How Does Potential Difference Relate to Faraday's Law?

[Nikhil Rajagopalan]

The potential difference between two points is given ans the negative of integral of E(vector) <dot product> dl(vector) from initial to final points.
Therefore, integral integral of E(vector) <dot product> dl(vector) from initial to final point should give the negative of potential difference between them.
In Faraday's law, closed loop integral of E(vector) <dot product> dl(vector) is given as ε- induced. Why is it not the negative of ε-induced. Should ε-induced not be treated like potential difference?

 

[Charles Link]

The voltage from an EMF with $V=+\int E_{induced} \cdot ds$ gets computed in just the opposite way of how the voltage from an electrostatic source gets computed as $V=-\int E_{electrostatic} \cdot ds$. It should be noted, the induced $E$ does not give rise to a potential and, in general, $\nabla \times E_{induced} \neq 0$, so that we cannot write $E_{induced}=-\nabla \Phi$. The $V$ from a Faraday EMF is a voltage, at least when it is observed in an inductor, but it is not a potential type function. $\\$ In a conductor, since $E_{total}=0=E_{induced}+E_{electrostatic}$, we have $E_{electrostatic}=-E_{induced}$. The argument can be made that this is why the voltage from an inductor is in fact $V=\mathcal{E}=+\int E_{induced} \cdot ds$.