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Homework Help: Finite Sheets of Charge Compared to Infinite Sheets of Charge

  1. Sep 10, 2007 #1
    1. The problem statement, all variables and given/known data


    Consider two thin disks, of negligible thickness, of radius R oriented perpendicular to the x axis such that the x axis runs through the center of each disk. The disk centered at x=0 has positive charge density n, and the disk centered at x=a has negative charge density -n, where the charge density is charge per unit area.

    For what value of the ratio R/a of plate radius to separation between the plates does the electric field at the point x=a/2 on the x axis differ by 1 percent from the result n/epsilon_0 for infinite sheets?

    2. Relevant equations

    E(disk_x) = n/(2epsilon_0) * (1-((x)/(sqrt(x^2 + R^2))))

    3. The attempt at a solution

    I have found the electric field at x=a/2 to be:

    E(disk_x) = 2*(n/(2*epsilon_0) * (1-((.5a)/(sqrt((.5a)^2 + R^2))))

    However, I'm not sure where to go from here. Should I work the above equation to get what R/a is equal to and then set it equal to .99*(n/epsilon_0)?
  2. jcsd
  3. Sep 11, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Set that expression equal to .99*(n/epsilon_0) and solve for R/a. (Hint: First rewrite that expression in terms of R/a.)
  4. Sep 11, 2007 #3
    So now I have:

    Sorry this is gonna be messy...

    R/a = sqrt(.25*((1-((epsilon_0*E)/n))^-2)-.25) = .99(n/(epsilon_0))

    How do I deal with the E? If I sub in the other equation won't that just negate everything? I'm also asked to give the answer to two significant digits, but I don't see how the R/a ratio is going to give me numbers without that E still in it.

    Help me out Doc please :)
  5. Sep 11, 2007 #4
    Okay I got it on my own.

    Using the %difference = (E_infinite - E)/(E_infinite)

    Since E = n/epsilon_0 * (1-((.5a)/(sqrt((.5a)^2 + R^2)))) approaces n/epsilon_0 as R/a approaches infinity you can plug in n/epsilon_0 for E_infinite and the original equation for E. Solving this for %difference = .01 you arrive at R/a = 50.

    Nasty little problem...
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