pi \in_{0} a R \qquad(b)\;4 \pi \in_{0} a R \qquad(c)\;2 \pi \in_{0} a R \qquad(d)\; \pi \in_{0} a R $

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Answer : (b $\;4\pi \in_{0} a R \;$
Explanation :
At any point P(x , y ,z) on the sphere $x^2+y^2+z^2=R^2$
and a unit vector perpendicular to the sphere radially outwards is
$\hat{n} = \large\frac{x}{\sqrt{x^2+y^2+z^2}} \hat{i} + \large\frac{y}{\sqrt{x^2+y^2+z^2}} \hat{j}+ \large\frac{z}{\sqrt{x^2+y^2+z^2}} \hat{k}$
$\hat {n} = \large\frac{x}{R} \hat{i} + \large\frac{y}{R} \hat{j} + \large\frac{z}{R} \hat{k}$
 
Let us find the electric flux through a small area dS at point P on the sphere
$d \phi_{e} = \overrightarrow{E} . dS \hat {n}$
$=(\large\frac{ax^2}{R(x^2+y^2)}+\large\frac{ay^2}{R(x^2+y^2)})$ dS
$d \phi_{e}=\large\frac{a\;dS}{R}$
$\phi_{e}=\oint\;d \phi_{e} = \large\frac{a}{R} \oint dS =\large\frac{a}{R}\;4 \pi R^2$
$\phi_{e} = 4 \pi a R $
From Gauss's Law
$\large\frac{q_{inc}}{\in_{0}}= \phi_{e}$
$q_{inc}=4 \pi \in_{0} a R$
$=\large\frac{aR}{k}\;.$

 

answered Feb 14, 2014 by yamini.v
edited Aug 16, 2014 by thagee.vedartham
 

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