Electric scalar potential?

Now we have a problem. We can only write the electric field in terms of a scalar potential (i.e., ${\bf E} = - \nabla\phi$) provided that $\nabla\times{\bf E} = {\bf0}$. However, we have just found that in the presence of a changing magnetic field the curl of the electric field is non-zero. In other words, ${\bf E}$ is not, in general, a conservative field. Does this mean that we have to abandon the concept of electric scalar potential? Fortunately, no. It is still possible to define a scalar potential which is physically meaningful.

Let us start from the equation

$$\nabla \cdot {\bf B} = 0,$$ (379)

which is valid for both time-varying and non time-varying magnetic fields. Since the magnetic field is solenoidal, we can write it as the curl of a vector potential:

$${\bf B} = \nabla\times{\bf A}.$$ (380)

So, there is no problem with the vector potential in the presence of time-varying fields. Let us substitute Eq. (380) into the field equation (376). We obtain

$$\nabla\times{\bf E} = - \frac{\partial \nabla\times{\bf A}}{\partial t},$$ (381)

which can be written

$$\nabla\times\left( {\bf E} + \frac{\partial {\bf A} }{\partial t} \right) ={\bf0}.$$ (382)

We know that a curl-free vector field can always be expressed as the gradient of a scalar potential, so let us write

$${\bf E} + \frac{\partial {\bf A} }{\partial t} = -\nabla\phi,$$ (383)

or

$${\bf E} = - \nabla\phi - \frac{\partial {\bf A} }{\partial t}.$$ (384)

This is a very nice equation! It tells us that the scalar potential $\phi$ only describes the conservative electric field generated by electric charges. The electric field induced by time-varying magnetic fields is non-conservative, and is described by the magnetic vector potential ${\bf A}$.