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Uniqueness Theorem, Concentric Equipotential Cylinders & Moving Charge

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider two electrodes 2 mm apart in vacuum connected by a short wire. An alpha particle of charge 2e is emitted by the left plate and travels directly towards the right plate with constant speed 106 m/s and stops in this plate. Make a quantitative graph of the current in the connecting wire, plotting current against time. Do the same for an alpha particle that crosses the gap moving with the same speed but at an angle of 45°. Suppose we had a cylindrical arrangement of electrodes with alpha particles being emitted from a thin wire on the axis of a small cylindrical electrode. Would the current pulse have the same shape?

    2. Relevant equations
    Uniqueness Theorem
    For a cylinder, E = 2kλ/r

    3. The attempt at a solution

    d = 2 * 10-3 m
    v = 106 m/s

    Let Plate 1 be positioned at x = 0; point charge at x = a; and Plate 2 at x = d.
    Also, Plate 3 will be positioned at x = a in different scenario;
    E1 is the electric field between Plates 1 & 3;
    E2 is the electric field between Plates 2 & 3.

    Application of Uniqueness Theorem: If I find a solution for specific charges Q1 on Plate 1 & Q2 on Plate 2 given a point charge located at a, then it is THE solution. (Did I miss any points? I'm still a little iffy on this.)

    Placing Plate 3 of total charge Q parallel to the other two plates and in the plane of x = a will distribute the same charges Q1 & Q2 as the a single point charge at x = a.

    E1 = σ1o
    E2 = -σ2o

    Because they are connected by wires, the potentials are equal.

    σ1o(-a) = -σ2o(d-a)
    σ2 = σ1a/(d-a)

    Q2 = Q1a/(d-a)
    Q1 + Q2 = -Q

    Q1 = -Q[itex]\frac{d-a}{d}[/itex]
    Q2 = -Q[itex]\frac{a}{d}[/itex]

    Now, considering that there is a moving charge, a = vt, Q = 2e
    Q2 = -2e[itex]\frac{vt}{d}[/itex]
    I = dQ/dt = -2ev/d

    The graph of current versus time for a plate capacitor is constant. Now, if the charge were moving at a 45 degree angle, it's just a vector consideration.
    I = dQ/dt = -2ev/d cos(45°) = -sqrt(2)ev/d

    Here's the hard part: The concentric cylinders

    Let the inner cylinder be of radius a with total charge Q1; and the outer cylinder be of radius b with total charge Q2.

    Application of Uniqueness Theorem: If I find a solution for specific charges Q1 on Cylinder 1 & Q2 on Cylinder 2 given a point charge located at vt, then it is THE solution. Now, let me consider Cylinder 3 of radius vt with total charge Q to satisfy the above conditions.

    E1 = 2kλ1/r
    E2 = 2kλ2/r

    Because they are connected by wires, the potentials are equal:

    2kλ1ln(vt/a) = 2kλ2ln(b/vt)

    Q2 = Q1[itex]\frac{ln(vt/a)}{ln(b/vt)}[/itex]
    Q1 + Q2 = -2e
    Q1 (1 + [itex]\frac{ln(vt/a)}{ln(b/vt)}[/itex]) = -2e
    Q1 [itex]\frac{ln(vt/a) + ln(b/vt)}{ln(b/vt)}[/itex] = -2e
    Q1 [itex]\frac{ln(b/a)}{ln(b/vt)}[/itex] = -2e
    Q1 = -2e [itex]\frac{ln(b/vt)}{ln(b/a)}[/itex]

    I = dQ1/dt = 2e[itex]\frac{1}{tln(b/a)}[/itex]

    ^This seems really off...
     
  2. jcsd
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