A cylinder is lying with axis parallel to uniform electric field E.what is total flux through the cylinder with area of cross section S and length l.can this example be used to verify gauss' law?
Answers
Answer:
he cylindrical Gaussian surface has three parts, the two flat ends and the cylindrical surface. So you have the field entering end 1 at right angles to it, passing parallel to the cylindrical surface CS, and then exiting end 2 again at right angles to it.
The corresponding flux through the curved surface would be zero since the outward pointing surface vector ∆S (CS) at all points on it would be perpendicular to the surface (and therefore the axis) while the field is parallel to the axis of the cylinder. Thus phi (CS) = 0.
If the area of each of the flat ends is A, the contribution to the total flux at end 1 would be phi (End 1) = - E*A (the negative sign arises because the surface vector A is outward and opposite to the field E ).
Similarly, the flux through end 2 would be given by phi (End 2) = E*A (the positive sign arises because the surface vector A is outward and in the same direction as the field E ).
Adding the three contributions we have
phi (CS) + phi (End 1) + phi (End 2)
= 0 - E”A + E*A = 0
Explanation: