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?
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$\begingroup$ Means, the field isn't vortex.. $\endgroup$– Earth is a SpoonMar 29, 2014 at 9:12
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$\begingroup$ Yes, electrostatic field lines don't form closed loops because $\vec{\nabla} \times \vec{E} = 0$, meaning it is a curl-free vector field. This is a property of a conservative vector field, as it can be expressed as the gradient of some function. (In this case, the electric field being $E = -\nabla V$. $\endgroup$– vs_292Jun 21, 2017 at 9:16
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9 Answers
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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.
answered Mar 29, 2014 at 8:22
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3$\begingroup$ 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? $\endgroup$ Mar 29, 2014 at 10:12
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2$\begingroup$ 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. $\endgroup$ Mar 29, 2014 at 12:19
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$\begingroup$ 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? $\endgroup$ Mar 30, 2014 at 6:49
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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.
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3$\begingroup$ 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. $\endgroup$– jvriesemJun 26, 2015 at 17:11
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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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$\begingroup$ 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. $\endgroup$– jvriesemJun 26, 2015 at 17:07
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$\begingroup$ @jvriesem: this is indeed a nice mathematical answer. But it does not address the "conservative force" aspect of the question does it? $\endgroup$– gatsuJun 27, 2015 at 19:25
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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. 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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$\begingroup$ 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. $\endgroup$– jinaweeMar 29, 2014 at 22:29
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$\begingroup$ Sorry friend jinawee, I didn't understand what you mean by open loops? $\endgroup$– SensebeMar 30, 2014 at 0:38
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$\begingroup$ 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) $\endgroup$ Mar 30, 2014 at 6:27
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1$\begingroup$ I am saying once again, if electric lines would have been closed loops, you can't assign starting and ending points and thus you can't define isolated charges. But we know that there exists isolated charges, this proves that some closed loops can't exist:) $\endgroup$– SensebeMar 30, 2014 at 16:53
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1$\begingroup$ @jinawee What are those "some closed loops" that you are stating? $\endgroup$ Mar 30, 2014 at 17:22
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A different route to the same result, which you may or may not find more intuitive, would be
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).
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.
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].
By definition field lines go in the direction of the E-field, which is the gradient of the total potential.
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].
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.
answered Mar 29, 2014 at 22:04
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$\begingroup$ "Increase monotonically"? Doesn't potential decrease when one moves along the field line?(Sorry if this is silly) $\endgroup$ Mar 30, 2014 at 6:33
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$\begingroup$ @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. $\endgroup$ Mar 30, 2014 at 10:47
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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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$\begingroup$ The conservation of energy is not guaranteed in any situation with time dependendency on the system. $\endgroup$– DavidmhMar 29, 2014 at 9:18
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1$\begingroup$ Excuse me...isnt the question about ElectroSTATIC field? Maybe we need not speak about CHANGING flux, time dependency and all? $\endgroup$ Mar 30, 2014 at 6:23
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$\begingroup$ I was just stating a fact. $\endgroup$– user42733Mar 30, 2014 at 6:36
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$\begingroup$ @parthvader Ok, fine. I didnt mean any offense. $\endgroup$ Mar 30, 2014 at 7:22
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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.
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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.
answered Feb 25, 2015 at 13:16
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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.
