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.

第八章静电场-2高斯定理_百度文库

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电场线示例负电荷点电荷的电力线正电荷+ 一对带等量异种电荷的电力线+ 一对等量正点电荷的电力线+ + 一对异种不等量电荷的电力线+ 2q q 带电平行板电容器中 ...

場線是由向量場和初始點設定的軌跡。在空間裏，向量場在每一個位置，都設定了一個方向。只要按照向量場在每一個位置所指的方向來追蹤路徑，就可以素描出正確的場線。更精確地說，場線在每一個位置的切線必須平行於向量場在那一個位置的方向。
在空間內，由於，伴隨著每一個點的向量，組合起來，構成了向量場，場線可以說是一個專為向量場精心打造的顯像工具，能夠清楚地顯示出向量場在每一個位置的方向。假若向量場描述的是一個速度場，則場線跟隨的是流體的流線。在磁鐵的四周灑散鐵粉，可以清楚地顯示出磁場的磁場線。靜電荷的場線稱為電場線，從正電荷往外擴散，朝著負電荷聚集。
對於一個向量場，假若我們能夠完整地描述其所有的場線，那麼，這向量場在每一個位置的方向已完全地被設定了。為了同時表示出向量場的大小值，我們必須控制場線的數量，促使場線在任意位置的密度等於向量場在那位置的大小值。
場線的圖案能夠用來表達某些重要的向量微積分概念。場線從某一個區域的往外擴散或往內聚斂可以表達散度。場線的螺旋圖案可以表達旋度。
雖然大多數時候，場線只是一個數學建構，在某些狀況，場線具有實際的物理意義。例如，在電漿物理學裏，處於同一條場線的電子或離子會強烈地相互作用；而處於不同場線的粒子，通常不會相互作用。
1851年，法拉第提出了場線的概念^[1

第八章静电场-2高斯定理_百度文库

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... 在无电荷的地方)电场线不会中断3）(在无电荷的地方电场线不会相交； ） 在无电荷的地方电场线不会相交； 在无电荷的地方)电场线不会相交4）静电场线不闭合。

 

第八章静电场-2高斯定理_百度文库

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... 1）电场线起于正电荷，终止于负电荷；电荷是电场线的“源”或“尾” 2）(在无电荷的地方)电场线不会中断； 3）(在无电荷的地方)电场线不会相交； 4）静电场线不闭合。

 

 

[DOC]真空中的静电场

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结论：静电场力作功与路径无关，静电力是保守力. 二． 静电场的环流定理. (). 说明：的环流为零，静电场力作功与路径无关，静电场是无旋场(有势场)，静电场线不闭合.

 

 

Why cant Electrostatic field lines form closed loops?

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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. electrostatics electric-fields share|improve this question edited Mar 31 at 16:10 asked Mar 29 at 7:57 Venki 1508

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

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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. share|improve this answer answered Mar 29 at 13:23 gatsu 2,276414

Ok. Point taken. –  Venki Mar 30 at 7:03 add a comment |

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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. share|improve this answer answered Mar 29 at 8:22 Robin Ekman 4,8971421

3   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 add a comment |

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The solution of Laplace's equation, ∇ 2 ϕ=0  , is a harmonic function, which has the property that it has no local minima or maxima. This implies that E ⃗ =−∇ ⃗ ϕ  can not be zero if ϕ  is not constant, hence it can be used to define a curve, the field curve with tangent vector E ⃗   pointing in the direction that ϕ  decreases. If we take the contour integral ∫ ∂A E ⃗ ⋅dl ⃗   , with A  an arbitrary surface and ∂A  its boundary (a closed curve), such that along it the inequality E ⃗ ⋅dl ⃗ ≥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 ∇ ⃗ ×E ⃗ =0 ⃗   , and, by the Stokes theorem, 0=∫ A (∇ ⃗ ×E ⃗ )⋅dA ⃗ =∫ ∂A E ⃗ ⋅dl ⃗ ,  hence we have a contradiction. Clearly, we must allow for points where the field is not continuous and the sign of E ⃗ ⋅dl ⃗   changes, which become the endpoints of the field curves that form if we break ∂A  in two pieces (one for each sign of E ⃗ ⋅dl ⃗   ), and these endpoints are the charges. share|improve this answer answered Mar 29 at 23:25 auxsvr 1,069137

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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. share|improve this answer answered Mar 29 at 9:15 Parth Vader 1,256114

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 add a comment |

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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. share|improve this answer answered Mar 29 at 10:53 Godparticle 2,3731735

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  | show 3 more comments

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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. share|improve this answer answered Mar 29 at 22:04 Henning Makholm 592312

"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