Maxwell's Correction to Ampere's Law I have not yet officially studied Electromagnetism but am trying to teach myself at the moment.  I understand Maxwell's equations in the context of Magneto- and Electrostatics: they are equivalent, along with appropriate boundary conditions, to the Biot-Savart and Coulomb's law, respectively.  In particular, they give the magnetic field due to a particular steady current distribution and the electric field due to a particular configuration of static point charges.  
I am, however, confused about the meaning of Maxwell's equations when all terms are involved.  In principle, I understand that a changing magnetic field can induce an electric field and a changing magnetic field can induce a magnetic field.
If both a changing magnetic field and a charge distribution is present will the ${\bf E}$ that we calculate from Gauss' Law be equal to the ${\bf E}$ we calculate from Faraday's Law?  (I suspect so because of the freedom coming from the fact that divergence and curl involve derivatives and physically, from the superposition principle, we would expect the total electric field to be the sum of the electric field due to static charges and that produced due to changing magnetic field).  
Griffiths, in his text on the subject, rearranges Maxwell's equations so that field sources ($\rho$ and ${\bf J}$) are on the right hand side of the equation while fields are on the left hand side.  By doing so, we expect a current to produce a changing electric field and a magnetic field.  If this is the case, why when we considered magnetostatics, could we neglect the changing electric field produced by the current? 
A couple of other questions:


*

*In Ampere's Law + Maxwell correction term: ${\bf J}$ produces a magnetic field and a changing electric field.  How does the magnetic field produced when the changing electric field term is present compare to the magnetic field produced when the Maxwell's correction term is not present?  i.e. is less current required to produce the same magnetic field?

*Potentials ${\bf E} = -\nabla V - \partial{\bf A}/ \partial t$, ${\bf B}= \nabla \times {\bf A}$ where ${\bf A}$ is the vector potential and $V$ is the scalar potential.  How does the $V$ when there is a changing magnetic field present and so ${\bf E}$ is due to both changing ${\bf B}$ field and static charges compare to the $V$ due to just the same static charges?
Thank you in anticipation of your help.        
 A: 0) In your 2nd paragraph, I presume you meant to write "a changing magnetic field can induce an electric field" (Faraday's law.)
1) Your first question concerns, I think, a static charge distribution plus a set of changing magnetic fields (produced, perhaps, by a set of dynamic currents somewhere off-stage).  In this case, 


*

*the static charges will produce a conservative (curl-less) electric field.

*the changing magnetic fields will induce a divergence-less (non-conservative) electric field


The total electric field will be the sum of these two.  (They obviously cannot be equal, since one has a divergence but no curl, and the other has a curl but no divergence.)
2) While a current is by definition charges in motion, magnetostatics studies cases where the currents do not change with time (and net charge density = 0).  In those cases the current density $\boldsymbol{j}$ is constant in time, the resultant fields are magnetic fields that are also unchanging in time, and no changing electric fields are induced.  So they're not being neglected, they just don't exist in these cases.
3) (your first bullet)  The displacement current (aka Maxwell correction term) is required to maintain the consistency of the equation.  The classic demonstration is of a capacitor being charged via wires.  Here the current is constant but the charge on the capacitor plates is increasing linearly with time.  
Imagine trying to calculate the line integral of the magnetic field around a path surrounding the wire, without the displacement current term.  


*

*If one constructs a surface, bounded by the path, that intersects the wire, you can apply Stoke's theorem and Ampere's law and get a non-zero result.  

*However, a second surface, also bounded by the path, that passes between the plates of the capacitor, has no current crossing it and so gives a zero result for the line integral. 


The displacement current term removes this inconsistency, since the charging capacitor has a changing electric field between its plates.
4) (your 2nd bullet)  


*

*In the static charge case, $V$ is just the normal Coulomb potential. 

*For the time-varying case, the potentials are not unique: various possibilities, all related by gauge transformations, give the same fields.  For example, the Coulomb gauge (in which $ \boldsymbol{\nabla \cdot A} = 0$) gives the same formula for the potential as for the static case.  Other gauge choices will give different results.  I'm sure your text discusses the gauge conditions.

