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Potentials from Equation for the Field - Independent Variables? I'm Lacking Intuition

  1. Aug 15, 2012 #1
    Hey guys and gals, this isn't actually an assignment of any sort, so I didn't want to put it in the homework section. This is also my first post, though I have been lurking for quite a while, reading the copious amounts of information available here. :p

    Anyhow, could somebody please elaborate on why it doesn't seem to work to treat the variables as a constant even though they are seemingly independent? As demonstrated below, I've tried it on two problems and both times it gives the wrong answer, though I am not sure why. I think I'm probably missing some intuition for these line integrals, and would very much appreciate a helping hand. Thanks!

    Problem 1
    For the field [tex]\vec{E}= 54y \mathbf{i} + 54x \mathbf{j} + 3 \mathbf{k}[/tex] find the potential function [tex]V_E(x,y,z)[/tex].
    [tex]V_E(x,y,z) = -\int \vec{E}\bullet \vec{ds}
    = - \int (54y \mathbf{i} + 54x \mathbf{j} + 3 \mathbf{k}) \bullet (dx\mathbf{i} + dy \mathbf{j}+ dz\mathbf{k})
    = - \int (54y\ dx + 54x\ dy + 3\ dz)[/tex]

    [tex]\therefore V_E(x,y,z)= - 108xy - 3z + C[/tex]

    But according to the book, the answer is actually [tex]V_E(x,y,z) = -54xy - 3z[/tex]

    ***

    Problem 2
    Suppose our field [tex]\vec{E} = \dfrac{F}{a^2}(yz \mathbf{i} + xz \mathbf{j} + xy \mathbf{k})[/tex], where F and a are constants. Calculate the potential difference [tex]\Delta V_E[/tex] between the origin and the point (1,1,1) along the straight line given by the formula (s, s, s), where the variable s runs from 0 to 1.

    Using the same method as above, [tex]V_E(x,y,z) = - \dfrac{F}{a^2}\int (yz\ dx + xz\ dy + xy\ dz) = - \dfrac{3F}{a^2}(xyz) + C[/tex]

    [tex]\Rightarrow \Delta V_E = V_E(1,1,1) - V_E(0,0,0) = - \dfrac{3F}{a^2}[/tex]


    However, this is wrong, according to the book, and is out by a factor of three. Using a different method, spotting that along the line, x=y=z, leads to a different integral for the potential: [tex]V_E(x,y,z) = - \dfrac{F}{a^2}\int (xx\ dx + xx\ dx + xx\ dx) = - \dfrac{F}{a^2}\int (3x^2\ dx) = - \dfrac{F}{a^2}x^3[/tex]

    [tex]\therefore \Delta V_E = - \dfrac{F}{a^2}[/tex]

    This is the correct answer, according to the book.
     
  2. jcsd
  3. Aug 15, 2012 #2
    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    Those line integrals can't, in general, be computed like a normal integral as you did there.

    In the first case, for instance, you ignored the fact that you need limits of integration. The potential is relative to zero potential at infinite distance-- you need to do the line integral from infinity to point (x, y, z). You can't simply integrate. Easier to solve it this: find the function whose gradient gives you the desired E-field. After some squinting, you see it must be the books solution.

    In the second case, your realization that x=y=z is correct. It's called parameterizing the integral, and you must always do it for a line integral.
     
  4. Aug 16, 2012 #3

    Philip Wood

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    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    Your mistake in the first question is in the last line. This is how the reasoning should go: using the product rule, –54ydx + –54xdy = –54(ydx +xdy) = –54d(xy), which integrates directly to –54xy.

    In the second question, your method is fine in principle (and, imo, very much nicer than the textbook's method), but you've made a mistake in the execution. Your factor of 3 simply shouldn't be there. Again, using the product rule,
    d(xyz) = yzdx + zxdy + xydz. I'll do a thumbnail to show this, if requested.

    [These examples are for 'conservative' E fields, arising from configurations of static charges. 'Conservative' means that the same amount of work is done by the field on a 'testing' charge going from point A to point B, by whatever path you choose. It is only because of this that you get these neat integrations' leading to a scalar field, V(x, y, z). Such a function of position does not exist for an E arising from changing magnetic flux, and different quantities of work done for different choices of paths between A and B.]
     
    Last edited: Aug 16, 2012
  5. Aug 16, 2012 #4
    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    Interesting, both failures are due to the same thing really aren't they? Though I see immediately that the right answer comes out when the product rule is used, I am confused as to why it doesn't work without this - as the two are equal, are they not? Nevertheless, thank you for that insight as I can now see where the answers came from, and I'm also glad to see that what I felt was the "natural" method seems to work for the conservative fields.

    If scalar fields can't be found, how does one calculate the energy changes for a non-conservative field, such as a changing magnetic field? Sorry for asking probably such trivial questions, as well, I just start my physics degree in October so haven't had any vector calculus training as yet.

    Thanks again!
     
  6. Aug 16, 2012 #5

    Philip Wood

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    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    Which two?

    You choose the path you're interested in, and integrate round that path, probably using a parameter.

    There's nothing trivial about your questions. Well done for doing some work in advance of your degree. I'm sure you'll do very well.
     
  7. Aug 16, 2012 #6
    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    I think working in terms of differentials can be tricky though, especially with the fact that things are generally parameterized and not integrated directly like that. It's why the gradient method is probably safer, as I described above.
     
  8. Aug 16, 2012 #7

    Philip Wood

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    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    I'd say it depended on the problem, and spotting 'exact differentials' is quick and rigorous - in cases like your two examples where there are exact differentials.

    Why I don't like the textbook method as much as JackDP's for the second problem is that integrating along a particular line doesn't show that you get the same answer whichever path you choose, that is it doesn't show that the field is conservative. [It's fine, of course, if you already know that the field is conservative, for example if you know that it arises from static charges.]
     
    Last edited: Aug 16, 2012
  9. Aug 16, 2012 #8
    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    I meant that [tex]x\ dy + y\ dx = d(xy)[/tex] and therefore I don't understand why you have need to use the exact RHS to get the correct answer out? And likewise for the other differentials.

    Ah right, so it's all line integrals then. Sounds fun. :P

    Thanks, I hope so! It will be tough, no doubt, but I love physics, so I will do what it takes to do well; I want to understand these beautiful things!
     
  10. Aug 16, 2012 #9

    Philip Wood

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    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    It's because d(anything) integrates to (that thing). So dx integrates to x. And d(xy) integrates to xy. But you can't integrate xdy by itself unless you know how x varies with y, that is unless you choose an integration path. The same goes for ydx. But if you have xdy together with ydx, then the combination is equal to d(xy), whose (indefinite) integral is xy, and which is clearly, therefore, path independent.

    No need to parametrise in such cases. Please also see my previous post.
     
  11. Aug 16, 2012 #10
    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    Ahhhh that makes perfect sense! So would it be accurate to say that (for conservative fields):

    - whenever you aren't integrating on a path, you have to use an exact derivative as you don't know the relationship between x,y,z?

    - when you choose a path, then you know how x,y,z vary wrt each other, and so can integrate without an exact derivative after substituting the variables (i.e. like x=y=z in "Problem 2")?

    I cannot thank you enough, my friends and I were deliberating this for hours last night and could come up with no reasonable explanation as to where we were making an error.
     
  12. Aug 16, 2012 #11

    Philip Wood

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    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    Yes!
    The concepts of exact and inexact differentials are important in studying fields, and also in thermodynamics. To have grasped them in advance puts you at a considerable advantage. Good luck!
     
  13. Aug 17, 2012 #12
    Re: Potentials from Equation for the Field - Independent Variables? I'm Lacking Intui

    Thank you for your help, kind sir! :smile:
     
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