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Electric Field Help SIMPLE problem

  1. Feb 23, 2008 #1
    Electric Field Help!! SIMPLE problem!!

    1. The problem statement, all variables and given/known data
    Two particles are fixed to an x axis: particle 1 of charge -3.50*10-7 C at x = 6.00 cm, and particle 2 of charge +3.50*10-7 C at x = 29.0 cm. Midway between the particles, what is their net electric field?

    2. Relevant equations

    Fe= KQ1Q2 / R^2

    3. The attempt at a solution

    Fe= (8.99E9)(-3.50E-7)(3.50E-7) / (11.5^2)
    Fe= 8.33E-6

    Okay just to let you know i got 11.5 for R by doing (29-6) / 2 because the question asks what is teh net charge between teh two particles, but when i put the answer in it came wrong even when i did and didnt put the negative sign.

    so then i decided to do (29+6) / 2 and i got 17.5 and i used that for R but the answer was still wrong. my answer was -3.597E-6. i typed this answer in with the negative sign and without it, but it was still wrong.

    I HAVE aTTACHED A picture of my Free-Body DIAGRAM!!

    i don't know what to do from here, can anyone please guide me/help me? THANKs!

    Attached Files:

    Last edited: Feb 23, 2008
  2. jcsd
  3. Feb 23, 2008 #2
    To find E at a given point, we use the concept of a 'test charge', I assume you are familiar. Now, if you were to place a positive test charge midway between these two particles and were to draw a free-body diagram of the forces acting ON the test charge it may make things clearer.
  4. Feb 23, 2008 #3
    Oh yes, okay so I will attach a picture of the free-body i drew. Will you look at it?
  5. Feb 23, 2008 #4
    Yes. If you like. You can just hack it since it is in 1-D. Like this ---->q<-----
  6. Feb 23, 2008 #5
    Well my physics teacher doesn't draw fere-body's like that, he said that it would go like downwards i think. i'm not really sure, but i attached what i thought was the free-body, but it is probaby wrong. can you explain to me how to do a correct free-body for not only this problem but any of these kinds of problmes in general. like if you are doing the positive test charge Q1 would attract to the Particle becuase Q1 is negative while Q2 would repel since it is positive. How would you go about drawing that? Anyone's help with this question would be greatly appreciated!
  7. Feb 23, 2008 #6
    The PF attachment system is VERY slow. We will not be able to see your FBD until a moderator approves it.

    But, in general, to draw a free body diagram:
    Draw the object in question as a simple dot or box free from all of its surroundings (hence the name).

    Then, draw all of the forces that ACT ON the particle (do not include forces that the particle exerts on its surroundings)

    Draw the forces as arrows pointing in the direction that they act. If you do not know the direction of the force you can ASSUME it.

    If I don't know a forces direction, I always assume it is in the positive direction. Then, if I solve for it and it comes back negative, i know that it is really in the direction opposite to what I assumed.
  8. Feb 23, 2008 #7
    oH OKAY well this is how i drew it:
    the first dot is Point P for the Positive test charge particle
    .-------------------> Q1
    ^ Q2 (this arrow should be facing the opposite way)
    Last edited: Feb 23, 2008
  9. Feb 23, 2008 #8
    Yes, this is correct and by naming the test charge Q0 you could simply draw it like this


    where F01 is the "force on 0 ue to 1" and
    F02 is the force on 0 due to 2.

    As you can see, i have drawn F01 as attracting and F02 as repelling.

    EDIT: I just saw your diagram. In the problem statement it says two particles on the x axis and a point that lies midway between them, so why would you draw any forces on the vertical direction?
    Last edited: Feb 23, 2008
  10. Feb 23, 2008 #9
    OOO okay thankyou, your diagram makes sense. but see i still dont understand how to do the problem now that i understand teh free body....like now what is the purpose of this free body. im sry i dont know much, our teacher is pretty bad...so how do you use this free-body for the question / equation?
  11. Feb 23, 2008 #10
    A lot of teachers are bad at teaching!

    Now, have you added and subtracted vectors before? Since these two lie along the same line, they do not need to be decomposed. You just need to find the electric field at O due to one particle. Then find the electric filed at O due to the other particle and add them together. This is known as the Principle of Superposition.

    Now how would you find E at O due to just one of the particles? Lets start with 1.

    First write down the formula of E due to a point charge.
  12. Feb 23, 2008 #11
    First, you should identify the midway point and how far away it is from both particles. The electric field is different from Fe (electric force). The electric field only depends on a source charge.
  13. Feb 23, 2008 #12
    Stay with me physicbhelp. What is E due to a point charge?
  14. Feb 23, 2008 #13
    E= kQ / R^2
  15. Feb 23, 2008 #14
    looks right
  16. Feb 23, 2008 #15
    So Gear300 can you further help me with this problem.
  17. Feb 23, 2008 #16
    sure...first, find the midpoint between the two charges.
  18. Feb 23, 2008 #17
    Its between 6cm and 26cm...to find the midpoint you subtract the two and divide by 2.
  19. Feb 23, 2008 #18
    i explained how i got 17.5 and 11.5 in my first post, but i trie dboth of them out and they were wrong in teh full equation's answer
  20. Feb 23, 2008 #19
    the question states it is between 29 and 6 so the midpoint would be 11.5
    which i put in for R
  21. Feb 23, 2008 #20
    Well...when you subtract the two and divide, you should get 11.5cm
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