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Coulomb's Law and charged particles

  1. Sep 29, 2007 #1

    UNG

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    1. The problem statement, all variables and given/known data
    Hi, I would appreciate it if someone could help me with this question, im a n00b here
    Please and thank you
    Okay heres the question.

    The particles have charges Q1 = -Q2 = 100nC and Q3 = -Q4 =200nC
    and distance a = 5.0cm.

    What are the x and y components of the net electostatic force on particle 3 ?

    So basically a diagram is given with the question showing the 4 particles equally seperated from each other at a distance of 5.0cm, making a square. And I have to find the force on particle 3.

    2. Relevant equations

    This is the equation I used.

    F1,3 = (8.854x10^-12) x (Q1 x Q3)/0.0707xR^2 (0.0707 comes from Phyagrous therom)
    H= (square root) 0.05^2 + 0.05^2
    F1,4 = (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2

    Fnet = F1,3 + F1,4

    = F1,3 + (8.854x10^-12) x (Q1 x Q3)/R^2 Cos 45 (45 degrees middle of square)
    + (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2 Sin 45

    3. The attempt at a solution

    I have a feeling this is seriously wrong and have made it more complicated than it is.

    I placed the values into the equation shown and came out with

    = -79.46 x10^3 + 79.46 x10^3

    Heres a site I found afterwards...
    http://physics.bu.edu/~duffy/PY106/Charge.html

    I think it might go a little something like that instead^^

    If you could help that would be great thanks.
     
  2. jcsd
  3. Sep 29, 2007 #2

    learningphysics

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    I'm confused by the "pythagorean theorem" part...

    What is the distance between charge 1 and charge 3?
     
  4. Sep 29, 2007 #3

    UNG

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    Doesnt say.
    But I worked it out by spliting the square in two (a triangle)
    and used pythagoras theorem to find the longest side (diagonal between Q 1 and 3)
     
  5. Sep 29, 2007 #4

    UNG

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    As in h^2 = a^2 + b^2.

    So the distance between 1 and 3 should be 0.0707
     
    Last edited: Sep 29, 2007
  6. Sep 29, 2007 #5

    learningphysics

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    Why do you have 0.0707*R^2, instead of just (0.0707)^2?

    The distance between 1 and 3 is... sqrt(0.05^2 + 0.05^2) = 0.0707

    so F1,3 =

    F1,3 = (8.854x10^-12) x (Q1 x Q3)/(0.0707)^2

    And also the distance between 1 and 4 is 0.05m right? so how is the 0.0707 coming into that part?
     
  7. Sep 29, 2007 #6

    learningphysics

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    yes. so why the 0.0707*R^2?
     
  8. Sep 29, 2007 #7

    UNG

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    Oh yes sorry, realised that a while ago but needed to edit it.
    Also I only put the 0.0707 bit in F1,4 not F1,3
     
  9. Sep 29, 2007 #8

    learningphysics

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    let's start fresh... what is the magnitude of the force that particle 1 exerts on particle 3?
     
  10. Sep 29, 2007 #9

    UNG

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    I worked it out as -71.92x10-3
     
  11. Sep 29, 2007 #10

    learningphysics

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    wait... you're making a mistake with the 8.854x10^-12.

    force = kQ1Q2/r^2,where k = 9*10^9.

    instead of k you can also use: 1/(4*pi*(8.854*10^-12))
     
    Last edited: Sep 29, 2007
  12. Sep 29, 2007 #11

    UNG

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    I see,
    it really should be 8.99x10^9 when using the 1/ 4x pi x8.854*10^-12
     
  13. Sep 29, 2007 #12

    learningphysics

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    not -12... k = 8.99*10^9.

    k=1/(4pi*8.854*10^-12) (the 8.854*10^-12 is in the denominator).
     
  14. Sep 29, 2007 #13

    UNG

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    My mistake, another typing error
     
  15. Sep 29, 2007 #14

    learningphysics

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    no prob. I'm getting the force between 1 and 3 as having a magnitude of 0.036N.
     
  16. Sep 29, 2007 #15

    UNG

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    I did,

    F1,3 = (8.99^9) x (100x10^-9)(200x10^-9)/ 0.05^2

    = -71.92x10-3 N

    Wrong Q 1 and 3 values I put into the equation perhaps?
     
  17. Sep 29, 2007 #16

    learningphysics

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    you should have 0.0707^2 not 0.05^2.
     
  18. Sep 29, 2007 #17

    UNG

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    Oh no.
    Its Q1 to Q3 are the charges on the same line.

    Q1 to Q4 are diagonal, being 0.0707^2
     
  19. Sep 29, 2007 #18

    learningphysics

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    Ah... ok. so the magnitude of the force of Q1 on Q3 is 71.92x10-3N. Is this in the y-direction or x-direction?

    Use [tex]\vec{i}[/tex] and [tex]\vec{j}[/tex] vectors... to write this force in vector form... careful about directions and signs.
     
  20. Sep 29, 2007 #19

    UNG

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    Well Q1 and Q3 are together on the y axis
    and the question asks for the force on Q3
     
  21. Sep 29, 2007 #20

    learningphysics

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    Is the force upward or downward?

    what are the positions of the 4 charges. Is Q1 the upper left, Q3 bottom left...
     
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