IIT JEE , AIEEE Forum - Physics , Electricity

kane

a square surface of side L metre is in the plane of the paper.a uniform electric field E also in the plane of the paper is limited only to the lower half of the square surface.the electric flux in S.I. unit associated with the surface is

a-EL²

b-EL²/2 epsilon not

c-EL²/2

d-zero

arpan1

ans is zero

rate me

ashish_banga

zero
E.ds = 0
ds is always perpendicular to the surface

ankitagg

i think the answer should be zero because acc. to gauss law electric flux is equal to charge enclosed divided by epsilen not. here charge enclosed is zero therefore net electric flux should be zero

amarswamy

well ans is zero

bcoz, E-flux=E-field . area vector

E-flux=E-field .area . cos90⁰(angle between positive normal of the surface and e-field is 90⁰)

therefore E-flux=0

plz rate if use ful

kane

can anybody explain me why the angle is 90 degrees and not 0 degrees & 180 degrees for the two opp. faces of cube.plzzzz

joyfrancis

area vector is taken to be normal to the surface ..coming out of it..

so the field lines are perpendicular to the area vector..

so.. e.da = 0

kane

yaar my doubt is that if we see left and right side of the square,then how come field lines are perpendicular to the surface

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