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Circulating electric fields

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider two “circulating” electric field configurations. Expressed in polar coordinates
    (s,φ,z) they are: (these expressions are not dimensionally correct)
    1. E = (0,s,0)
    2. E = (0,1/s,0)

    a. Calculate ∇×E for both configurations.
    b. Note that ∫E⋅dl≠ 0 in either case. Explain the answer to part a in light of this fact.
    c. Is either configuration a valid electrostatic field? Why or why not?


    2. Relevant equations
    Curl in cylindrical:
    just the φ since the rest are 0
    ∇×V=[tex]\frac {\partial E_s}{\partial x} - \frac {\partial E_z}{\partial s}[/tex]


    3. The attempt at a solution
    ∇×V=[tex]\frac {\partial E_s}{\partial x} - \frac {\partial E_z}{\partial s}[/tex]

    I believe these should be zero for both since φ is not a part of either of these.

    So the curl is zero but ∫E⋅dl≠ 0. I just don't really understand what this means. I understand that as it pertains to Stoke's Theorem it does not jive but further than that I'm at a loss.
     
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  3. Sep 10, 2013 #2

    TSny

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    (I think you meant to write z instead of x here.)

    Double check to see if the ##\hat{z}## component of the curl is zero.
     
  4. Sep 10, 2013 #3
    Shoot, okay messed this up then.
    for [itex]\hat {\phi}[/itex]
    [tex]\frac {\partial E_s}{\partial z} - \frac {\partial E_z}{\partial s}[/tex]=0
    but for [itex]\hat {z}[/itex]
    [tex]\frac {1}{s} [\frac {\partial s E_\phi}{\partial s} - \frac {\partial E_s}{\partial \phi}][/tex]

    1. [tex]\frac {1}{s} [\frac {\partial s E_\phi}{\partial s^2} - \frac {\partial E_s}{\partial \phi}]=2s \frac{1}{s}=2[/tex]

    2. [tex]\frac {1}{s} [\frac {\partial \frac {s}{s}}{\partial s} - \frac {\partial E_s}{\partial \phi}]=0[/tex]

    So for 1. (0,s,0) the electric fields makes sense. However, 2. (0,1/s,0) is not a valid field because it disagrees with the fields line integral. That is, there is some curl around the the object creating the field for 1 but not 2.

    So, what does it mean physically that this is the case? I picture in my head an infinitely long wire with some charge on it and I would expect there to be some curl of E along the z axis (with the wire as the z axis). But for 2 I cannot simply explain the lack of curl. I suppose I can't explain exactly why I suspect a curl around 1 either though.
     
  5. Sep 10, 2013 #4

    TSny

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    This looks good, except that the answer for (2) is not valid for certain points of space. Can you identify those points?

    I'm not quite sure what you mean when you say that field (0, 1/2, 0) disagrees with the field's line integral.

    If the z-axis is charged, would that produce a "circulating" field in the ##\hat{\phi}## direction?

    It is possible to create electric fields of the form (0,s,0) and (0,1/s,0) using time-dependent magnetic fields, but I don't know if you've gotten that far in your study of electromagnetism.
     
  6. Sep 10, 2013 #5
    First, your second question, I assume that Stoke's theorem should be true here that is:

    [tex]\int( \nabla × E) \cdot da = \int E \cdot dl[/tex]

    Since for both both 1 and 2 the line integral is nonzero then Stoke's disagrees with the result of 2. Or, at the very least, raises questions to it's legitimacy.

    The third question, I guess when I think of this curl I'm thinking of the field circulating in ##\hat{\phi}## around ##\hat{z}##

    For the first question I guess I just don't understand curl well enough to answer. I understand that it describes the "swirl" of a vector around some point. But, honestly, having a value of 2 for the curl is meaningless to me.
     
  7. Sep 10, 2013 #6

    TSny

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    In order for Stokes theorem to be valid, the curl of E should be well-defined at all points of the area enclosed by the closed path. For case 2, there are points where the curl of E is undefined.

    But a uniformly charged infinite line (z-axis) produces a radial field, not a circulating field.
     
  8. Sep 11, 2013 #7
    Then, I guess I have no idea what the curl would be describing at 1. As for 2 it seems that is not defined anywhere.
     
  9. Sep 11, 2013 #8

    TSny

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    Roughly speaking the z-component of the curl tells you how much the field at that point circulates about an axis parallel to the z-axis. An analogy with fluid motion is often used to describe the curl, where you replace the E field vectors with velocity vectors of a fluid. If you then imagine placing a little paddle wheel at a point in the field with the axis of rotation of the paddle wheel in the z-direction, the paddle wheel will tend to rotate with a strength proportional to the z-component of the curl at that point. In part 1, you found the z-component of curl to have a constant value of 2. So, the little paddle wheel would have the same tendency to turn no matter where you placed the wheel (as long as the axis of the wheel is parallel to the z-axis).

    In part 2 you found the curl to be 0 at all points except on the z-axis. At points on the z-axis, s = 0 and the curl is undefined for part 2. So, the paddle wheel would not tend to rotate no matter where you placed it as long as the axis of the wheel is not at the z-axis. The fact that the curl is undefined on the z axis means that you can't evaluate the integral of the curl over an area that contains a point of the z-axis. So, you can't expect Stokes' theorem to hold for such areas. As long as you select a closed path that does not enclose the z axis, you would find Stokes' theorem to hold with the area integral and the line integral both equal to 0.
     
  10. Sep 11, 2013 #9

    vanhees71

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    Just do what the problem asked you to do and calculate the curl first!

    The think about the line integral along a closed curve. I don't know, if you gave us the complete problem text. If not, I suggest to use a circle in the [itex](x,y)[/itex] plane around the origin as an example.

    There's no big surprise for case (a), but there is one for case (b). Just try to figure out, what's going on. In case (b) think about, where the given vector field is defined!
     
  11. Sep 11, 2013 #10
    I'm just not getting how the curl describes anything about the E part of the field. What exactly is circulating in phi? I understand that curl speaks of rotation about some point but, in the case of the electric field, I don't understand *what* is rotating. And since I don't understand that I can't decide how the line integral of would have anything to do with curl.

    It seems like 1 is a valid electrostatic field and 2 is not but I cannot explain why or why not.
     
  12. Sep 11, 2013 #11

    TSny

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    Make sure you understand http://en.wikipedia.org/wiki/Curl_(mathematics [Broken] in terms of a limit of the line integral over a loop as the area enclosed by the loop shrinks to a point. The line integral of a field around a loop is fairly easy to understand in terms of the net "accumulation" of the field around the loop.

    Nothing "physical" is circulating in the E field when the curl of E is nonzero. But you could take a little hollow plastic tube and bend it around into a closed hollow ring (kind of like a doughnut). Imagine a small charged particle that can move without friction inside the tube. If you placed the loop in the E field at a point where the curl is nonzero, then the field would accelerate the charged particle around the loop and the particle would continue to gain more and more kinetic energy (like a tiny betatron or particle accelerator).

    An electrostatic field is a "conservative field". It cannot add energy to a particle that goes around a closed loop. So, the line integral of an electrostatic field around a closed loop must be zero. This means that the curl of E should be zero at every point of space. In part 2, you are dealing with a singular field that is not defined on the z-axis. This is causing the curl to be undefined on the z-axis. Stokes' theorem does not apply to paths that enclose the z axis.
     
    Last edited by a moderator: May 6, 2017
  13. Sep 11, 2013 #12
    I had to turn in my paper today. I spoke with the TA and he cleared some things up for me (at least as much that was acceptable for the homework).

    Both of the fields were invalid because of the existence of the curl for 1 and because 2 was undefined at all points but along the z axis. I didn't make the connection between the curl and divergence of a 1/s function. So there was a delta function for the second field. This discussion and the discussion with my TA helped me understand quite a bit more than I did before.

    Thanks for the help!
     
  14. Sep 12, 2013 #13

    vanhees71

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    Here are a lot of things mixed up in this thread! So let's clarify a few of them.

    First of all an electric field can have a curl, if there is a time-dependent magnetic field present, because due to Faraday's Law, which is one of the fundamental Maxwell equations of the electromagnetic field:
    [tex]\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.[/tex]
    In static or stationary situations, i.e., when only time-independent fields are present, thus the curl vanishes.

    Now you have given vector fields given in terms of cylindrical coordinates, [itex](s,\varphi,z)[/itex]. The naming [itex]s[/itex] is uncommon, but let me keep your notation. Let's first clarify the notation a bit. The components in cylinder coordinates are given in terms of the coordinate's basis vectors, i.e., your two fields are

    1. [itex]\vec{E}_1=s \vec{e}_{\varphi}[/itex],
    2. [itex]\vec{E}_2=(1/s) \vec{e}_{\varphi}[/itex].

    Next you have to think about the curl, which is a vector operator, and you have to express it in terms of the coordinates used. For the cylinder coordinates, it looks as follows:
    [tex]
    \begin{split}
    \vec{\nabla} \times \vec{E} =& \vec{e}_s \left (\frac{1}{r} \frac{\partial E_z}{\partial \varphi} - \frac{\partial E_{\varphi}}{\partial z} \right ) \\
    &+ \vec{e}_{\varphi} \left (\frac{\partial E_s}{\partial z}-\frac{\partial E_z}{ \partial s} \right ) \\
    &+ \vec{e}_z \frac{1}{s} \left (\frac{\partial(s E_{\varphi})}{\partial s}-\frac{\partial E_s}{\partial \varphi} \right ).
    \end{split}
    [/tex]
    Now, luckily your vectors are pretty simple, and you can easily plug them in and take the derivatives. You finally get
    [tex]\vec{\nabla} \times \vec{E}_1=2 \vec{e}_z, \quad \vec{\nabla} \times \vec{E}_2=0.[/tex]
    This shows you that [itex]\vec{E}_1[/itex] cannot be an electrostatic field, because it's curl is not vanishing. It also cannot be a gradient field.

    For [itex]\vec{E}_2[/itex] it's more tricky, because apparently is looks as if it is a gradient field, because its curl vanishes. In fact it is locally a gradient field, namely in any simply connected neighborhood of a regular point of the field, but as a whole the region, where this field is defined is not a connected region in [itex]\mathbb{R}^3[/itex], because obviously there is trouble for [itex]s=0[/itex], which is the [itex]z[/itex] axis.

    Now you must become even more alarmed, because cylindrical coordinates are not covering the entire [itex]\mathbb{R}^3[/itex]! The coordinates are singular precisely for [itex]s=0[/itex], because there the Jacobi determinant, which is just [itex]s[/itex] between the cylinder and Cartesian coordinates vanishes. Thus, we have to check, whether the apparent singularity is only due to these trivial coordinate singularities or whether there are true singularities.

    In this case, the most simple way to check this is to go back to Cartesian coordinates. To that end we just need
    [tex]s=\sqrt{x^2+y^2}, \quad \vec{e}_{\varphi}=-\vec{e}_x \frac{y}{s} + \vec{e}_y \frac{x}{s}.[/tex]
    So there is indeed a real singularity along the [itex]z[/itex] axis. The [itex]\mathbb{R}^3[/itex] with the [itex]z[/itex] axis taken out obviously is not simply connectec, because any closed curve that encircles the [itex]z[/itex] axis cannot be contracted in any continuous way to a single point, because you somehow always have to cross the [itex]z[/itex] axis, which is fobidden, because it's taken out of the region, where the field is defined.

    You can also check this in another way. Just calculate a line integral along a closed line around the [itex]z[/itex] axis. It's simple to prove with help of Stokes's integral theorem that the result only depends on the number of windings around the [itex]z[/itex] axis, because the curl of the field vanishes everywhere except along the [itex]z[/itex] axis, where the field is singular and undefined. So we can choose any curve we like, and here it's most easy to choose a circle parallel in [itex]xy[/itex] plane with the origin of the coordinate system as its center. We can again switch to cylinder coordinates for that calculation, because we are working off the [itez]z[/itex] axis. This gives for a counter-clockwise oriented circle
    [tex]\int_{C_r} \mathrm{d} \vec{s} \cdot \vec{E}_2=\int_0^{2 \pi} \mathrm{d} \varphi s \frac{1}{s}=2 \pi.[/tex]
    If we had wound the circle [itex]n[/itex] times around the [itex]z[/itex] axis, we'd had gotten [itex]2 \pi n[/itex].

    Finally we can ask, how to realize such a field as a electrostatic field, because except along the [itex]z[/itex] axis its curl vanishes, and thus there is may describe an electrostatic field!

    To that end let's find the local potential, i.e., a scalar field [itex]\Phi[/itex] such that
    [tex]\vec{E}_2=-\vec{\nabla} \Phi.[/tex]
    The gradient in cylinder coordinates reads
    [tex]\vec{\nabla} \Phi=\vec{e}_s \frac{\partial \Phi}{\partial s} + \vec{e}_{\varphi} \frac{1}{s} \frac{\partial \Phi}{\partial \varphi} + \vec{e}_{z} \frac{\partial \Phi}{\partial z}.[/tex]
    Thus you get for the potential of [itex]\vec{E}_2[/itex]
    [tex]\frac{\partial \Phi}{\partial s}=\frac{\partial \Phi}{\partial z}=0, \quad \frac{1}{s} \frac{\partial \Phi}{\partial \varphi}=-\frac{1}{s}.[/tex]
    From the first equation you see that [itex]\Phi[/itex] can depend only on [itex]\varphi[/itex], and the last equation tells you that
    [tex]\Phi=-\varphi.[/tex]
    Now, this clearly shows, what's going on! The potential is not simple valued. You can define [itex]\varphi[/itex] to take values in any interval of length [itex]2 \pi[/itex] to cover the whole [itex]\mathbb{R}^3[/itex] with the [itex]z[/itex] axis taken out. Let's choose the interval [itex](-\pi,\pi)[/itex]. This covers the whole [itex]\mathbb{R}^3[/itex] with the half-plane [itex]x<0, \quad y=0[/itex] taken out, and on the "upper" boundary the potential takes the value [itex]-\pi[/itex] and on the "lower" boundary it takes the value [itex]+\pi[/itex].

    Thus you have a voltage difference along this half-plane, leading to this curly electrostatic field. Of course, this is a highly academic interpretation, because of course you cannot realize this configuration in practice, because you cannot set up an infinitely large half-plane with two metal foils separated by an infinitesimally thin insulator between them and hooking it up on a battery at infinity.
     
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