# [Physics] Why in electrostatics is $dV=-E.dr$ but in electromagnetic induction, $EMF=+E.dS$

conventionselectromagnetismelectrostaticsinductionmagnetic fields

In electrostatics we learned that $$dV=-E.dr$$.
I understood the derivation which was used to derive this.

Now when I have come to Electromagnetic Induction,I see that when there is a time varying magnetic field then the EMF caused by the induced electric field is given by $$\int E.dS$$.
So why is there not a negative sign in the second equation?

Well I understand that EMF and potential are not exactly the same thing but their difference shouldn't cause the difference in the signs in this situation and I can probably interchange them in this situation(Am I right?I am not too sure).
I have understood the derivation of the second one as well.

So what is it that I have not understood properly?
I can't seem to understand .

In $$dV=-\vec E\cdot d\vec r$$ $E$ is electrostatic electric field and the negative sign implies that the electric potential $V$ decreases with increasing electric field $E$. This is the significance of this negative sign.

In $$\epsilon=\oint \vec E\cdot d\vec l$$

E is not electrostatic, but induced electric field. This field is induced due to time-varying magnetic fields or motional electromotive force causing a change in magnetic flux.This electric field is not a conservative field like the electrostatic field. Work done by this induced electric field around a closed loop is not zero.

Also this induced electric field does not necessarily decrease in the direction of increasing induced emf. Rather it increases. Hence the question of having a negative sign in this second equation is meaningless.

To explain the last paragraph, let us consider a conducting loop of radius R placed in a time-varying magnetic field $\vec B$ directed perpendicular to the plane of the loop. Now there will be a change in magnetic flux causing an induced emf $\epsilon=-\frac{d\phi}{dt}$.

Think of an electron in this loop.The induced electric field E is directed tangential to each point on the loop since all points on the loop are equivalent.

On one hand, from point of view of induced emf $\epsilon$, work done to move a test charge $q$ once around the loop is $q\epsilon$. Again from point of view of electric field $E$, same work done as above is $qE(2πR)$.These two work must be same.

So,$$q\epsilon=qE(2πR)$$

and combining above results, we have $$E=-\frac{1}{2\pi R}\cdot\frac{d\phi}{dt}$$

Therefore, E decreases with increase in rate of change of magnetic flux , that is, E decreases with decrease in induced emf. So,induced electric field increases in the direction of increasing induced emf.