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Electric Field and an object

  1. Jan 14, 2006 #1
    Pretty much, an object is moving at a certain velocity towards an "infinite" sheet with uniform charge density of sigma=4.4 nC/m^2. The object has a mass of m=62 ug and a charge of q=3.5 nC. It also has a velocity given in a vector v=(3.90i-4.80k) m/s. I am to find a bunch of stuff, but the first and foremost being the acceleration of the object.

    I've drawn the FBD. I'm not sure if I have the correct equation but I know that a=qE/m in general. However, I can't seem to find E. My teacher explained that when the distance r of an object is r<<length*width then to use E=sigma/(2*epsilon-not) (epsilon-not being the permittivity of space or 8.845*10^-12). But if I were to use that, I'd get E=248 N/C. Popping that into a=qE/m gives me the wrong answer however. Should I be accounting for the force the velocity has? Am I even using the right equation to find E? Using the answer I know for the acceleration, I can calculate that the E should be 74400. But then if I were to backtrack even further, that would mean that sigma would have to be some really off number. This is why I'm speculating that E is wrong. Unless I'm not accounting for another force to determine the acceleration.

    Ok, I'm rambling now. Any tips out there? Thanks in advance.
     
  2. jcsd
  3. Jan 14, 2006 #2

    lightgrav

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    Your E = 248.5 N/C looks okay to me.
    What does that give you for a?
    How do you know that this acceleration is wrong?

    If you use the acceleration that you calculate,
    what other results do you obtain?

    Have you treated the magnetic field yet?
     
  4. Jan 14, 2006 #3
    The answer sheet the professor provided lists the acceleration to be 4.2 m/s.

    So am I coming up with the wrong equation to use? I tried qE+a=ma but the problem is I know that this is not in equalibrium so I don't know how a FBD can help me figure this out in this instance.

    I didn't use my calculations any further since everything else is based off this acceleration. Nice little stepping stone. And the magnetic field is beyond my scope atm. I don't need to worry about that.
     
    Last edited: Jan 14, 2006
  5. Jan 14, 2006 #4

    lightgrav

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    The Electric Field provides a Force on a charge that is immersed in it.
    This kind of Force is just like any other Force ... gravity, for instance.
    It is the SUM of Forces (including gravity, springs, ...) that causes ma .

    That's why we use a FBD ... to help us add the Forces vector-wise.

    F_electric + F_gravity = ma .

    NO, a is NOT equal to qE/m , in general situations.
     
    Last edited: Jan 14, 2006
  6. Jan 15, 2006 #5
    I totally understand your reply and already knew that. What I'm figuring my problem is, is since it's not in equilibrium (I know I have to figure it's lowest height before it starts on an upswing before hitting this "infinite" sheet), I don't think I can use F_electric + F_gravity = ma. I know this because this also does not equal the correct answer.

    I'm starting to think that maybe the teacher wrote down the wrong answer. He's done it before...
     
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