- #1
kaotak
Consider a gaussian surface of arbitrary size and a point charge located outside of the gaussian surface at an arbitrary distance.
Gauss's law states that the flux through the gaussian surface is zero, since there is no charge enclosed by that surface. From this we can deduce that the electric field must be zero everywhere on the surface, since the flux is equal to the integral of the dot product of the electric field and dA. But from Coulomb's Law we know that the E = kq/r^2 at any point on the surface, where r is the distance from the point charge.
It seems to matter what gaussian surface you choose and whether or not it encloses the point charge. If you choose a spherical gaussian surface centered around the point charge, you can easily derive E = kq/r^2. So why does this contradiction occur if you choose a gaussian surface that does not enclose the point charge? Are you simply not supposed to or allowed to do that?
Gauss's law states that the flux through the gaussian surface is zero, since there is no charge enclosed by that surface. From this we can deduce that the electric field must be zero everywhere on the surface, since the flux is equal to the integral of the dot product of the electric field and dA. But from Coulomb's Law we know that the E = kq/r^2 at any point on the surface, where r is the distance from the point charge.
It seems to matter what gaussian surface you choose and whether or not it encloses the point charge. If you choose a spherical gaussian surface centered around the point charge, you can easily derive E = kq/r^2. So why does this contradiction occur if you choose a gaussian surface that does not enclose the point charge? Are you simply not supposed to or allowed to do that?