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Electric Field Help!

  1. Jan 17, 2006 #1
    Can anyone help me with this question? thank-you

    An electron and a proton are each placed at rest in an external uniform electric field of 540 N/C. Calculate the speed of each particle after 46 ns.

    I changed the protons and electrons, I'm not sure why its just a step my instructor told us to do, I'm assuming it is becoming the acceleration? I know I have to incorporate the mass somewhere. Its hard tryin to remember kinematics since we did them in September. But i changed the electrons (540x-1.6e-19) and protons (540x-1.6e-19) and the time is 46e-6 seconds. Would I use vf=vi+at? If so I know vi=0 and the time amd I would be solving for the final velocity
     
  2. jcsd
  3. Jan 17, 2006 #2
    You *almost* have it, your thinking is definitely on the right track. You can solve this problem in this series of steps.

    Step 1: Find the force acting on each particle.
    Step 2: Find the acceleration for each particle.
    Step 3: Plug the acceleration and time into the formula that you mentioned above and solve for vf.

    For completeness sake, I will work out the steps below. Don't look at them unless you are absolutely stuck even after trying the steps given above (as i think you can do this without the help!).
    .
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    Step 1:
    --------------
    The force acting on each particle is the same because the electric field is uniform everywhere and the charge on each particle is the same. However, the direction is *opposite* because positive and negative charges are either attracted to the source of the electric field or repulsed by it.

    Electron
    ---------------
    F = EQ = (540 N/C)(-1.60x10^-19 C) = - 8.64x10^-17 N

    Proton
    ----------------
    F = EQ = (540 N/C)(1.60x10^-19 C) = 8.64x10^-17 N

    Step #2:
    Note that even though the force is the same on each particle, the acceleration will NOT be the same. This is because the proton is much more massive than an electron (it is easier for the force to speed up the electron). The masses of the proton and electron are constants so be sure to memorize them because they will be needed in problems a lot (and you usually are not given them in the text!). You can find the acceleration for each particle via Newton's law:

    Electron
    ---------------
    F = ma; (- 8.64x10^-17 N)= (9.11x10^-31kg)a
    a = - 9.48x10^13 m/s^2

    Proton
    ---------------
    F = ma; (8.64x10^-17 N)= (1.67x10^-27kg)a
    a = 5.17x10^10 m/s^2

    Step #3
    ---------------
    Assume that the initial veloctity is zero and solve. Note that nanoseconds are 10^-9 s not 10^-6 like you have above.

    Electron
    ---------------
    Vf = Vi +at
    Vf = (0 m/s) + (-9.48x10^13m/s^2)(46x10^-9s) = - 4.36x10^6 m/s
    (the negative sign simply means that the velocity is in the opposite direction for the electron when compared to the proton)

    Proton
    ---------------
    Vf = Vi +at
    Vf = (0 m/s) + (5.17x10^10m/s^2)(46x10^-9s) = 2.34x10^3 m/s
     
    Last edited: Jan 17, 2006
  4. Jan 17, 2006 #3
    e

    Can anyone help me with this problem?

    [​IMG]

    Three positive point charges of q1 = 4.5 nC, q2 = 6.0 nC, and q3 = 2.0 nC, respectively, are arranged in a triangle, as shown in Figure 17-25. Find the magnitude and direction of the electric force on the 6.0 nC charge. I looked this up but im pretty sure there are some errors in my work. After finding the force of F1,2 (1.21e-7) and F2,3 (5.89e-8) with coulombs law, I then broke up into components and I got for F1,2,y (1.21e-7 x sin45), is that correct? And I got 8.58e-8. I did the same for F2,3,y and I got 6.36e9. For Fi,2,x which is (1.21e-7 x cos45), I got 8.58e-8, and for F2,3,x I got 3.81e-9. After adding the x and y components I got the sum of x=8.96e-8, and y=6.36e-9, I did a2 + b2=c 2 and I got 6.39e9. Can someone check my work?
     
    Last edited: Jan 17, 2006
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