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Simple Coulomb's Law Problem

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations
    F = | kqQ / r*r |


    3. The attempt at a solution
    Magnitude:
    F_net = F_1on3 + F_2on3
    F_1on3 = (9E9)(70E-6)(15E-6) / ( (sqrt(5))^2 ) = 1.89N
    F_2on3 = (9E9)(-36E-6)(15E-6) / ( (2)^2 ) = -1.215N ~ -1.22N
    F_net = 1.89 - 1.22 = 0.67N

    Direction = arctan(F_y/F_x)
    In component form the vectors for r_13 and r_23 are:
    r_13 = 2i + 1j && r_23 = 2j

    Therefore the force in component notation would be:
    Fvec_13 = 1.89(2i + 1j) = 3.78i + 1.89j
    Fvec_23 = -1.22(2j) = -2.44j
    Fvec_net = 3.78i - 0.55j

    Thus angle must be:
    theta = arctan(-0.55/3.78) = -8.13 degrees

    I entered 0.675 for the magnitude and was told it was incorrect, although I'm positive I did the work correctly. Am I wrong? (I am a little unsure about the angle part)

    Thanks in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 8, 2008 #2
    One mistake I see is that, for example, r_13 must be converted to a unit vector before you can use it to write Fvec_13. r_23 as a unit vector is just j.
     
  4. Sep 8, 2008 #3
    So r_13 as a unit vector would be squareroot(5), and r_23 would be squareroot(2), no?
     
  5. Sep 8, 2008 #4
    I'm not sure what you mean.

    r_13 as a unit vector: (2i + 1j) / sqrt(5)

    r_23 as a unit vector = j
     
  6. Sep 8, 2008 #5

    LowlyPion

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    Are you sure that's the magnitude?
     
  7. Sep 8, 2008 #6
    Are you saying that I should have entered 0.67N? Either way, I would like to ensure that my magnitude calculations are correct, regardless of what I entered on my homework software.

    If they asked for the Force vector, then I would have to multiply the magnitude by r^ (r hat), no?
     
  8. Sep 8, 2008 #7

    LowlyPion

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    What I'm asking is given your statement of the force vector is that it's magnitude?
     
  9. Sep 8, 2008 #8
    I take Fvec_net as the Force vector 3.78i - 0.55j (with magnitude 0.67N) which I wrote in component notation
     
  10. Sep 8, 2008 #9

    LowlyPion

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    The problem is that this is not the net magnitude. To properly calculate the magnitude of the result you need to have done a vector addition as these magnitudes are not acting along the same line.
     
  11. Sep 8, 2008 #10
    Ok, here is my retry.

    r_vec31 = -2i - j
    r_vec32 = -2j

    r_hat31 = (-2i - j) / sqrt(5)
    r_hat32 = (-2j) / 2 = -j

    Fvec_31 = 4.23 * r_hat31 N = -3.78i - 1.89j
    Fvec_32 = -1.22 * r_hat32 N = 1.22j
    Fvec_net = (-3.78i - 0.67j)
    Fvec_mag = 3.84 N



    So the magnitude would be 3.84N, NO?
     
  12. Sep 9, 2008 #11
    And since the components are -3.78 and 0.67, then the direction is:
    arctan(-.67/-3.78) = 10 deg, no?

    thanks in advance
     
  13. Sep 9, 2008 #12

    LowlyPion

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    Yes, that looks good.
     
  14. Sep 9, 2008 #13

    LowlyPion

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    10 degrees with respect to what? x-axis?
     
  15. Sep 9, 2008 #14
    yes, the x-axis
     
  16. Sep 9, 2008 #15

    LowlyPion

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    OK. Looks good then.
     
  17. Mar 4, 2009 #16
    hello sir ..........can u help me i have a problem.........two point charges placed at d distance .two mediums of dielectric constants k1 and k2 are filled between them.one upto distance d1 and other upto d2 distance.what is the force acting between the charge.d1+d2=d.
     
  18. Mar 4, 2009 #17

    LowlyPion

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    I think I would use the equivalent εeq in place of εo

    ε1 = k1*εo
    ε2 = k2*εo

    which I think intuitively can be determined similarly to capacitors in series by

    d/εeq = d1/(k1*εo) + d2/(k2*εo)

    You may want to satisfy yourself that this is correct.
     
  19. Mar 4, 2009 #18
    sir thanks for replying. sir i have another question also if these mediums are placed one above the other between two point charges .the lengh of mediums is same as the distance between charges .what will be force between the charges .
    _____________________________________
    q1|________________k2_____________________| q2
    |________________k1_____________________|
     
  20. Mar 4, 2009 #19

    LowlyPion

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    That seems to be more complicated because the thicknesses are unknown. If the thicknesses are large and the boundary bisects the point charges then by symmetry I would suppose that

    εeq = 1/2*(k1 + k2)εo

    BUT for thicknesses less than d between the charges, ... that looks more complicated than I want to consider.
     
  21. Mar 4, 2009 #20
    sir both the mediums have thickness same as that of d distance between the chages.
     
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