# Why cant Electrostatic field lines form closed loops?

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 cant quite understand. Can anyone elaborate?

PS:- This question is answered. Please refer both the accepted answer and also the answer and the comments in Robin Ekman's answer for a complete understanding.

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Means, the field isn't vortex.. –  Sachin Shekhar Mar 29 at 9:12

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 $A \equiv B$ 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.

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Ok. Point taken. –  Venki Mar 30 at 7:03

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.

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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? –  Venki Mar 29 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 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? –  Venki Mar 30 at 6:49
Yes, that's it. –  Robin Ekman Mar 30 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.

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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.

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

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.

Field lines of a bar magnet.

Field lines of an electric dipole.

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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 at 22:29
Sorry friend jinawee, I didn't understand what you mean by open loops? –  Godparticle Mar 30 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) –  Venki Mar 30 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. –  Godparticle Mar 30 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 at 13:48

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.

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"Increase monotonically"? Doesn't potential decrease when one moves along the field line?(Sorry if this is silly) –  Venki Mar 30 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 at 10:47
Right. Point taken. –  Venki Mar 30 at 17:01