Ask Questions, Get Answers

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

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 . $x=l\;,x=2l\;,y=0\;,y=-l\;,z=0\;,z=l\;.$


Can you answer this question?

1 Answer

0 votes
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 $\;ABFE|_{x=l}=(\alpha\;l^2)\;l^2$
Flux entering face $\;CDHG|_{x=2l}=(\alpha\;4\;l^2)\;l^2$
net flux passing through cube
answered Feb 4, 2014 by yamini.v

Related questions

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