Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My physics textbook says "Electrostatic field lines do not form closed loops. This is a consequence of the conservative nature of electric field." But I can't quite understand this. Can anyone elaborate?

share|cite|improve this question
Means, the field isn't vortex.. – Evil Angel Mar 29 '14 at 9:12
up vote 8 down vote accepted

A force is said to be conservative if its work along a trajectory to go from a point $A$ to a point $B$ is equal to the difference $U(A)-U(B)$ where $U$ is a function called potential energy. This implies that if $A=B$ then there is no change in potential energy. This fact is independent of the increase or not of the kinetic energy.

If a conservative force were to form loops, it could provide a non zero net work (because the direction of the force could always be the same as that of the looping trajectory) to go from A and then back to A, while at the same time its conservative character would ensure that this work should be zero; which is a contradiction.

Hence, "conservative force" and "forming loops" are two incompatible properties that cannot be satisfied at the same time.

share|cite|improve this answer
For those interested in the mathematics of this, check out @auxsvr's answer below. Note that only electrostatic field lines cannot form closed loops. If the magnetic flux is changing, it will induce a current via closed loops of electric field. – jvriesem Jun 26 '15 at 17:11

If there was a closed field line a particle following that line would eventually return to the same place but having a different energy so the field would not be conservative.

share|cite|improve this answer
The particle would have gained a lot of kinetic energy while it moved along that closed loop and when it stops (so that its displacement became zero) its kinetic energy would have become potential energy. This potential energy would be greater than the initial potential energy. Is this what you mean when you say that the particle's energy would be different? – VenkiPhy6 Mar 29 '14 at 10:12
The particle doesn't have to come to a stop, the point is that the kinetic energy is different when it comes back to where it started. This means that you can't define a potential energy. For motion in a conservative field the total energy, kinetic plus potential, is conserved. Since potential energy depends only on the particle's current position, the change in kinetic energy when the particle moves between A and B doesn't depend on how the particle gets to B. In particular when A and B are the same point, the kinetic energy can't change at all. – Robin Ekman Mar 29 '14 at 12:19
So the particle will differ in (ie would have gained) Kinetic energy when it comes back to the same point, in case of a closed loop. But Kinetic energy at the same point should not differ as displacement and so velocity and so Kinetic Energy gained would be zero when the particle comes back to the initial point. This is a contradiction and so there cant be closed loops. Right? – VenkiPhy6 Mar 30 '14 at 6:49
Yes, that's it. – Robin Ekman Mar 30 '14 at 23:29

The solution of Laplace's equation, $\nabla^2 \phi =0$, is a harmonic function, which has the property that it has no local minima or maxima. This implies that $\vec{E} = -\vec{\nabla}\phi$ can not be zero if $\phi$ is not constant, hence it can be used to define a curve, the field curve with tangent vector $\vec{E}$ pointing in the direction that $\phi$ decreases. If we take the contour integral $\int_{\partial A} \vec{E}\cdot d\vec{l}$, with $A$ an arbitrary surface and $\partial A$ its boundary (a closed curve), such that along it the inequality $\vec{E}\cdot d \vec{l}\geq 0$ is satisfied, the integral must be $>0$. This is equivalent to the statement that the work done is positive for a positively-charged particle moving along the field line. However, for the static case Maxwell's equations yield $\vec{\nabla}\times \vec{E} = \vec{0}$, and, by the Stokes theorem, $$0=\int_A (\vec{\nabla}\times \vec{E}) \cdot d \vec{A} = \int_{\partial A} \vec{E} \cdot d \vec{l},$$ hence we have a contradiction. Clearly, we must allow for points where the field is not continuous and the sign of $\vec{E} \cdot d \vec{l}$ changes, which become the endpoints of the field curves that form if we break $\partial A$ in two pieces (one for each sign of $\vec{E}\cdot d \vec{l}$), and these endpoints are the charges.

share|cite|improve this answer
This is the best mathematical answer, and ought to be included with the accepted answer, above. I would add, however, that the curl of electric field is only zero iff the magnetic field is time-independent. Because the question is asking about a static case ("electrostatic field lines"), then this condition is satisfied. – jvriesem Jun 26 '15 at 17:07
@jvriesem: this is indeed a nice mathematical answer. But it does not address the "conservative force" aspect of the question does it? – gatsu Jun 27 '15 at 19:25

A different route to the same result, which you may or may not find more intuitive, would be

  1. By definition the electrostatic field is the sum of the Coulomb fields of all of the source charges. (In case of a continuous charge distribution, we can either consider the limit of a collection of ever smaller point charges, or replace the sum with an integral, but I will not worry about such mathematical pedantry here).

  2. Each Coulomb field for a point charge happens to be expressible as the gradient of a scalar potential field. We don't need to know much about the Coulomb field to know this, only that it is rotationally symmetric about the point charge.

  3. Therefore the electrostatic field (which is the sum of the point-charge Coulomb fields) is also the gradient of the sum of the point-charge potentials. [More mathematical pedantry swept under the carpet here].

  4. By definition field lines go in the direction of the E-field, which is the gradient of the total potential.

  5. Therefore, as we move along a field line, the potential increases monotonically. [Hmm... there's a sign convention going wrong here -- the potential I'm talking about is minus the usual meaning of potential, but never mind that].

  6. However, if there were a closed-loop field line, when we got back to the starting point, we would now be at the same potential we started out at, but all the way around the loop it has been increasing all the time. This is absurd, so it can't happen.

share|cite|improve this answer
"Increase monotonically"? Doesn't potential decrease when one moves along the field line?(Sorry if this is silly) – VenkiPhy6 Mar 30 '14 at 6:33
@Venik: Yes, that's the "Hmm..." comment immediately afterwards. In my train of thought I had considered the potential to be something whose gradient is the field, but in the usual physical convention the field is minus the gradient. That makes no difference for the high-level argument sketched here. – Henning Makholm Mar 30 '14 at 10:47
Right. Point taken. – VenkiPhy6 Mar 30 '14 at 17:01

Robin is right in stating that if Electric Fields form closed loops, they wouldn't be conservative. But keep in mind that non-conservative Electric Fields can also be produced in some situations, like changing magnetic flux.

share|cite|improve this answer
The conservation of energy is not guaranteed in any situation with time dependendency on the system. – Davidmh Mar 29 '14 at 9:18
Excuse me...isnt the question about ElectroSTATIC field? Maybe we need not speak about CHANGING flux, time dependency and all? – VenkiPhy6 Mar 30 '14 at 6:23
I was just stating a fact. – user42733 Mar 30 '14 at 6:36
@parthvader Ok, fine. I didnt mean any offense. – VenkiPhy6 Mar 30 '14 at 7:22

If I assume electric field lines form closed loop that would mean electric field has non zero curl. So I cant write electric field to be gradient of some scalar function.

That would imply work done by the electric field will depend on the path. we know thats not the case really.

Another way to see this: Closed electric field lines would mean number of field lines passing thru a closed surface is zero and thus divergence of electric field is zero which cannot be the case since electric monopole always exist.

share|cite|improve this answer

If electric field lines would have been closed loops, then there would have been no isolated electric charge as like there exists no isolated magnetic pole. So, this is another reason why electric field lines can't form closed loops.

The magnetic field lines of a magnet form continuous closed loops, this is unlike electric dipole where the field lines begin from a positive charge and end on the negative charge or escape to infinity.
enter image description here

Field lines of a bar magnet.

enter image description here

Field lines of an electric dipole.

share|cite|improve this answer
If electric field lines would have been closed loops, then there would have been no isolated electric charge as like there exists no isolated magnetic pole. There could be some open loops, so the following reasoning is wrong. – jinawee Mar 29 '14 at 22:29
Sorry friend jinawee, I didn't understand what you mean by open loops? – Immortal Player Mar 30 '14 at 0:38
How is the existence of a closed loop going to affect the existence of an isolated charge? Can you elaborate?(Sorry if this is silly. But I am an amateur) – VenkiPhy6 Mar 30 '14 at 6:27
If field lines were not closed loops and they began and ended at one of the charges, you can have isolated charges, as you have some end point and beginning point. If you have closed loops and they didn't begin and ended at some pole, you can't have isolated poles, as there is no point where you can assign them to start and end. If there were to be isolated magnetic poles, you should define it to begin or end, which is not true in case of magnetic field, as they neither end or begin at any point, they are continuous. – Immortal Player Mar 30 '14 at 9:10
I mean, you have proved that open electric lines must exist. But you haven't proved that some closed loops couldn't exist. The same applies for the magnetic field. We've only seen closed loops, but if there were magnetic monopoles, we would have closed and non closed loops. – jinawee Mar 30 '14 at 13:48

If the lines of electric feild form closed loops, the Guass Law won't be suitable. Therefore you can get the conclusion that the conservation of electric charge doesn't exist anymore. The lines of magnet feild form closed loops because there is no such thing like magnetic monopoles.

share|cite|improve this answer

An essential feature of a conservative vector field is that the line integral of the field is path indendent, i.e., the value of the line integral between two points depends only on the two points, not the path taken between the points. This is why we can associate a scalar potential function with a conservative vector field.

If the field forms a closed loop, the above cannot hold. This is most easily seen in the degenerate case that the two points are the same point.

In the case of a conservative field, the line integral must be zero since one of the possible paths is the zero length path and the result must not depend on the path.

However, if the field forms a closed loop, then one of the possible paths is along the closed field line which must give a non-zero result.

Thus, if the vector field is conservative, there can be no closed field loops.

share|cite|improve this answer

protected by Qmechanic Oct 2 '15 at 19:50

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.