Electric Field from a Uniformly Charged Disk As I was reading the solution of this problem the author gave the electric field in the point P as follows:
$$ \vec{E} = \sigma /(2 \epsilon ) [1-x/(x^2+R^2)^{1/2}]\ \hat \imath$$
Where:


*

*$\sigma$ is the surface charge density on the disk

*$x$ is the distance from the center of the disk to the point P

*$R$ is the radius of the disk


Here the question comes: for $R \rightarrow 0$ with keeping $Q$ constant (the total charge) why is it that $E$ should go to a point charge?
According to my knowledge $E$ goes to zero as we calculate the limit of $E$ while $R$ tends to zero. Even if the absolute value of $x$ gives $-x$ we still get another constant. 
 A: First, I think a little intuition could help. If you imagine a disk with a charge $Q$ get smaller, but all the while keeping the charge Q intact, shouldn't it geometrically approach a point with charge Q? So, in theory, we should expect the field to approach that due to a point charge.
Now, intuition aside, let's go to the mathematics. While the factor $\left[ 1-x/(x^2 + R^2)^{1/2} \right]$ does go to zero as $R\to0$, the charge density $\sigma = Q/(\pi R^2)$ goes to infinity. This is the source of your problem, because you end up with an indeterminate form $\infty \times 0$ while calculating the limit, and not 0.
To evaluate the limit, you could use l'Hôpital's rule, after rewriting your formula as an appropriate fraction (and substituting $\sigma$ for its expression in terms of $R$).
As a bonus, a quick way to do this would be to use this handy approximation, which works for small $y$ values:
$$
(1+y)^n \approx 1+ny.
$$
You can use this if you factor $x^2$ from the square root:
$$
\frac{x}{ \left(x^2 + R^2\right)^{1/2} } = 
\frac{x}{ \lvert x \rvert \left(1 + (R/x)^2\right)^{1/2} } =
\frac{x}{\lvert x \rvert}\left(1+\left(R/x\right)^2 \right)^{-1/2}\approx\frac{x}{\lvert x \rvert} \left( 1 - \frac{R^2}{2x^2}\right).
$$
Here I used $y=R/x$ and $n=-1/2$. If $R$ goes to 0, then $y$ goes to 0, and this approximation gets better. Replacing in the expression you have given, we end up with
$$
\vec E = \frac{Q}{4 \pi \epsilon x^2} \frac{x \hat \imath}{\lvert x \rvert} = \frac{Q}{4 \pi \epsilon x^2} \frac{\vec x}{\lvert x \rvert},
$$
which is the expression for a field due to a point charge. (Notice that the term $\vec x / \lvert x \rvert$ only gives you the direction of the field, but doesn't change its magnitude.)
Edit: if you try to do the calculations for $x<0$ you'll end up in trouble. The actual formula for the electric field should be
$$
\vec E = \frac{\sigma}{2\epsilon}\left[ \frac{x}{\lvert x \rvert} - \frac{x}{\left(x^2 + R^2 \right)^{1/2}} \right] \hat \imath,
$$
which you can see if you follow the derivation of the equation.
A: I think the mystery comes from the fact that as $R \rightarrow 0$ you will find that $\sigma$, the area charge density, will become infinitely great. The two effects "cancel out" such that the resulting electric field is that of a single point charge. 
Note that $dq=\sigma dA$, and because we are preserving $Q$, as $R$ shrinks, the charge on the disk will become more and more compactly assorted. Namely, $dq$ will grow larger for the given strip of area. Consequently $\sigma$ will grow. 
In other words, (roughly) think of $\sigma=\frac{dq}{dA}$, where we are decreasing $dA$ while keeping the charge dq constant. Consequently $\sigma$ increases. 
