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Electric field problem

  1. Jan 28, 2005 #1
    Hi,
    Yesterday the professor was explaining an electric field problem. An electron with an initial velocity v0 is travelling to the right, and there's an electric field pointing up. The question was, what distance does the electron travel after 3s? After finding the acceleration due to the electric field, he showed us this:

    s(t) = v0(t) + (1/2)(a)t^2

    and simply plugged in the values for v0 and a. But I was a bit confused. If v0 is to the right and a is downwards, doesn't that screw up the equation? I thought this equation only worked for calculating distance travelled in one dimension. What exactly is this distance - the distance travelled in x, y, the arclength of the electron's path, or what?

    Thanks for any help.
     
  2. jcsd
  3. Jan 28, 2005 #2
    Which distance are you talking about?

    If it is the range ( horizontal distance ) then when givin the time, the equation that your proffessor used should have had the following values.

    let s(t) = sx (distance travelled in the horizontal direction)

    v0=v0
    t = 3
    a = 0

    If it was the verticle distance travelled, then the following values should have been used:

    let s(t) = sy (distance travelled in the verticle direction)

    v0 = 0
    t = 3
    a = a

    These are the two components of motion.

    The magnitude of the distance travelled is:

    sqrt((sx)^2+(sy)^2)
     
  4. Jan 28, 2005 #3
    The distance travelled is, I think, the length of the trajectory traversed in 3 sec.
    The displacement vector equals (x^2 (t) +y^2(t))^(1/2) in magnitude.
    I'm, with great respect,
    Einstone.
    P.S.- 'Velocity' of an electron? May God pardon the man who says(& means)this!
     
  5. Jan 28, 2005 #4
    Hmmmm. You see, I don't know what this "distance" is supposed to be myself. When we questioned him about he just said it was the "distance" and left it at that. I think he just didn't want to admit he was wrong. :D
     
  6. Jan 28, 2005 #5
    You need to work in two dimensions here. Suppose that the initial velocity is along the x-axis and the electric field is upward along the y-axis. Then you need to apply the general formula for position and project it onto each direction. [tex]\vec{r} =\vec{r_{0}} + \vec{v_{0}t} + \frac{\vec{a}t^2}{2}
    [/tex]

    Now, suppose the electron of mass [tex]m_{e}[/tex] starts out at the origin so that the components of [tex]r_{0} = (0,0)[/tex]

    For the velocity you have that [tex]v_{0}[/tex] is along the x-axis
    For the acceleration you have that [tex]m_{e}a = -eE[/tex] in the y-direction.

    Thus the position is :

    [tex]x = v_{0}t[/tex]
    [tex]y = \frac{-eEt^2}{2m_{e}}[/tex]

    i think you can move on from here, but it is very important that you get this way of thinking... Beware that e is the elementary charge and e itself is positive. That is why i took -e for the electron...

    marlon
     
  7. Jan 28, 2005 #6
    Thanks for your explanation, marlon. So what exactly do you think he meant with s = v0t + (1/2)at^2? It gave us some number, but I have no idea what this represents. Does this number actually represent anything?
     
  8. Jan 29, 2005 #7

    Well this is the general formula that represents the distance as a function of time of a particle with initial velocity v and acceleration a. The trick is to realize that this formula is always valid in classical physics BUT it is a vector-equation. This means that in order to attain any practical results, you will have to apply this equation for each direction of frame. You know, x and y and z - direction. This is why, in solving such problems, you will first need to write down the components of all forces acting on the particle and the components of the initial velocity. Then you fill these components in the equation for the position.

    A little exercise maybe, can you write down the equation for position (in both the x and y direction) if for example the initial velocity is a vector that makes an angle of 45 degrees with the x-axis (pointed to the right) and that has magnitude 100. E-field remains the same

    Second exercise, now same initial velocity as in the previous exercise and the E-field (same direction : upward the y-axis) makes an angle of 20 degrees with the y-axis. WHAT IS THE POSITION OF THE PARTICLE AFTER 2 SECONDS IN THESE TWO SITUATIONS...


    regards
    marlon and enjoy...
     
  9. Jan 29, 2005 #8
    I think you guys are mixing me up with the professor. I know that you're supposed to break it up like that, but I was wondering what you guys thought he might mean with that. Trust me, this is not a "A good friend of mine..." situation - this actually happened, it's not me who's confused about this! He took the equation I showed in the first post, then just plugged in v0 and a although they are in different directions. You see, I want to salvage any last bit of confidence I have in this guy, since I'm stuck with him for the rest of the semester. So I was hoping that he was at least right about it somehow, or else I have no idea what he wants us to do on his exams - the right way, or his screwed up way.

    In the x direction it would be s = (100 cos 45)t and in the y direction it would be (100sin45)t + (1/2)at^2 where a is given by a = Eq/m, with q being -e and m being the mass of an electron. Am I right?

    Well, you'd have to take the x and y components of the E-field, but I'm too lazy to do it right now. I understand fine how this is supposed to be done, I was just curious if my professor was actually right about it in some way.
     
    Last edited: Jan 29, 2005
  10. Jan 30, 2005 #9
    you are right with your solution and your professor is also right

    marlon
     
  11. Jan 30, 2005 #10
    How is he right? What is this distance that he found?
     
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