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Particles with a charge passing through electric field

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Particles with the mass 2.6*10^(-21) kg and the charge 4 times the electron charge are fired into a homogenous electrical field made by two plates. The plates are 7.5 cm apart from each other and the voltage is 60V. Which is the least speed they need to pass through the field if it is 8 cm long? (I will illustrate down below)

    2. Relevant equations
    E = U/d
    E = F/Q
    E means electric field strength (I think)
    F = ma
    s = v*t + (at^2)/2

    3. The attempt at a solution
    MnrmyKQ.png

    You can see my illustration above. Okay, so let's get some things out of the way first. I assume there is either an upward or downward acceleration toward the plates caused by the voltage. I think the question assumes they are fired into the middle, so 7.5cm/2. Since there is either an upward or downwards acceleration, the speed in the horizontal line needs to be fast enough so that it gets through before hitting a plate.

    So what do I need to figure out? The time it takes for the plates to draw the particle either upwards or downwards, and then just a speed in the horizontal line that gets it through the field in less than that time.

    So I begin with E = U / d. d is distance between two plates. U is voltage. E is the electric field strength. I insert U (60 V) and d (0.075m) and I get: 800 (I don't know which unit, V per meter? Long time since we did anything with electrics).

    Now I can get F right? Because I have both E and Q. So I insert what I have into E = F / Q => E * Q = F. E is 800. Q? Four times the electron charge. This gives F = 1.28174 * 10^(-16).

    Now we know F. We also have the mass! So we can calculate the acceleration through F = ma => F/m = a. This gives: 49297.7 m/s^2. I suspect this is wrong, or does it really get that big? Anyways. Lets continue.

    We have the formula s = v*t + (at^2)/2. We know the distance (s), that is 7.5cm/2 (or 0.075/2). v means initial velocity, since it is 0 in the upwards direction we get: s = (at^2)/2. We don't know t (time). We break that out:
    2s = at^2 => 2s/a = t^2. And then the square root of it all. Lets put it all together and get t.

    t = 0.0012334 seconds.

    Now we know it needs to pass the field which is 8 cm. v = s/t. All this gives v = 64.86m/s. So it needs to be fired into the field with a speed higher than 64.86m/s.

    Asides from eventual calculation errors (I might have forgotten multiplying the electron charge by 4), am I correct?
     
  2. jcsd
  3. Nov 9, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    I question that assumption. If you're looking for the least possible horizontal velocity then you want to maximize the "falling" time of the particle. I'd assume that the particle enters the plates close by one of them.
    Yes, looks good. V/m or N/C for electric field strength.
    You might want to check that force calculation. What exactly did you use for the value of Q?
    Your method looks fine. Redo your force calculation as you suggest: you did forget to multiply the charge of the electron by four.
     
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