Does electric field reaches maximum exactly at the surface ($r = a$) or slightly above ($r = a+0$) in example with charged conducting sphere? A sphere of radius $a$ is in the vacuum and it's evenly charged with charge quantity $Q$.
Does an electric field has maximum exactly at the surface ($r = a$) or slightly above the surface ($r = a+0$) of this sphere?
$$E=\begin{cases}
 0 &(r < a) \\
 \frac{Q}{4\pi\epsilon_0 r^2} &(r \ge a)
\end{cases}$$

In my book it says that maximum field is at the surface of the sphere E(r = a)


In another book(University Physics with Modern Physics by Hugh D. Young, Roger A. Freedman, A. Lewis Ford) it says E for electric field at the surface of a charged conducting sphere

 A: Using classical approach to solve your problem, there will be an an ambiguity, because of the simple modelling used for the calculation of electrostatic force. In addition, when Gauss' law applied with spherical Gaussian surface for the conducting sphere that you asked result will be ambiguous and inconclusive where the singularity comes into the calculations. On the other hand with more modern approaches, ambiguity can be solved and for this I suggest this pdf where the modelling has altered to overcome ambiguity.
Reference to the external link:
Lima, F. (2018, November). What Exactly is the Electric Field at the Surface of a Charged Conducting Sphere? https://www.ias.ac.in/article/fulltext/reso/023/11/1215-1223
A: There is charge exactly at the surface of the shell, so you can't unambiguously define the E-field at that point. One reasonable convention would be to define the electric field at that point by the electric force that is exerted on the charge there. In this case it is a charge per unit area so the electric field would be defined as $E=P/\sigma$.
Since charges do not exert a force on themselves, to find the force on an infinitesimal area, the electric field of the infinitesimal area itself should be excluded. In other words, this result can be obtained by cutting a small hole out of the shell and finding the electric field due to the rest of the shell, excluding that hole.
The E-field exactly at $r=a$ found this way comes out to $E=\frac{Q}{8\pi\epsilon_0 a^2}$, or half the maximum value. The calculation is left as an exercise to the reader.
This is gives the pressure due to the electric force that is felt by the shell as
$$ P=\sigma E=\sigma\frac{Q}{8\pi\epsilon_0 a^2} = \frac{Q}{4\pi a^2}\frac{Q}{8\pi\epsilon_0 a^2}=\frac{Q^2}{32\pi^2\epsilon_0a^4}$$
which agrees with the pressure as calculated with other methods.
The more typical convention is that the electric field is simply undefined at that point. In any case, regardless of convention all the physical measurables (force, pressure, etc) are the same.
