If I have 30 squared meter area CUBIC surface enclosing a point charge, and the question ask me to find the Flux generated on the surface, is it possible for me to use a 30 squared meter area SPHERICAL surface in my calculation instead because it is easier to integrate and taken the fact that the Flux generated by this one point is a constant?
I know you can also do Q/epsilon to get the Flux, but would the above method get the same answer too?
2007-09-15 10:23:54 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
You already have the answer if you are using Gauss's law. The flux through the entire closed surface is q/E0. Due to symmetry, the flux through ONE of the faces of the cube is
q/(6E0) (1/6 of the total flux through each side.)
2007-09-15 13:09:37 · answer #1 · answered by supastremph 6 · 0⤊ 0⤋
So we have a cube right? Since a cube has square side, it's dimensions are the same. A cube is a symmetric. Suyrely, you can find the surface area of a cube. It is different that a spherical case, but not by much maybe.
The reason we apply Gauss' Law to a symmetrical surfaces is so that we don't have to solve the integral on the left hand side. It simply becomes the electric field times area. This is true for all symmetrical surfaces.
2007-09-15 10:37:52 · answer #2 · answered by Anonymous · 0⤊ 0⤋