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Determine the Electric Field

  1. Jun 20, 2016 #1
    1. The problem statement, all variables and given/known data
    An electric field with a magnitude of 2.5N directed to the left, acts on a negative charge of -5.0C. Determine the electric field in which the charge is located.

    Give. FE=2.5N
    q=-5.0C


    2. Relevant equations
    FE
    =qε

    3. The attempt at a solution
    ε
    =2.5N/5.0C=0.5N/C. I don't know how to find the direction. The answer is 0.5N/C,
    but I don't know how to get that direction. I guess we have to assume the positive test charge is to the right of q and figure out the direction it would move as a result of FE applied on it by q. That doesn't really make sense though.

    Edit: I don't know why the formatting is weird. Also, please help- my exam is tomorrow!!​
     
    Last edited: Jun 20, 2016
  2. jcsd
  3. Jun 20, 2016 #2

    blue_leaf77

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    Use vector equation ##\mathbf F = q \mathbf E##. With ##\mathbf F = -2.5 \hat x## and ##q = -5\hspace{2mm} C##, solve for ##\mathbf E##.
     
  4. Jun 20, 2016 #3
    I did, but I don't know how to find the direction
     
  5. Jun 20, 2016 #4

    blue_leaf77

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    You did, then please write out your answer for ##\mathbf E##, note that the bold symbol denotes vector notation so your answer must consist of a number and the unit vector.
     
  6. Jun 20, 2016 #5
    Ok, and how do you find the direction?
     
  7. Jun 20, 2016 #6

    blue_leaf77

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    If you really have calculated ##\mathbf E##, not just ##|\mathbf E|##, then show it in your next comment.
     
  8. Jun 20, 2016 #7
    My teacher said you're not supposed to plug in - signs into the equations for fields because it implies direction, so we only use magnitudes
     
  9. Jun 20, 2016 #8

    blue_leaf77

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    I don't know in which manner your teacher wants you to find the direction, but the formal way is to use the vector equation. Alternatively, you can use the definition of negative charge. A negative charge is defined to be the condition such that the field the particle is exposed to exert a force in an opposite direction to that field.
     
  10. Jun 20, 2016 #9
    I'm confused about the wording of the question. Usually we use a positive test charge along with the given charge. Is the test charge applying the force on q? Then it would be to the right of q since the force is directed to the left. And then q would move to the right as it is attracted by the positive test charge?
     
  11. Jun 20, 2016 #10

    blue_leaf77

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    That's indeed the definition of negative charge when applied to the current problem at hand.
    No, the charge will move in the same direction as the force, not the field.
     
  12. Jun 21, 2016 #11
    But the answer is
     
  13. Jun 21, 2016 #12

    blue_leaf77

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    You already arrived at the correct answer there.
     
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