Does it make sense to speak of the value of electric field at points containing continuous charge distribution? Before getting into the question, here are some remarks:


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*Given a single point charge, the value of electric field at the position of the charge is singular/undefined, which makes sense, since a particle cannot interact with itself.

*Given a discrete charge distribution, the value of the field at an empty point(i.e. no particles reside at that point), is the field contribution from all the charges. However, if the point of interest contains a point-charge, then the value of the field is the field contribution from all the charges except the charge that resides at that point.
Now here's my inquiry: In the case of discrete charge distribution, it makes sense to speak of the value of the electric field at a given point, whether that point contains a point-charge or not. What about continuous distributions?
My intuition says yes, we can; since one can think of continuous distributions as an extension of discrete ones, with the difference being that the former contains uncountably infinite charges. Therefore, to calculate the field value at any given point on the continuous charge distribution, one has to consider the field contributions from all the charges except the one residing at that point.
Example: The value of the electric field at any point on an infinite sheet (plane) of charges should be zero. Since at any given point on the sheet, one can think of that point as being surrounded by infinite concentric rings, where the field contribution from each ring (by symmetry) is zero.
So does it make sense to speak of the value of electric field on continuous charge distribution? Or is it not defined? 
 A: As I understand your question, you wonder how we can get a well-defined electric field of a continuous charge distribution if, when forming it as a limit of discrete sets of point charges with increasing density, we'd get a function undefined at every point.
This is indeed a valid concern when viewed in a rigorous mathematical sense. But the trick is as usual in physics: when something doesn't work starting with the naïve, simple definitions, try generalizing these definitions. In particular, instead of using functions to describe charge densities, we can switch to distributions, also known as generalized functions.
One of such generalized functions, perfect for description of point charge density, is the Dirac delta. It's the distribution you'll get on the RHS of the Poisson's equation if you pay careful attention when evaluating Laplacian of the Coulomb potential.
Now, if you build your continuous charge distribution as the limit of a sequence of a delta-forest (Dirac comb) of charges, you'll naturally arrive at the distribution where there're no singularities, and every point is well-defined — in the distributional sense, so we can then consider the resulting charge density as a usual (piecewise-)smooth function.
In particular, consider a 1D charge distribution $\rho(x)$ that we want to obtain, and choose a sequence of distributions like
$$\rho_n(x)=\rho(x)\frac1n\sum_{k=-\infty}^{\infty}\delta\left(x-\frac kn\right),$$
which represents the sequence of sets of point charges, converging to a continuous charge distribution.
If you choose an interval $x\in(x_0-\frac\varepsilon2,x_0+\frac\varepsilon2)$ and average $\rho_n(x)$ over it, you'll get that the sequence of integrals (total charges in the interval)
$$Q_n=\int\limits_{x_0-\varepsilon/2}^{x_0+\varepsilon/2}\mathrm dx\,\rho_n(x)$$
converges to some value, and this value, when divided by $\varepsilon$ and considered in the limit of $\varepsilon\to0$, is nothing other than $\rho(x_0)$.
Note that this averaging, while being natural for manipulation of distributions, is actually exactly what we do in physics when approximating multitudes of real discrete charges (electrons, atomic nuclei etc.) with continuous charge densities. So using distributions for this purpose should be be expected to yield good results.
A: 
So Does is make sense to speak of the value of electric field on continuous charge distribution? or is it not defined?

Yes, it does make sense to have field at point where charge density is finite and non-zero - at least mathematically. This kind of situation is the basic one of macroscopic electromagnetic theory.
A: It is well-defined. One way to think about this is:-
Continuous charge distributions can be thought to be made up of many small point charges $dq$ Now for a discrete point charge $q$ the field at zero distance diverges to infinity but for $dq$ the "value" of the field would be:
$E = kdq/r^2$ with both $dq$ and $r$ tending to zero. Thus as we get a 0/0 limit form the answer can be a real number. The contributions from the other infinitesimal charges will also add up but decrease the further they are.
