Why in electrostatics is $dV=-E.dr$ but in electromagnetic induction, $EMF=+E.dS$?

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.

• can you please explain your last paragraph or maybe write more on it?what is the direction of increasing induced emf? – Karan Singh Aug 25 '15 at 18:00

$dV$ is the work done by an external force $q\vec E'$ in moving a unit charge $q$ over a displacement $\vec {dr}$ in another electric field $\vec E$. It's done in a way such that the kinetic energy of the charge remains zero so that the external force $qE'$ is equal but opposite to the force $q\vec E$ of the electric field $\vec E$ being probed; hence the minus sign.

The EMF isn't the work done by an external force to probe the induced electric field, but the work done by the electric field being probed itself. So there's no need for a change in sign as in the first case.