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The electric field in a region is given by $\;\overrightarrow{E}=\alpha\;x^2 \hat{i}\; $ Here $\;\alpha\;$ is a constant of proper dimensions . Find the total flux passing through a cube bounded by the surfaces . $x=l\;,x=2l\;,y=0\;,y=-l\;,z=0\;,z=l\;.$

$(a)\;\alpha\;l^4\qquad(b)\;2\;\alpha\;l^4\qquad(c)\;3\;\alpha\;l^4\qquad(d)\;4\;\alpha\;l^4$

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Answer : (c) $\;3\;\alpha\;l^4$
Explanation :
Electric field is along +ve x- axis . Therefore field lines are perpendicular to faces ABFE and CDHG . At all other faces field lines are tangential . So net flux passing these four faces will be zero .
Flux entering face $\;ABFE|_{x=l}=(\alpha\;l^2)\;l^2$
$=\alpha\;l^4$
Flux entering face $\;CDHG|_{x=2l}=(\alpha\;4\;l^2)\;l^2$
$=4\;\alpha\;l^4$
net flux passing through cube
$=4\;\alpha\;l^4-\alpha\;l^4$
$=3\;\alpha\;l^4\;.$
answered Feb 4, 2014 by yamini.v
 

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