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3 basic questions on electricity an magnetism

  1. Feb 7, 2006 #1
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    I tried many methods but i keep getting a different final answer
    can anyone help? Please!!!
     
  2. jcsd
  3. Feb 7, 2006 #2

    Doc Al

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    To get help, you need to show your work.

    Start with the first problem and describe what you did, step by step. If you tried multiple methods, describe them all! That's the only way to pinpoint what the problem is.
     
  4. Feb 7, 2006 #3
    in the 1st question answer a i got 0 because the j component is 0 and the formula is E = -Eoj and 0kind of seems odd for a solution i need someone to confirm it for me
     
  5. Feb 7, 2006 #4
    my formula for the magnitude of the acceleration is -qEoj/m and j is equal to 0 which will also give me 0 this all seems kinda odd
    in c) of the 1st question the formula is (2dm/qEo)^0.5 butall that is given is d
     
  6. Feb 7, 2006 #5

    Doc Al

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    I'm not sure what that formula is supposed to be, but clearly the field is not zero. It's given as 500 N/C.

    (Depending upon how you define your coordinates, the j-component might be zero--but so what?)
     
  7. Feb 7, 2006 #6

    berkeman

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    Question 63d seems a little odd to me. We can certainly say which way the electron will accelerate, but how are we supposed to know which end of the field it started in?
     
  8. Feb 7, 2006 #7

    Doc Al

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    Yes, oddly worded. But if the electron started on the wrong end, its journey wouldn't be very exciting.:smile: (They are checking whether you know which direction the electron would start moving.)
     
  9. Feb 7, 2006 #8
    "E = -E0j" and "-qEoj/m" sound like something you picked up from an example problem. That is definitely not the general expression for an electric field or acceleration of a particle in a field.

    E is an arbitrary vector. It can be just about anything.
    When they ask for the magnitude, this translates into math as the length of the vector.

    Fully re-read, or read the chapter in your book about this stuff. I strongly suspect you are glancing through the chapter for equations you can plug stuff into.

    After you've read, here are some more general equations that are relevant to the first problem (but you're going to need to do some thinking and reading to understand how to use them):

    [tex]\vec{F} = q\vec{E}[/tex]
    This tells us the force experience by a particle of charge q in an electric field E.

    [tex]\vec{F} = m\vec{a}[/tex]
    This is a special form of Newton's second law, and it is useful in this context to find the acceleration of a particle subject to an electric force.

    [tex]\vec{v}(t) = \vec{v}_0 + \vec{a}t[/tex]
    From introductory mechanics, which can help us find the velocities of things subject to constant acceleration.
     
    Last edited: Feb 7, 2006
  10. Feb 8, 2006 #9

    nrqed

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    We can tell because of question c). They say that it started at rest and that it traveled some distance in the field. So there is only one possible answer to which end of the field it started in.
     
  11. Feb 8, 2006 #10

    berkeman

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    Not to nit-pick, but what if it started in the middle?
     
  12. Feb 8, 2006 #11

    nrqed

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    The question clearly states " Indicate at which END OF THE FIELD DID THE ELECTRON BEGAN its journey?"

    Unless I am mistaken, this seems to imply that it started at one end of the field. :wink:

    Pat
     
  13. Feb 9, 2006 #12
    q63 turned out to be easy because the unknowns to me were e and mass of e which turned out to be a fixed variable given at the beggining of the book and 63d it was from right to left i think and in no. 76 the sketch was very easy and it turned out to be zero and in the last question i had to calculate r(the distance between the electron and p) then i found the formula which results in a cancellation cuz the distance of +e an - e from p are the same then multiply it by 1/4(pi)Eo which gives zero
    it wasnt so difficult but i am having headaches with this subject cuz the formulas are way too much
     
  14. Feb 9, 2006 #13
    Well, another way to look it at is this: When you calculate the dipole moment, you find it to be in the x direction.
    r (pointing to your observation point) is in the y direction.

    The electric potential of a dipole moment is:
    [tex]\frac{\vec{p}\cdot\vec{r}}{r^3}[/tex]
    and
    [tex]\vec{p}\cdot\vec{r}\ goes\ as\ \hat{x}\cdot\hat{y} = 0[/tex]
     
  15. Feb 9, 2006 #14

    siddharth

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    What happened to the [tex] \frac{1}{4 \pi \epsilon_0} ?[/tex]

    [tex] V = \frac{1}{4 \pi \epsilon_0} \frac{\vec{p}.\vec{r}}{r^3} [/tex]
     
  16. Feb 9, 2006 #15
    Yea, sorry.. If you want to be pedantic, in MKS you need that. (To be even more pedantic though, my expression is perfectly valid in cgs ;)

    But either way it illustrates the point that r and p are orthogonal, resulting in a zero electric potential.
     
  17. Feb 9, 2006 #16

    nrqed

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    Isn't it simpler to use the equation for the electric potential of a point charge and to see that
    [tex] {e \over 4 \pi \epsilon_0 d} + {-e \over 4 \pi \epsilon_0 d} = 0 [/tex] ???

    It's clear from the questions and the comments posted by hussain that he/she is learning introductory E&M. The answer should be adopted to the level of the student.

    To be pedantic.....:devil:

    Pat
     
  18. Feb 9, 2006 #17
    Don't get me wrong, I'm wasn't trying to say that was a better or easier way.. just another way.

    Personally, I hate trying to remember equations. I just naturally remember the ones I've used often enough, and only make a point to memorize ones that are general and used to derive the more specific ones.
    So to for me, the first thing I thought of when I looked at the picture was "well, r and p are orthogonal," without having to calculate anything or look up any equations for specific problems.
     
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