Gauss\'s law says that the flux through a closed surface which contains neither
ID: 1264880 • Letter: G
Question
Gauss's law says that the flux through a closed surface which contains neither a sink nor a source will be zero.
It's quite clear that all field lines will have to exit somehow, but the strength of the E-field is also proportional to the inverse of the distance squared.
Where is my misconception? Thank you.
Gauss's law says that the flux through a closed surface which contains neither a sink nor a source will be zero. It's quite clear that all field lines will have to exit somehow, but the strength of the E-field is also proportional to the inverse of the distance squared. So if, for example, we have a cube, and the E field is perpendicular to one of the sides, the electric flux through that one side will be A?E = A * kq/r^2 But on the opposite side, the distance from the source of the E field will be larger, so the magnitude of the E field should be smaller. Where is my misconception? Thank you.Explanation / Answer
The OP wants an intuitive answer to an intuitive obstacle to seeing its truth. Well, the intensity of the flux is like how many lines we draw per unit area. No one line
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