Why doesn't a gaussian surface pass through discrete charges?

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 $E \rightarrow \infty$ as, $r\rightarrow 0$

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 (as in pic and even agreed by @JoshuaBarr).

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.

• But i have seen a question where they specified that gaussian surface cuts the charge in half then what is the flux. Answer was still can't be told. WHy is it so if the ambiquity is just that we can't tell amount of charge inside the gaussian surface – ABC Mar 22 '13 at 15:53
• @exploringnet If the charge is truly cut in "half" then you can just treat this by allowing the charge to have some extent (e.g. a sphere) and then taking the limit where its size goes to zero. The flux would simply be (1/2) Q / epsilon_0. – Joshua Barr Mar 22 '13 at 16:01
• Please see my extended question. – ABC Mar 23 '13 at 6:04
• The limit of the flux will converge in the picture you drew. This is so because for any finite value of the radius, Gauss' law will hold just fine. If you take the limit in such a manner that the sphere is always split in half, then the flux will always be (1/2) Q / epsilon_0. Remember that the limit depends on the behavior of the flux with the radius near R = 0 but not on the value at R = 0. – Joshua Barr Mar 24 '13 at 3:55

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. If the gaussian surface is made to pass through a discrete charge itself then the significance of gauss law is lost as at the point where the discrete charge itself is present the value of electrical field becomes undefined(inverse square law). And hence field cannot be calculated.

Secondly,to make a gaussian surface passing through a continuous charge distribution is always fine as long as the point at which field is required is not present on the intersection of the Gaussian surface and the charge distribution.

The gaussian surface is the one where you compute the product $\vec{E}\cdot d\vec{S}$

$d\vec{S}$ is the normal unit vector multiplied by the differential area: $d\vec{S}=S\cdot \hat{n}$. This vector will always exist, BUT the electric field is not defined on the charge itself, so you cannot compute the flux in that point, hence you cannot know the value of the integral so the formula is absurd, as one of the sides is undefined.

The electric field of a point charge is defined everywhere except the very point where it is placed. It is infinitely small (a point, 0 dimension). The closest environment has it defined, but not the concrete point.

Now I read the second part of your question. A continuous distribution has the same problem anytime you find a point with undefined electric field. Where you don't know $\vec{E}$ you cannot use this. The thing is that many times we do KNOW the electric field because of simmetries.