My textbook mentions that electric field produced by electric charge and that produced by changing magnetic field are different in nature. After searching from various sources I found that electric field produced by changing magnetic field is not conservative while that produced by electric charge is conservative. Another statement I found out was line integral of electrostatic field between any two points is potential difference while line integral of induced electric field between any two points is electromotive force

Now combining the two can I say that potential difference is conservative while electromotive force is not conservative? Is it a right conclusion or statement written above are wrong ? Also it would be great if more difference between the two types of electrical field are pointed out. Thanks in advance for your help!!


Short answer:

Now combining the two can I say that potential difference is conservative while electromotive force is not conservative?

No, you can't say that. "Conservative" is an attribute of the electric field and not of its line integrals.

Long answer:

It is true that some electric fields are conservative while others are not. In particular, all electrostatic fields (that is, fields generated by non-accelerating charges) are conservative. A conservative field is such that its line integral on a closed path is always zero:

$$ \oint \vec{E} \cdot d\vec{s} = 0. $$

This relation can be also written in a local form by means of Stokes' theorem:

$$ \vec{\nabla} \times \vec{E} = 0. $$

A consequence of being conservative is that it is possible to define a potential difference between two points $A$ and $B$ which is independent of the path chosen to connect them. In other words, the quantity

$$ \Delta V_{AB} \equiv \int_A^B \vec{E} \cdot d\vec{s} $$

can be written as the difference between a function evaluated in $A$ and $B$, that is

$$ \Delta V_{AB} = V(A) - V(B). $$

Here is the important bit: the line integral can always be computed, even when the electric field is not conservative. However, in this case the final result depends on the path over which the line integral is computed, and hence no function $V$ (the electrostatic potential) can be defined.

The electromotive force, on the other hand, is defined as

$$ f_{\rm em} = \oint \vec{E} \cdot d\vec{s}, $$

which, by definition, can be different from zero only if $\vec{E}$ is not conservative. However, note that the closed path integral above might be zero for some paths even when $\vec{E}$ is not conservative.

  • $\begingroup$ > "the potential difference can always be computed as the line integral of the electric field, even when the latter is not conservative." the line integral can be indeed always computed, but if the field is not conservative, it is incorrect to call it potential difference (or voltage). $\endgroup$ – Ján Lalinský Feb 20 '19 at 22:40
  • $\begingroup$ Yes, you are of course right. Thanks for the comment! I have updated my answer. $\endgroup$ – lr1985 Feb 21 '19 at 7:50
  • $\begingroup$ You're welcome. There is still a minor pet peeve of mine: you write that no function $V$ can be defined. However, in practice $V$ is used even then. This is possible because it is defined through the electrostatic part of the electric field, not total electric field. For example, we talk about voltage on a coil/solenoid in AC circuits, and this can be measured, although the integral $\int \mathbf E\cdot d\mathbf s$ has no single value. $\endgroup$ – Ján Lalinský Feb 21 '19 at 13:25

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