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Homework Help: Electric field problems literally

  1. Mar 9, 2004 #1
    There are 2 questions I am totally stuck on.

    Problem 1 ----- Uniform electric field problem
    A uniform electric field of magnitude 640 N/C exists between 2 parallel plates that are 4.0 cm apart. A proton is released from the positive plate at the same instant an electron is released from the negative plate.

    a) Determine the distance from the positive plate at which the two pass each other (ignore the elecric attraction between the proton and electron) b) What if? Repeat part a for a Sodium ion (Na+) and a chloride ion (CI-).

    Here is the work I have on this one

    accel = qE/m
    q = particle charge
    E = Electric field charge
    m = mass of respective particle
    p = proton
    e = electron

    The equations are similar in nature in a way to the kinematics. I get that. So....

    xfp = (qpE/2mp) * t^2
    xfp = (.04m - xfe)
    xfe = (qeE/2me) * t^2

    Now I get completely lost on the math.... 3 equations 3 unknowns, with t being equal in both. Solve for t....

    problem 2 ----- Electric Flux & Gauss Law
    In the air over a particular region at an altitude of 500m above the ground the electric field is 120 N/C directed downward. At 600 m above the ground the electric field is 100 N/C downward. What is the average volume charge density in the layer of air between these two elevations? Is it positive or negative?

    And in this problem I am just unclear as to which specific concept is being addressed!
  2. jcsd
  3. Mar 9, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    It's a breeze. You have three equations. Plug the first and third into your second one. Then you can solve for t. Etc.
    Consider a gaussian surface in the shape of a box (say) with the top at 600m and the bottom at 500m. You know the fields, so apply Gauss's law to find the charge per unit volume. (Use an arbitrary area, A, for your gaussian box.) If you don't know what I'm talking about, look up Gauss's Law.
  4. Mar 9, 2004 #3
    Here is a step by step of what I did to get an answer, but it doesn't match the answer in the back of the book which is 21.8*10^-6 m.

    So I pluged it in

    ((qpE/2mp) * t^2) = .04 - ((qeE/2me) * t^2)

    THEN added the right accel to both sides

    ((qpE/2mp) * t^2) + ((qeE/2me) * t^2) = .04

    take out common factor of t^2

    (t^2)* ((qpE/2mp) + (qeE/2me)) = .04

    t^2 = (.04) / ((qpE/2mp) + (qeE/2me))

    Then use the time to put back into the xfp formula to see what the distance is at the time they cross.

    xfp = 1/2 * accel of proton * t^2

    any idea where I went wrong?
  5. Mar 9, 2004 #4

    Doc Al

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    Staff: Mentor

    I see nothing wrong in your method. My answer agrees with the book's, so I suspect you made an arithmetic error. A suggestion: don't plug in numbers until the latest possible second. (The final expression for Xp depends only on the ratio of the masses, not on E.)
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