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Grade Upto college levelElectrostatics

This question is for aspirants:

The electric field strength depends only on the x and y coordinates according to the law


align=absMiddle


where a is a constant, i and j are unit vectors along x and y axes respectively. Find the flux of the vector E through a sphere of radius R with its centre at the origin of coordinates.
its a simple problem.

Profile image of Rahul askIITiansExpert.IITR
16 Years agoGrade Upto college level
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2 Answers

Profile image of aman walia
16 Years ago

 respected sir

 

 i m little confused about this question. actually electric field is 2-d only but for anwsering flux we have to consider total surface area.

 

well i am presenting my solution:;

 

E at (R,0,0) is a/R. similarly E at (0,R,0) is also a/R. so this could be taken as the a simple charge system such that a charge q is placed at origin.

 

now

kq/R^2 =a/R

thus q=1/k* a*R

so flux is 4pi*a*R.

i am not sure this is right . if u find it wrong plz correct it.

Profile image of Akash Verma
16 Years ago

pleeeez tell d answer b'coz the answer coming is radius or size dependent which is not possible