Gauss' Law and Electric Field Close to a Ball So I've learned about Gauss' law and I have something in my head. Why does electric field that is very close to a ball is not close to infinity. Take a look at this image:

As we can see, if we make a partition of the shell, we get a particle with positive charge. The distance between this charge and its test point is very close to zero so that $dE$ is infinity, another partition will do the same thing with a sum of the same direction (without cancelling out each other). However, if we apply the Gauss' law, we can then assume that the ball is just a particle with the distance from the test point is being measured from its center.
So, what is the explanation for this question?
 A: To clear your confusion, let us try to find the field contributed by a ring section close to the test charge P as shown in the figure.


In the figure, $\angle AOP=\theta$ and $AB=R$. Since the ring we have taken is close to $P$, we can say that $\theta \to 0$. 

Now, the area element of the ring is $dA=2\pi R^2 \sin(\theta) d\theta = 2\pi R^2 \theta d\theta$ and the distance $PA= R\theta$. Thus the field contributed by this element is ($\sigma$ is the surface charge density):
$$dE= \frac{1}{4\pi\epsilon_{0}}\frac{\sigma dA}{(PA)^2} \cos(\angle APO)$$
$$= \frac{1}{4\pi\epsilon_{0}}\frac{\sigma 2\pi R^2 \theta d\theta }{R^2 \theta^2} \cos(\frac{\pi}{2}-\frac{\theta}{2}) $$
$$=\frac{\sigma d\theta}{4\epsilon_{0}}$$
This is a constant and is not infinite.

Now its understandable why you have that doubt. As the ring gets smaller and smaller, the distance tends to zero. But since the ring is getting smaller, the charge in the element also decreases and since the field is proportional to the ratio of the charge to distance squared, the 2 effect sort of cancels each other and makes the field constant.
