# E-induced lines

In electrostatics, × E = 0 so E that is a conservative field and there must be sources of E from which E flows. We know that this sources are the electrical charges given by Gauss' Law.

But when B changes in time, × E = - ∂ B / ∂t. Now the Gauss' Law no longer applies and if there are not net charges anywhere, there are no sources of E, so ∇ ⋅ E = 0.

So how are the lines of an induced E? Are they like B lines in magnetostatics? They just "turn" around something and they don't have any start or end?
And if they are, since Lenz's Law says that ε = - ∂φ / ∂t, are the lines of this E induced exactly the opposite of the B that induces it?

Please let me know if i'm not making my self clear, my english is not that good.

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In electrostatics, × E = 0 so E that is a conservative field and there must be sources of E from which E flows. We know that this sources are the electrical charges given by Gauss' Law.

But when B changes in time, × E = - ∂ B / ∂t. Now the Gauss' Law no longer applies and if there are not net charges anywhere, there are no sources of E, so ∇ ⋅ E = 0.

So how are the lines of an induced E? Are they like B lines in magnetostatics? They just "turn" around something and they don't have any start or end?
And if they are, since Lenz's Law says that ε = - ∂φ / ∂t, are the lines of this E induced exactly the opposite of the B that induces it?
Yes, without charges, but with changing magnetic field, the electric field lines have neither start nor end. But Lenz's Law states that the electromotive force ε is exactly opposite to the change of B that induces it. The time-dependent B is related to the curl of the electric field: curl E = -∂B / ∂t, or in integral form: $\oint Eds = -\partial φ / \partial t$ (the line integral of the tangential component along a closed curve is equal to the negative of the flux across the enclosed area).