Gaussian surfaces and charge

[pringless]

A positive test charge is placed at the center of a spherical Gaussian surface. What happens to the net flux through the Gaussian surface when the surface is replaced by a cube of the same volume whose center is at the same point? or When the sphere is replaced by a cube of one third the volume centered at the same point?

 

[Chi Meson]

same

same

same


See Gauss' Law. Net flux through ANY closed surface depends only on...

 

[M98Ranger]

The net flux through the Gaussian surface will remain the same in both cases. This is because the electric field lines passing through the center of the spherical Gaussian surface will also pass through the center of the cube, regardless of its volume. This is due to the fact that the electric field is spherically symmetric around the positive test charge, meaning the direction and magnitude of the field will be the same at any point along the surface, regardless of its shape. Therefore, the net flux through the Gaussian surface will not change when it is replaced by a cube of the same volume or a cube with one third the volume.