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Purcell 3.19

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data

    I'd like to preface this by saying I'm not doing this as "homework" per se, but rather reviewing my physics coursework from my undergraduate years, prepping for grad school.

    So w/ that brief introduction, I'm working through Problem 3.19 in Purcell's famous textbook on E&M. The problem involves two conducting cylinders, radii r=a and r=b, w/ a < b. The inner cylinder (call it cylinder A) has a negative charge and the outer cylinder (call it cylinder B) has a positive charge, so they form two "plates" of what is essentially a capacitor. We cut these cylinders in half, longitudanally, so that when I look at them "end-on" (i.e., looking down the z-axis, if the z-axis measures the "length" of the cylinder, as in traditional cylindrical coordinates) I'm looking at two half-circles, w/ a < b.

    Into this circular space between the two cylinders (and in which these is an electrical field), I shoot a charged particle w/ potential V(0). The task is to determine the radius, r(0) that the charged particle will follow as it travelling, in a circular-motion, through the half-circle of the cylinders.

    2. Relevant equations

    The problem statement essentially gives you the answer, but asks you to prove it, which is that the two 'shells' of the cylindrical capacitor must be kept at potentials:

    V(B) = 2*V(0)*ln(b/r(0))
    V(A) = 2*V(0)*ln(a/r(0))

    Note that since a < r(0) < b, we necessarily have:

    V(A) < 0 < V(B)

    3. The attempt at a solution

    As we're modeling a capacitor, w/ a positive charge +Q on the outer cylinder and a negative charge -Q on the inner cylinder, Gauss gives us that E=0 for all r > b and r < 0, and correspondingly that V(r) = const in these regions.

    The potential difference between the two plates of a cylindrical capacitor, w/ inner radius a and outer radius b, is given by:

    V(diff) = 2*Q*ln(b/a)/L

    where Q is the charge on either plate, and L is the length of the cylinder.

    Knowing this and knowing that we can set the "zero" of potential to be anywhere, we set the potential of this system to be zero at an arbitrary radius r(0) w/ a < r(0) < b. Then the potential of the two cylindrical shells will be:

    V(b) = 2*Q*ln(b/r(0))/L
    V(a) = -2*Q*ln(r(0)/a)/L = +2*Q*ln(a/r(0))/L

    This of course preserves the condition that V(diff) between the two plates is 2*Q*ln(b/a).

    It's at this point that I kinda get stuck. The answer seems to be "right" there .... but... I'm not sure how to proceed. If there were some way to show that:

    Q = V(0) * L

    then the problem would be solved. From a dimensional standpoint, this makes sense. V(0) is measured in charge/meters, and that multipled by meters should give a measure of Q, charge. But I can find no outstanding reason *why* we should have Q=V(0)*L.

    Any help would be appreciated!
  2. jcsd
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