how can we find electric flux due to dipole when the surface is a plane, let say xy plane or yz plane or xz plane??```````````````````````````````````````````````
Answers
Nothing has been said about the plane except that it is z=20 plane. So we can take z=20 plane of infinite extent. Now we can take an another plane z=-20 also of infinite extent. Then at infinity we can take x=+infinity and - infinity. Similarly we can take y =+ and - infinity planes and complete the box . Since z= 20 and -20 planes are of infinite extent there is/are no special points on (x,y ) plane. So, positions (1,1,0) and (-1,-1,0) are immaterial .
Now, there is no flux through x and y =+ and - infinity, because field on those planes is zero. Then, flux is through two z=+ and -20 planes. That will be total flux through surface of the box imagined above.
According to Gauss' theorem this flux is total enclosed charge(=30C)/(epsilon zero.)
By symmetry , flux through both z planes is same. Therefore flux through plane z=20 is (15C)/epsilon zero).
Answer:
Let us consider some OA and BC be any area which is lying on the yz plane
Therefore
Area of vector A=10i
Hence vector E=10i+3j+4k
We know that the flux=vector E x vector A
We also know that the area of the vector is perpendicular to the area of plane.
Therefore the area of vector E x vector A = 0
Now the net electric flux is due to the 10i
Net flux= vector E x vector A
Vector E=10i+3j+4k vector A=10i
Net flux=10i+3j+4k x 10i
=100+0+0
=100 units