Are the $V$ in electronics and the $\phi$ in physics the same? Is the electric scalar potential $\phi$ in physics the same thing as 'voltage', $V$, in electronics?
 A: The potential $\phi$ has more general meaning: at any point in an electric field, it is the energy acquired by a unit of charge residing at that point. This involves the following:

*

*if the field is conservative, the difference $\Delta \phi$ between any two points is the $V$ you see in electronics. In that case, $V$ (or $\Delta \phi$) is well-defined and path independent. Then in electronics, since all "points of interest" are on a closed path called "circuit", it is reasonable to assign a common reference point called ground "GND" if we wanted to talk about the absolute potential $\phi$. Otherwise, we usually use $V$ which is just the difference.

*if the field is non-conservative, the difference $\Delta \phi$ is not well-defined between any two points. That is, it is path dependent. To avoid such situation in circuits dealing with time varying fields and to keep the useful well-defined $V$ concept, we assume all fields to be totally contained in circuit elements and deal with the terminal voltages only. This view of fields inside a "black box" keeps us able to use KVL in these cases also.

In short: $V = \Delta \phi$ under suitable conditions.
A: The confusion around the concepts of voltage and scalar potential stems from the fact that so many engineering books use the same symbol (namely, V) to denote the path integral in the conservative and nonconservative cases. There is a simple way to make the two terms coexist with each other in a way that is consistent with Helmoltz's decomposition theorem and is also shared (apart from a sign convention) by the International Electrotechnical Commission (IEC):
IEC definition of voltage
IEC definition of (scalar) potential difference
IEC definition of induced voltage
If we define voltage V as the path integral of the electric field along a path
$$
V_{\gamma_{A->B}}= - \int_{\gamma_{A->B}}\vec{E}.\vec{dl}
$$
we have a quantity that is in general path dependent, meaning the work done per unit charge along a path joining point A to point B depends on the particular path chosen.
Conservative fields
When the electric field is conservative, the path integral of E becomes path independent, and we can introduce a potential function $\phi$ such that
$$\vec{E}= \vec{E_c} = -\nabla \phi$$
and the above path integral will be a function of the endpoints alone and not the path joining them.
$$
V_{\gamma_{A->B}}= - \int_{\gamma_{A->B}}\vec{E_c}.\vec{dl}=- \int_{A}^{B}-\nabla\phi.\vec{dl}=\int_{A}^{B}\nabla\phi.\vec{dl}=\phi_B-\phi_A
$$
In the conservative case, voltage (the path integral of E along a path joining two points) becomes a potential difference (the difference in the values assumed by the electric scalar potential phi  at the endpoints.) In this sense, the scalar electric potential is a specialization of the more general concept of voltage.
In lumped circuit theory, where we deal with conservative electric fields (and we hide the non-conservative part inside the lumped inductive and magnetic components - more on that later) it is customary to (ab)use the same symbol V (which denotes voltage) for the scalar potential phi, so that the above scalar potential difference can be written $V_B-V_A$ or, in short, $V_{BA}$ . In this context there is no difference between $V$ and $\phi$ and the (scalar) potential difference is the same thing as the voltage along any path joining A and B: since it does not depend on the path, we can use the term 'voltage difference' to denote the work per unit charge done on a charge that goes from point A to point B. The path independence implies that, for a conservative field $\vec{E_c}$
$$
\oint_{any \ closed \ path}\vec{E_c}.\vec{dl}=\int_{A}^{A}\nabla\phi.\vec{dl}=\phi_A-\phi_A=0
$$
Nonconservative fields
But the above relation is not true in general. In fact, for a generic nonconservative electric field E in a non-motional context, the relevant Maxwell equation in integral form is
$$
\oint_{\Gamma} {\vec{E}.\vec{dl}}=-\frac{\partial}{\partial t}\iint{\vec{B}.{dS}}
$$
Here E is the total electric field. We can invoke Helmoltz's decomposition theorem to decompose it into its irrotational and solenoidal parts $\vec{E} = \vec{E_{irr}} + \vec{E_{sol}} = \vec{E_c} + \vec{E_{nc}}$, and use the previously established relation to show that
$$
\oint_{\Gamma} {\vec{(E_c}+\vec{E_{nc})}.\vec{dl}}= \oint_{\Gamma} {\vec{E_c}.\vec{dl}} +\oint_{\Gamma} {\vec{E_{nc}}.\vec{dl}} \\
= 0 +\oint_{\Gamma} {\vec{E_{nc}}.\vec{dl}} = -\frac{\partial}{\partial t}\iint{\vec{B}.{dS}}
$$
or, by invoking Stokes' theorem
$$
\oint_{\Gamma} {\vec{E_{nc}}.\vec{dl}} = -\frac{\partial}{\partial t}\iint{\nabla\times \vec{A}.{dS}}=-\frac{\partial}{\partial t}\oint{\vec{A}.{dl}}
$$
which, in a non-motional context where the closed path $\Gamma$ does not change its shape gives us Enc = -dA/dt.
Now watch what happens to voltage when we substitute the expressions for the conservative and non-conservative parts of a generic electric field
$$
\vec{E}=\vec{E_c}+\vec{E_{nc}}=-\nabla\phi-\partial{\vec{A}/\partial{t}}
$$
...in the definition of voltage as path integral of the total field along a path gamma (in a non-motional context):
$$
V_{\gamma_{A->B}}= - \int_{\gamma_{A->B}}\vec{E}.\vec{dl}=- \int_{\gamma_{A->B}}\vec{E_c}.\vec{dl}- \int_{\gamma_{A->B}}\vec{E_{nc}}.\vec{dl} \\ =\int_{A}^{B}\nabla\phi.\vec{dl}+\int_{\gamma_{A->B}}\partial{\vec{A}}/\partial{t}.\vec{dl}
$$
Voltage in general
We conclude that, in general:
$$
V_{\gamma_{A->B}}=(\phi_B-\phi_A)+\int_{\gamma_{A->B}}\partial{\vec{A}}/\partial{t}.\vec{dl}
$$
That is, voltage (the path integral of the total electric field along a specific path) is the sum of scalar potential difference (which is path-independent) and the contribute to the emf (which is path-dependent). This decomposition of voltage can be found, for example, in Popovic & Popovic's introductory EM textbook.
If you want to recover the emf, specify a closed path - the potential difference will go to zero.
If you are interested in the scalar potential difference you need to remove the non-conservative component a from the total electric field E and compute the path integral of
$$
\vec{E_c}=\vec{E}-\vec{E_{nc}}=\vec{E}-(-\partial{\vec{A}}/\partial{t}) \\
=\vec{E}+\partial{\vec{A}}/\partial{t}
$$
in accordance with what specified in the IEC definition for scalar potential difference (in a non-motional context). That plus sign before the time derivative of A makes it seem that the Ec field is somewhat some sort of resultant field, while in reality it is just a component of the total field E.
To summarize: voltage as the path integral of the total electric field is the more general quantity (and it comes from the definition of work done on a charge), while the (scalar) potential difference is a specialization that is valid only where the electric field is conservative. In lumped circuit theory we can afford the luxury of confusing these two quantities (there is a way to treat the emf contribute of a coil as if it were a voltage drop), but there are unlumpable circuits - like the Lewin ring - where this is no longer the case.
A: In electronics, voltage refers to the potential difference between two points in a circuit. The scalar potential $\phi$ while being similar in concept, is usually used in the context of fields. It is the change in energy per unit of charge while moving the charge. Also note that one could still ask of the difference in $\phi$ between two points in space (where the electric field resides). i.e., Electric potential difference.
