Question

I have read that Gaussian surface cannot pass through discrete charges. Why is it so? I have even seen in application of Gauss' Law when we imagine a Gaussian Surface passing through a charge distribution, e.g. in case of infinite plane charge carrying sheet .

If it cannot pass through discrete charges how do we use it in continuous charge distributions as same 'objection' must be there for it also.

Please explain the reason.

Here E?? as, r?0

If this is ambiguity then this must be same in continuous charge distribution , otherwise please state it more clearly because we can define charge to be a spherical ball and half charge can be considered inside surface

Answer 1

If there is a point charge located on the Gaussian surface, then it is ambiguous how much of that charge should be counted as inside the surface. The same ambiguity does not arise for a three dimensional charge distribution, or when a Gaussian surface intersects a two dimensional charge distribution only over a one dimensional region.

The same ambiguity would arise if a two dimensional charge distribution coincided with the Gaussian surface over a two dimensional region.

Answer 2

Gaussian surface cannot pass through discrete charges because the significance of Gauss law is to calculate electric field at a point. This is easily done by making a gaussian surface to pass through the point at which field is required and simultaneously enclosing the charge due to which the field is required inside the gaussian surface.

To make a gaussian surface passing through a continuous charge distribution is always finite as long as the point at which Electric field is required which not present on the intersection of the Gaussian surface & the charge distribution.

For a spherical symmetric charge distribution, the magnitude of E can only depend on the radial coordinate 'r' and on the charge 'Q'. To determine E as a function of r, we use Gauss' law.

here, E $\rightarrow$ infinite & r $\rightarrow$ 0