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

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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

$=4\;\alpha\;l^4-\alpha\;l^4$

$=3\;\alpha\;l^4\;.$

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