Mass energy of electron is 0.510 998 9461MeV/ c2
1) Does it include energy of electron due to electric field too?
2) If yes, how much?
I am more interested in second part.
Links:Here John Rennie says:
Nevertheless, you could use the above reasoning to claim that a charged electron actually has a lower energy than an uncharged one would. Now there's an unexpected result :-)
Can anyone tell how much exact less energy it will have?
Or is it just, we can't because we don't have uncharged electron.
There is the concept of the self energy of the electron that can be described by its electric field. The model, which goes back to classical physics, can be described as a spherically distributed charges (that equal the charge of the electron) at infinity and the work required to bring the charges to the radius of the electron. The calculation would show the total energy of the electron due to its own charge and ensuing electric field. The standard model describes the electron as a point charge. If you do this calculation to a zero point, the electron would have infinite mass, however we know what it should be. The calculated radius however is much larger that the upper limit of the electron radius as measured in high energy probe experiments into subatomic particles. The electron should have much more mass due to its own self energy of its own electric field. This is a big puzzle and renormalization techniques are used to arrive at the mass of the electron. Feynman did a lot of work in this area and spends quite some time describing it in his Nobel Prize speech.
To begin with, I can't think of any reason why an increase in the mass of an electrically charged object should affect the magnitude of the electric field it produces. You can easily check that $E=kq/r^2$ doesn't depend on $m$. Moving charges do produce magnetic fields, but this is a different matter (and the magnitude of the magnetic field has nothing to do with the mass of the object, neither).
On the other hand, as it was noted in the comments, mass is a Lorentz scalar, then a relativistic invariant quantity (i. e., it is the same on any reference frame). In fact, the Minkowski norm (length) of the 4-momentum is $p^{\mu} p_{\mu}= -m^2c^2$. Only the components of a 4-vector are modified when subject to Lorentz transformations, never its norm.
Edit: To sum up, as far as I know charge/electric field doesn't affect to the mass/energy of any charged object, is it moving or at rest. By "electric energy" you might be referring to the electric potential energy or to the electric field density, you can see that both concepts are unrelated with mass.
2) Does it include energy of electron due to electric field too?
We do not know. If the electron was composed of charged elementary parts, its apparent inertial mass would be greater than sum of inertial masses of the parts. This is because such parts will act on each other with electromagnetic forces and in accelerated motion, sum of these internal forces is not zero, but can be expressed as $-m_{em}\mathbf a$, where $m_{em}$ is a positive number. This force can be put next to the expression $m\mathbf a$ in the equation of motion to get the approximate equation
$$ (m+m_{em})\mathbf a = \mathbf F_{ext}. $$
Hence $m_{em}$ is called electromagnetic mass(for slow motions of charged sphere, this was derived by Lorentz). The apparent mass $m+m_{em}$ will be greater than sum of individual masses $m$.
If the electron was a point and therefore not composed of parts, there would be no such effect, for there are no internal forces due to any parts.
It is necessary that at least part of apparent mass is non-electromagnetic; for electron made of parts, non-electromagnetic forces need to be present to keep the parts of electron together and these forces probably have some similar effect on the apparent mass; for point electron, there are no mutual electromagnetic forces at all, so they cannot result in any change in apparent mass.
