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Thin, Bent Rod, Continuous Charge

  1. Jul 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin rod bent into the shape of an arc of a circle of radius R carries a uniform charge per unit length lambda = 2.20×10-9 C/m. The arc subtends a total angle 2 theta0, symmetric about the x axis, as shown in the figure below. theta0 = 28.0° and R = 0.28 m
    Hint: hard to avoid integrating for this one but at least the integral isn't too bad. Remember that the problem asks for the magnitude of the electric field.


    2. Relevant equations
    ive tried:
    E=[tex]\frac{kq}{r\sqrt{\frac{L^2}{4} + r^2}}[/tex]


    3. The attempt at a solution
    [​IMG]
    I found the length of the rod and the charge of the rod using simple circle calculations:

    C=2*pi*r
    then using C and the 56[tex]^{o}[/tex]:304[tex]^{o}[/tex] ratio, i found the length of the rod. Then using the charge/length given, i found the charge. After i had all my info, i filled it into the equation but got the wrong answer. Any Suggestions?
     
  2. jcsd
  3. Jul 4, 2008 #2
    im not sure if i should be using gauss's law or if i should be trying to integrate here..
     
  4. Jul 4, 2008 #3
    What exactly are you being asked to calculate?
     
  5. Jul 4, 2008 #4
    Oops,

    "Determine the magnitude of the electric field E at the origin 0."

    that part was under the picture, i must have missed it when i copied the question over.
     
    Last edited: Jul 4, 2008
  6. Jul 4, 2008 #5
    does that help any?
     
  7. Jul 5, 2008 #6
    anyone have any insight on this one?
     
  8. Jul 5, 2008 #7

    Defennder

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    Where did this come from? Are you told to use this?

    This is based on the equation given above, right? Why not start with what you're given, instead of that equation? From the diagram, you can see that the electric field in the y-direction is 0 by symmetry. So you need only find the x-component.

    So [tex]dE_x = dE \cos \theta[/tex]

    And make use of this formula:
    [tex]dE = \frac{\lambda ds}{4\pi \varepsilon R^2} [/tex]
    ds is simply the differential arc length of the charged rod. Find an expression which expresses that in terms of [itex]\theta[/itex] and perform the integration from 0 to theta. But this accounts for only the top half. You should be able to see how to get the electric field due to the bottom half easily once you found the above.
     
  9. Jul 5, 2008 #8
    That first equation was one which was derived in the textbook, and i tried it although i wasnt too sure if it would work properly. I integrated the second formula you gave, and came up with:
    [tex]

    E = \frac{\lambda s}{4 \pi \epsilon R^2}
    [/tex]

    correct?
     
  10. Jul 5, 2008 #9

    Defennder

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    What does L and r in the equation refers to? And does the derivation in the textbook have the same geometrical setup as this problem?

    As for your integration, you're not supposed to do it with respect to s. s is the differential arc length, and you should note in such problems, the integration should always be performed with respect to a variable which ranges across a certain interval defined by the parameters of the problem. Over here, either x or theta is supposed to be the variable, not s. Another problem with your integration is that you haven't quite found the expression for E_x. Remember that E_y is zero by symmetry. Integrating with respect to E only without taking E_y is 0 into account doesn't give you the right answer.
     
  11. Jul 5, 2008 #10
    The first equation is from a problem very similar(done by the instructor, not in the text, my mistake); it has a straight rod instead of a bent one. He derives dECostheta and comes up with that. Where L is the length of the rod and r is the value of the hypotenuse if you were to draw a line from the point where the feild is acting to any point on the rod. It doesnt matter where on the rod, since symmetry cancels out the x-values. It should be exactly the same thing here, only i dont know if i should use the actual length of the rod, or the length of the arc.
     
  12. Jul 5, 2008 #11

    Defennder

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    I don't think you can simply use s in place of l here. For one thing, the r in your instructor's derivation is a variable since we're talking about a straight line of charge with respect to a point. But in this case, r is a constant since every charged differential length of the arc is at the same distance from the origin.
     
  13. Jul 5, 2008 #12
    Oops, that formula is copied down wrong. where i have r, i should have the distance between the center of the rod and the point p, which is constant. In this case, it should be x, which is the distance from the origin to the center of the rod.
    [tex]
    E = \frac{kq}{x\sqrt{\frac{L^2}{4} + x^2}}
    [/tex]
     
  14. Jul 6, 2008 #13

    Defennder

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    Firstly do you have any reason to believe that the formula you quoted, which was derived for a charge configuration of a straight line would also work here? This problem is concerning a circular arc of uniform charge. It appears as though you're confusing between this thread and your other thread on E-field due to a line of charge. If that formula can be applied (I'm assuming it's correct since I didn't check it), it should be applied there and not here.

    Why not start from first principles and find [itex]dE_x[/itex] in terms of [itex]dE[/itex] first? Then, find [itex]dE[/itex] due to a differential charged segment of the arc. And then perform the integration with respect to theta once you've changed variables to theta.
     
  15. Jul 6, 2008 #14
    OK. I'm getting pretty confused, so lets start from scratch using what we're given and what we know (and what i don't know). We know the length of the rod (the ACTUAL length of the rod. this would be the same if the rod was straight or bent in any other shape). We know the charge on the rod. We know the radius and that the radius is always constant. We know that due to symmetry, y values cancel out and there will only be an electric field acting in the x-direction. We know that we only need to perform one integral and that is for the E-field in the X direction from [tex]+\theta o[/tex] to [tex]-\theta o[/tex]. We know that this can be done by doing the integral of dExCostheta. Is all of this correct?
     
  16. Jul 6, 2008 #15

    Defennder

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    Yes, but in this case knowing the length of the rod doesn't matter so long as we take the length of the rod into account by integrating with respect to theta.
     
  17. Jul 6, 2008 #16
    OK. So then, [tex]Cos \theta[/tex] is s/R where s is the distance from the origin to an imaginary line made by drawing a line from one tip of the rod to the other?
     
  18. Jul 6, 2008 #17

    Defennder

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    Are you integrating with respect to x or theta? It's easier to do it with respect to theta, so you should be expressing the differential charged segment [itex]\lambda ds[/itex] in terms of [itex]\theta[/itex] instead of changing variables from theta to x.
     
  19. Jul 6, 2008 #18
    I can't integrate since i dont even know what the expression is supposed to look like. But if you say that doing the integral in terms of theta is easier, then ill do it that way when i get there.

    First things first; i need to figure out what the integrated formula looks like. Is Costheta s/R? If it is, then i need to integrate the following:

    [tex]
    dEx = \frac{\lambda ds}{4\pi \varepsilon R^2} * \frac{s}{R}
    [/tex]

    how am i supposed to put this in terms of theta?
     
  20. Jul 6, 2008 #19

    Defennder

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    You got the expression partly right. First note that [tex]dE_x = \cos \theta dE[/tex].

    [tex]dE = \frac{dq}{4\pi \varepsilon_0 R^2}[/tex]
    [tex]dq = \lambda ds[/tex] where ds is a small segment of arc length. Use this expression in the previous one.

    There is a formula relating arc length s to theta and radius r. Use that formula to express [itex]ds[/itex] in terms of [itex]d\theta[/itex].

    Once you're done with that, you have the expression for [itex]dE_x[/itex] ready to integrated. Perform the integration from [tex]-\theta_0 \ \mbox{to} \ \theta_0[/tex]. That gives you E_x.
     
  21. Jul 6, 2008 #20
    Problem Solved!
    [tex]
    E = \frac{k\lambda(Sin \theta - Sin- \theta)}{R}
    [/tex]
    E = 66.3 N/C

    Thank you very much for your efforts and immense patience!
     
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