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Uniform Electric Field:finding Voltage

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Ok, i have this problem i am trying to solve and i am stuck.

    Q. A uniform Elec. field @ 75N/C is at an angle, 33degrees, relative to the positive X-axis. Find the voltage between the points (.5meters,0) and (0,0).

    I started the problem off by using the equation V=Ed; i thought that i could also write the equation as V=Ed(cos theta) due to the field lines not being parallel to the direction of movement.

    Im not even sure that is the right way to do taht.
    Then in part B is asks to find the value of Y such that the V between (.3 meters, Y) and (0,0) is equal to the same value from A.

    I know it would have to be on a perpendicular electric field line running through the first point. Do i just use trig functions to get the unknown sizes for the triangles that can be drawn?

    Any help or pointers would be appreciated, thanks.


    2. Relevant equations

    V=Ed; V=Ed(cos theta)

    3. The attempt at a solution
    I started the problem off by using the equation V=Ed; i thought that i could also write the equation as V=Ed(cos theta) due to the field lines not being parallel to the direction of movement.
     
  2. jcsd
  3. Feb 4, 2007 #2

    Dick

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    You are right that it is only one component of the E field which will count in computing the voltage difference. It's the component parallel to the direction between the two points (since normal forces don't do work). So it is V=d*E*cos(theta) where the angle theta is the angle between the E field and the path. For the second part, do it just like the first part - leaving Y as an unknown. Then solve for Y.
     
  4. Feb 4, 2007 #3
    Thanks for the help!
    How would i set that 2nd equation up? There would be 2 unknowns now though, right? The Y value and theta will change becuase it will no longer be along the x-axis.
     
  5. Feb 4, 2007 #4

    Dick

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    You should be able to write theta in terms of Y (or vice versa) - reducing it to only one variable.
     
  6. Feb 4, 2007 #5
    Setting it up like that i get:

    V=EdCostheta

    -31.5=(75)*[(.03^2+y^2)^1/2)]*[Cos(arctan y\.3)]

    Is that set up right at all??
     
  7. Feb 4, 2007 #6

    Dick

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    Offhand it looks about right. But remember the angle needs to be the angle between the E field and the path direction. You will probably get an equation that you can't solve algebraically - but you can always do it numerically.
     
  8. Feb 4, 2007 #7
    I cant for the life of me figure out part B. Are there any other ways other than numerically to figure out the answer; we havent gone over anything in class that would have us figure out such a crazy equation like that..
     
  9. Feb 4, 2007 #8

    Dick

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    You can't solve it cleanly. Period. But you don't have to use any sophisticated numerics either. Just guess an answer and see if it's too big or too small. Use that answer to take a better guess and keep refining the guess. You can get pretty close pretty fast. But you have realized that the last equation you posted has a couple of problems? One in the angle department and I'm a little worried about a sign.
     
  10. Feb 4, 2007 #9
    ahh the last portion would need to be:

    [180-Cos(arctan y\.3)]

    ?
     
  11. Feb 4, 2007 #10

    Dick

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    Nooo. Remember the angle is the difference between the path direction and the field direction. I don't see the field direction in there.
     
  12. Feb 4, 2007 #11
    gotcha, so this is how it would look then:

    [cos(180-33+arctanY/.3)]
     
  13. Feb 4, 2007 #12

    Dick

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    Yes. I assume you stuck the 180 in there just to get a sign to match the -31.5. That's ok. But I'd just drop the 180 and the - sign.
     
  14. Feb 4, 2007 #13
    Ok, gotcha, well i got the answer now! Thanks so much for your help! I really appreciate it!

    A:
    y=.309 w/ theta=192.8 (or 12.8)
     
  15. Feb 4, 2007 #14

    Dick

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    That's what I get. Hey, how did you conquer that nasty equation anyway?
     
  16. Feb 4, 2007 #15
    i actually plugged the equation into my TI, used the table function to narrow the y-values down to what X i was looking for.

    how did you do it?
     
  17. Feb 5, 2007 #16

    Dick

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    I typed the formula into Maxima and called 'find_root'. This 'numerical methods' thing is not so bad...
     
  18. Feb 5, 2007 #17
    whats that? a program??
     
  19. Feb 5, 2007 #18

    Dick

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    Yes. Sorta like Matlab et al. Except it's free.
     
  20. Feb 5, 2007 #19
    i just downloaded it and am playing with it, i cant get it to work right though; what did you type into the screen..

    (if you dont still have it, dont worry about it ill figure it out.)
     
  21. Feb 5, 2007 #20

    Dick

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    f(y):=75*(0.3^2+y^2)^(0.5)*cos(atan(y/0.3)-float(%pi)*33/180)-31.5;
    find_root(f(y),y,0,1);
     
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