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Homework Help: Electric field, magnetic flux , potetial difference

  1. Jul 23, 2012 #1
    1. The problem statement, all variables and given/known data
    As shown in the figure below, potential difference V exists between two parallel plates(a,b) that have a small hole. Uniform electric field E and magnetic flux density B exist in the region above the plates, and are perpendicular to each other. The directionof E is parallel to this page and plate b. The directionof B is perpendicular to this page from back to front. A positiely charged particle of charged q and mass m, initially at res in the hole of plate a, is accelerated by potential difference V so that it enters te region above the plate perpendiculary to E and B and travels straight through the region.

    What is the ratio E/B?


    Please tell me how i put images in the post.
    Last edited: Jul 23, 2012
  2. jcsd
  3. Jul 23, 2012 #2

    Simon Bridge

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    Great - so where did you get stuck?
    Do you know the equation that relates the force on a charged particle due to electric and magnetic fields?
    Do you know how to relate the kinetic energy a charge gains when it falls through a potential difference of V?
  4. Jul 23, 2012 #3
    you mean lorentz force?
  5. Jul 23, 2012 #4
    No, i dont know. please tell me.

    this is really important. help please.
  6. Jul 23, 2012 #5

    Simon Bridge

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    Are you not doing a course in elecromagnetism that you have notes for? Go read them?

    The voltage is the electric potential energy per unit charge ... when the charge travels between the plates, then, it loses potential energy [itex]qV[/itex] and gains kinetic energy - what's the equation? You can do this.

    The electric field strength is the force per unit charge ... so [itex]\vec{F}_E = q\vec{E}[/itex]

    For the magnetic field, only moving charges experience a force: [itex]\vec{F}_B = r(\vec{v} \times \vec{B})[/itex]

    Do you know how to evaluate a cross product? (Hint - v and B are at right angles: it's the right-hand slap rule.)

    Now all you need is Newton's Laws.
  7. Jul 23, 2012 #6
    i get it, thanks a lot.
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