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Energy in Field of Circular Distribution

  1. Aug 23, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the energy stored in the field of a circular charge distribution of constant charge density.


    2. Relevant equations
    I know that the energy stored in a field is the same as the potential energy of the system.
    dU= [itex]\varphi[/itex] dq = [itex]\varphi[/itex](σ dV)
    Though my book also says that dU= (1/2) ρ [itex]\varphi[/itex] dV
    So I'm not sure which equation to use. (I'm using Purcell.)


    3. The attempt at a solution
    If I integrate without the 1/2, I get the correct answer (8Q^2)/(3πa).

    But why don't I use the 1/2?

    Also, I understand that dq=σ dA, and dA=2πr^2 dr, but why do I not use the expression with the 1/2 to get the correct answer when the book says that this formula gives the energy of a system? Where does the 1/2 even come from in doing from dq to dV (or dA) as the quantity being integrated?
     
  2. jcsd
  3. Aug 24, 2012 #2
    φ what does this denote here ?
     
  4. Aug 24, 2012 #3
    I'm sorry not to specify! That's the potential. I didn't want to use V since I was writing that as a volume element.

    Essentially, my question is really where the 1/2 comes from in the book's (general) definition, and why it's not in the definition I "derive" in thinking about the potential that a layer experiences as it is added to the circular distribution.
     
  5. Aug 25, 2012 #4

    gabbagabbahey

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    Your first equation should be [itex]U=\frac{1}{2}\int \varphi dq = \frac{1}{2}\int \varphi \sigma dA[/itex]

    The factor of 1/2 is is best seen by first looking at the discrete case of a collection of point charges. The work done when assembling the distribution can be calculated by first positioning one charge at its final location (this takes no work since there are not yet any other charges/fields present), then moving a second charge in from infinity to its final location, while holding the first charge fixed (so work is done against the electric field of the first charge), then bringing in a third charge while holding the first two fixed (so work is done against the fields of the first two charges), and so on. You end up with a sum like

    [tex]W=\frac{1}{4 \pi \epsilon_0} \sum_{i=1}^{n} \sum_{j=1}_{j > i}^{n} \frac{q_i q_r}{ | \mathbf{r}_j - \mathbf{r}_i |}[/tex]

    where the condition [itex]j>i[/itex] ensures that you don't double-count interactions. You can simplify this by intentionally double-counting interactions and then dividing by 2 to get

    [tex]W=\frac{1}{2} \sum_{i=1}^{n} q_i \left( \sum_{j=1}_{j \neq i}^{n} \frac{1}{4 \pi \epsilon_0} \frac{ q_r} {|\mathbf{r}_j - \mathbf{r}_i|} \right) = \frac{1}{2} \sum_{i=1}^{n} q_i \varphi( \mathbf{r}_i )[/tex]

    where [itex]\varphi( \mathbf{r}_i )[/itex] is the potential at the location [itex]\mathbf{r}_i[/itex] of the charge [itex]q_i[/itex] due to all the other charges.

    Generalizing to a continuous distribution with [itex]\sum \to \int[/itex], [itex]q_i \to dq[/itex] gives you the correct formula.

    No, [itex]dA= r dr d\phi[/itex]. Try your integration again.
     
  6. Aug 25, 2012 #5
    I believe my initial statement was correct. dA is the area of a little circular ring being added.

    So I did accidentally raise r to the 2nd power, but other than that I was right (think of it has starting after the angular integration of your suggestion, if that helps you understand the math more easily).

    It still varies by 1/2 if I use the book's definition.
     
  7. Aug 25, 2012 #6

    gabbagabbahey

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    Sure, [itex]\int_0^{2\pi}\int_0^r dA = \int_0^{2\pi}\int_0^r r d\phi dr = 2 \pi \int_0^r rdr[/itex], but that does not mean [itex]dA=2\pi r dr[/itex]. ([itex]dA[/itex] is a 2D differential, while [itex]dr[/itex] is only a 1D differential, so it makes no sense to say [itex]dA=2\pi r dr[/itex])

    What does? Your final answer, or the formula for the electrostatic energy?
     
  8. Aug 25, 2012 #7

    TSny

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    In this expression for dU, [itex]\varphi[/itex] stands for the total potential at the location of dV due to all of the charge on the entire disk. The ½ is to avoid double-counting as explained by gabbagabbahey. In the method that you used to get the correct answer, you are building up the charge step by step from the center outward and calculating the work for each step. So, for your method, [itex]\varphi[/itex] is not the potential due to all of the charge on the disk , rather it’s the potential due to only the charge within the value of r at which you are adding the next infinitesimal ring of charge.
     
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