# Does changing a closed surface in the manner outlined contradict Gauss' Law?

As I understand it, Gauss' Law states that the electric flux on any arbitrary closed surface is equivalent to the sum of all charges enclosed within the surface times a constant. Mathematically, this is its expression:

$$\oint \overset{\rightarrow }{E}dA = \frac{q}{\varepsilon _{0}}$$

Ok, now let's consider a sphere of radius $r$ as the closed surface surrounding a point charge $P$ located at the center of the sphere. I will be simplifying images with drawings of a circle rather than a sphere, but I expect it should be obvious how I mean to extend this into 3d. It should look like this:

Now how Gauss' Law holds should be obvious - the electric field at all points on the surface is the same, so one may multiply the electric field at radius $r$, $E_0$, by the surface area of a sphere to get the total electric flux. Thus the flux of Sphere 1 is $E_04πr^2$, and the charge enclosed is thus:

$$E_0\varepsilon _{0}4πr^2=q=P$$

Now on a sphere we will use four points to make a plane, but to continue with our drawing let us take two points $A$ and $B$ on the surface of the sphere. The line $AB$ must meet two criteria: 1) It is perpendicular to the radius of the circle. 2) Its length is less than $r$.

Thus the line will look like this:

Now let us take the chunk on the side opposite the line and reflect it over the line to make a dimple on the circle that looks like this:

On a sphere this should look just like a dimple on a ball.

Now every point that was not reflected has the same electric flux going through it as the first sphere's corresponding point. Further, the portion that was reflected avoids complications with the area through which the flux goes through by maintaining the same curvature as before it was reflected. However, because the reflected surface is now closer to the point charge, that portion of the surface has a greater value of electric field than the same portion prior to its reflection. Thus we can conclude that the net electric flux is greater when we reflect a dimple onto our sphere. In our equation: $$\oint \overset{\rightarrow }{E}dA = \frac{q}{\varepsilon _{0}}$$ This would imply that the enclosed charge is greater because the permittivity of free space is constant, but that can not be the case because we did not change the enclosed charge $P$ at all.

I am asking for an explanation of why this contradiction comes about, or of what I am misunderstanding. :P

• You are multiplying vectors with a scalar product, not scalars, which should be obvious if you look at the integrand and the right hand side, moreover, the are of the surface is getting smaller. – CuriousOne May 25 '16 at 20:04
• @CuriousOne Is the issue you're referring to one that arises because of the angle of the electric field vector with the small area? Also, the area of the surface is not getting smaller. That was avoided by using the reflection to maintain the same curvature on every small area as before the reflection. – Striker May 25 '16 at 20:06
• When the surface is a sphere, the electric field (which is radial) is everywhere normal to the surface and the dot product is a maximum. When you deform the surface in the manner that you suggest, you have created regions which although closer to the point charge no longer have the normal to the surface coinciding with the direction of the electric field. – Evan Rule May 25 '16 at 20:08
• There is no issue, you are merely not doing the multiplication correctly. You know the field of a point charge, you can calculate your area elements and you know how to integrate. Do it explicitly and you will see that the result will be the same. – CuriousOne May 25 '16 at 20:09
• Oh alright, thanks @CuriousOne. And thanks a ton Evan, I can't believe I missed that. – Striker May 25 '16 at 20:16