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Ranking force on q via multiple charge density vs. theta graphs

  1. Aug 26, 2014 #1

    BiGyElLoWhAt

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    1. The problem statement, all variables and given/known data
    A point charge +q is placed near a curved, charged, insulating rod as shown at the left below. (I'll just draw it, I don't have access to a camera at the moment) The charge is placed near the center of the curvature of the curved rod. For each of the five cases A-E, the charge density on the rod varies according to the graphs, but the total charge is the same.


    2. Relevant equations

    ##F= K\frac{q_1q_2}{r^2}##
    Not really sure, that's the only one I've used on this one.

    3. The attempt at a solution

    First off, I'm not really sure what's expected on this, we've only had one class, and this is due tomorrow (if I could get help in the next few hours I'd love you long time), maybe I'm going too far with this, I don't know.

    It looks like i have an arc of a circle radius d, and the arc goes from -60deg to 60deg with a charge at the middle (see diagram)

    Due to symmetry, all the y components cancel so that's nice.

    It doesn't really say what the charge density function yields, so I assumed that the area under the curve from a to b was the total amount of charge from a to b measured in coulombs. Is that too big of a jump?

    So given ## \rho (\theta)##,

    We can say that (my assumption)

    ##\text{number}_{\text{charges from a to b}}= < \rho (\Delta \theta)>(\Delta \theta)
    = \frac{\rho (a) +\rho(b)}{2}(b-a)##

    and therefore my approximate force would be

    ##F = K\frac{q}{d^2}\Sigma \frac{\rho (a) +\rho(b)}{2} cos(\frac{a+b}{2})\Delta \theta##

    1) Is my logic correct thus far? I mean, I think it makes pretty good sense, but... (continued after #2)

    2) Can I simply jump from this riemann sum to an integral form of this equation? I think the best way I'd be able to "explain my reasoning" on this homework would be through the math.

    ...1 ctd.) ...[but]... I have my doubts about using the average charge density in my function... Can I do this as long as my function is continuous? I'm pretty sure I'm going to have to do some of these piece wise, but I don't know. I think I'm missing something here.

    Let me provide an example:

    Graph a is linear from -60deg to 0 and from 0 to 60. From -60 to 0 my density as a function of theta is ##\frac{1}{30}(\theta + 60)## and from 0 to 60 it just mirrors it back down to 60.

    Don't ask me what the 2 at theta = 0 is, it's not labeled, it just goes up 2 ticks on the charge density (the graphs make me think this is potentially a qualitative analysis question, but I want to do it as quantitatively as possible.)

    Using my logic and my function, the integral form of the force for this density would be:

    ##(2)(\frac{1}{30})K\frac{q}{d^2}\int_{-60}^0 \frac{\rho(0)+ \rho(-60)} {2}(\theta +60)cos(\theta)d\theta##

    Here's my issue with this: I don't feel like my theta in the cosine is lining up with the other theta's, in my approximation, I was using the average charge density over an interval, and for the cosine theta I was using the angle smack dab in the middle of that interval, which is not exact, I know, but I'm not sure what to do about that.

    Fact of the matter is I don't like my function, but I'm not really sure what to do about it. Any pointers? This is due tomorrow morning (it's 8:40pm, class is at 9am) $1 trillion to the first post haha =]
     

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    Last edited: Aug 26, 2014
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  3. Aug 26, 2014 #2

    BiGyElLoWhAt

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    actually my number of charges function should be n*q (total charge) from a to b, not the number of charges (unless all the charges happen to be 1 coulomb)
     
  4. Aug 27, 2014 #3

    haruspex

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    You don't seem to have attached the charge distribution graphs. Without those, it's hard to be sure, but I suspect that you don't need to do any integration as such, just apply a bit of logic.
    But wrt your integral, it doesn't look right to me. If the charge density at angle θ is ρ(θ), what is the X component of the field at q due to the charge in the arc (θ, θ+dθ)?
     
  5. Aug 27, 2014 #4

    BiGyElLoWhAt

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    I know that I don't HAVE to, (this might sound weird) I WANT to integrate. I've been trying to do this thing lately where I try to derive formulas for certain situation that I haven't really seen before. I'm sure there's a formula for this somewhere, but I thought it would be a good mind practice. I didn't attach the graphs. I'll do so.

    As for your question,

    The field would be ##\frac{K}{d^2} \Sigma Q_{\text{on the bar}}cos(\theta)##
    where ##Q_{\text{on the bar}}## would be ##(\rho(\theta +d\theta) - \rho(\theta))d\theta##

    I have to go to class now. I'll be back.
     

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  6. Aug 27, 2014 #5

    BiGyElLoWhAt

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    Well I ended up just doing a weighted average. It would've been nice if I had realized (as one of my classmates pointed out to me after class) that the area under the curve was the same for all the graphs... Which means that every unit area carried the same amount of charge... I hate it when I miss something that simplifies things a lot:cry:
     
  7. Aug 27, 2014 #6

    haruspex

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    Now that I see the graphs, I see that it will not be hard to express the distributions algebraically, so integration is a reasonable approach.
    Yes.
    No. ρ is a charge density. Why take a difference in nearby densities?
     
  8. Aug 27, 2014 #7

    BiGyElLoWhAt

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    was ##(\theta ,\theta + d\theta)## not an interval?
     
    Last edited: Aug 27, 2014
  9. Aug 27, 2014 #8

    haruspex

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    Yes, and its length is dθ. What is the density on the interval (remember, it is going to change very little over an infinitesimal interval)? So what is the total charge on it?
     
  10. Aug 27, 2014 #9

    BiGyElLoWhAt

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    Do you mean ##\rho(\theta)d\theta##?
     
  11. Aug 27, 2014 #10

    BiGyElLoWhAt

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    If that's the case, then my original setup wasn't too far off, I just made things too complicated. Sound about right? BTW thanks for responding :smile:
     
  12. Aug 27, 2014 #11

    haruspex

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    Yes. So can you figure out the algebraic expression for ρ(θ) for each graph, and execute the integrations?
     
  13. Aug 28, 2014 #12

    BiGyElLoWhAt

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    Yea they're all pretty easy. Otherwise I would have probably tried a different approach. I already showed the one for A.

    So I think I figured out where my confusion came from. I wasn't looking at charge density as charge density -.-

    I was thinking total charge from one end to theta + d theta - total charge from the same end to theta .... haha

    Thanks again, hopefully I won't make the same mistake again. :smile:
     
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