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Total Electric Charge on an annulus

  1. Sep 14, 2005 #1
    I am getting stuck trying to find the total electric charge on the annulus. The Question reads: "A thin disk with a circular hole at its center, called an annulus, has inner radius R_1 and outer radius R_2. The disk has a uniform positive surface charge density, sigma, on its surface. Determine the total electric charge on the annulus." I realize that I need to intergrate dE_mag(r) from R_2 to R_1, but I'm not sure how to do it using the correct variables :frown: .
     
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  3. Sep 14, 2005 #2

    Physics Monkey

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    You don't need to integrate the electric field to solve this problem. There is a much simpler way to obtain the total charge. You don't even know what the electric field is, think about the known quantities ...
     
  4. Sep 14, 2005 #3
    Well you are given R_1, R_2, sigma, k, Q and based on the formula E = (Q/(4*pi*E_0*r^2))..... r^2= (R_1 + R_2)^2 ???
     
  5. Sep 14, 2005 #4

    Physics Monkey

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    Tcat, the formula you quoted, [tex] E = \frac{Q}{4 \pi \epsilon_0} \frac{1}{r^2} [/tex] is only valid for point particles of charge Q (and uniformly charged spheres). The electric field produced by the annulus is actually much more complicated than the above formula. Since the electric field isn't known, all you really have is the radii and the surface charge density. How is the surface charge density defined?
     
  6. Sep 14, 2005 #5
    surface charge density = E_o*E
     
  7. Sep 14, 2005 #6

    Physics Monkey

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    Try not to think in terms of formulas. Let me give an example. Suppose I give you a wire with uniform linear charge density [tex] \lambda [/tex], charge [tex] Q [/tex], and length [tex] L [/tex]. How are the three related? What does linear charge density mean? It is the charge per unit length, right? So how do I find it? I divide total charge by total length (since the system is uniform), [tex] \lambda = Q/L [/tex].
    Now try to ask similar questions about the surface charge density. What does surface charge density mean?
     
  8. Sep 14, 2005 #7
    surface charge density is the charge per surface area, which is Q/A
     
  9. Sep 14, 2005 #8

    Physics Monkey

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    Right! How can you use this to solve the problem?
     
  10. Sep 14, 2005 #9
    I'm not sure, do you find the SA of the disk first?
     
  11. Sep 14, 2005 #10
    Surface charge density=Q/A=((Q/pi*((R_2)^2 - ((R_1)/2)^2)
     
  12. Sep 14, 2005 #11
    I mean Q/A=((Q/pi*((R_2)^2 - (R_1)^2) Does that sound right? What do I do from here?
     
  13. Sep 14, 2005 #12

    Physics Monkey

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    You are almost there. Just solve for the unknown charge in terms of known quantities.
     
  14. Sep 14, 2005 #13
    Q= surface charge, but I'm not sure what quantites are related with it
     
  15. Sep 14, 2005 #14
    I know the radii and the surface charge density, sigma... does Q=sigma*((R_2)^2 + (R_1)^2)^1/2
     
  16. Sep 14, 2005 #15

    Physics Monkey

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    Nope, try again. But you are very close. What you have multiplying sigma is not the area ...
     
  17. Sep 14, 2005 #16
    Q=sigma*pi*((R_2)^2 + (R_1)^2)^1/2
     
  18. Sep 14, 2005 #17
    Why take the square root? [tex]\pi\cdot r^{2}[/tex] gives the surface area of a disk without any problems. [tex]\pi\cdot(r_{o}^{2} - r_{i}^{2})[/tex] works for an annulus.

    Also, since the charge is uniformly distributed over the surface of the annulus, don't you need to take into account the bottom surface as well as the top surface?
     
  19. Sep 14, 2005 #18
    So now using this formula how do I get the total electric charge
     
  20. Sep 14, 2005 #19
    You're almost there! You have the charge per unit area, and you can calculate the total area. Combine the two (how?), and voila.
     
  21. Sep 14, 2005 #20
    Well the charge per unit area = sigma*pi*((R_2)^2 - (R_1)^2) so I add that to the total area which is.... pi*((R_2)^2 - (R_1)^2) and I get sigma*2(pi*((R_2)^2 - (R_1)^2) for the final answer of total electric charge
     
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