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Homework Help: Heya Help with Electro magnitisim Question Please

  1. Aug 20, 2008 #1
    Heya everyone could u help me with these questions plz



    a charge +2nC is located at (0, 2 m). Another +2 nC charge is
    located at (0, -2 m). A third charge, +3 nC, is located at (2 m, 0).
    Calculate the total force (in vector form) acting on charge +3nC
    from both +2nC charges.

    Using the result obtained in part above, calculate the electric field at
    the position of the + 3nC charge due to both +2nC charges.

    plz guys thanks alot
     
  2. jcsd
  3. Aug 20, 2008 #2

    tiny-tim

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    Hi vip_uae! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help you. :smile:
     
  4. Aug 20, 2008 #3
    well i used the normal formula of F13 = (Q1 . Q3)/ 4 pi E0 r13 multiply by Rhat 13 ... and i knw wats rhat13 and r13 thats for the first part of the question i add the F13 AND F23 to give me FTOTAL... is that right?
     
  5. Aug 20, 2008 #4

    tiny-tim

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    Well, it looks ok …

    but it would be more convincing if you showed us some figures. :wink:

    erm … since you've got that far, what did you want help with? :smile:
     
  6. Aug 20, 2008 #5
    its just i am getting a weired answer for the total force i am getting Zero in the end of it...

    when i add F13 AND F23 = 0

    and i dont understand what to they mean by VECTOR form
     
  7. Aug 20, 2008 #6

    tiny-tim

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    Vector form means as a multiple of rhat, which you've done.

    I don't understand how your F13 and F23 can add to zero, if you're adding them as vectors. :confused:

    F13 is a multiple of the unit vector rhat13, while F23 is a multiple of rhat23.

    Since rhat13 and rhat23 are not parallel, how can any combination of them add to zero?
     
  8. Aug 20, 2008 #7
    awwwwww right sorry i actually missed out one part of it but it gave me a really small number for Xhat 9.537x10^-9

    i just found somthing that if i want to have to calculate it in a vector form

    F13 = Q1 X Q3 / 4 pi E0 (r13)^2 multiply by Rhat13 + (u0 X Q1 X Q3)/ 4 pi (r13)^2 Multiply [V3 [tex]\Delta[/tex](V1 [tex]\Delta[/tex] Rhat13 )]

    is that formula looks right it said that if a vector form needed Fe which is Electric Force + Magnetic force = the force
     
    Last edited: Aug 20, 2008
  9. Aug 20, 2008 #8

    tiny-tim

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    Is that supposed to be a magnetic force?

    Magnetic forces only act on moving charges.

    These charges are stationary … so forget magnetic forces!

    You're vector-adding a multiple of rhat13 to the same multiple of rhat23.

    Use components … what is the resultant? :smile:
     
  10. Aug 20, 2008 #9
    see i dont understand what you meant by Multiple of Rhat i knw r hat = r13/moduls r13
     
  11. Aug 20, 2008 #10

    tiny-tim

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    Yes, you're multiplying the vector rhat13 by (Q1 . Q3)/ 4 pi E0 r13,

    and adding to it the vector rhat23 multiplied by (Q2 . Q3)/ 4 pi E0 r23 …

    aren't you? :confused:
     
  12. Aug 20, 2008 #11
    yes thats wat i am doing exactly
     
  13. Aug 21, 2008 #12

    tiny-tim

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    ok, so you've got a sloping-down vector plus an equal sloping-up vector …

    so their sum (resultant) is … ? :smile:
     
  14. Aug 21, 2008 #13
    ya i have got Fnet = 9.536 x 10^ -9 Xhat ... does this answer looks right?
     
  15. Aug 21, 2008 #14

    tiny-tim

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    I haven't done the figures, but yes it certainly is a multiple of Xhat. :smile:
     
  16. Aug 22, 2008 #15
    Thats gr8 thanks alot i have an another question please

    when an electron beam shoots through a horizantal line in the Z direction carying 50 A .. how would the beam will deflect ... like the direction of the deflection how would it be :)
     
  17. Aug 22, 2008 #16

    tiny-tim

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    ok, but same thing … show us how far you've got!

    what sort of field is produced? which direction is it in? what is the equation that relates field strength and force? :smile:
     
  18. Aug 22, 2008 #17
    Hey tiny tim lol i found the answer some where basicly it will deflect upwards due to the way that the current is rotating ... soo o i got it straight away
     
  19. Aug 22, 2008 #18

    tiny-tim

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    he-he … now you see one of the reasons why we like to see some work first! :biggrin:
     
  20. Aug 22, 2008 #19
    hahahahahaha ohhhhhhhh i have a problem tooo :P see i have one of the questions asking me about 2 wires carying 10K A in different directions i need to see the force of one wire exerting on the other using the Motor Equation which i knw

    F = L ( I upside down triangle B)

    I is current L is lenght and B is magnetic Flux density

    B = u H

    H = I / 2 pi R *I guess*

    should i replace each unit and see wat i can get?
     
  21. Aug 22, 2008 #20

    tiny-tim

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    oooh … I always get confused by questions about induction. :confused:

    Can you start a new thread on this one, so that some other people will answer? :smile:
     
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