0
$\begingroup$

A point charge of $\rm+5.00 μC$ is located on the $x$-axis at $x=\rm5.00m$ , next to a spherical surface of radius $x=\rm4.00m$ centered at the origin.

According to Gauss's law, the net flux through the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at $x=\rm4.00m$) than on the far side (at $x=\rm-4.00m$). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)?

My Thoughts:

I know that the flux should be zero. I drew a picture, with the charge to the right of the sphere and drew lines radially from the point. However, I can't seem to wrap my head around why the net flux is zero. If I go through adding up all the $\rm\vec{E}.d\vec{A}$'s, does it have to do with the $\rm d\vec{A}$ being the opposite direction of the field on the close side to the particle, so flux is negative, but on the top, bottom, and left (front and back as well) sides of the sphere, the field is in a similar direction so flux would be positive?

$\endgroup$
1
1
$\begingroup$

It is true that it is much stronger on the side the flux is into the sphere, however if you imagine drawing a cone whose tip is at the charged point and such that the cone is tangent to the sphere (imagine an ice cream ball in a cone) then you will realize that the closer the charge is to the sphere and the stronger it there fore is on the near side, the smaller the portion of the sphere facing the charge becomes, and most of the sphere is facing away from the charge, thus negative flux is on the majority of the spheres surface.

This is equivalent to the standard way of describing gauss law by saying that as long as there are no sinks or sources in side the sphere, every electric field flux line that enters the sphere should leave it somewhere somehow!

$\endgroup$
1
  • $\begingroup$ Okay, so I'm thinking in the right frame of mind. Thanks! $\endgroup$
    – xSpartanCx
    Sep 17 '15 at 2:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.