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Electric Force Help!

  1. Jan 13, 2006 #1
    Can anyone help me with these problems? Thank-you
    1.) The point of fission, a nucleus with 92 protons is divided into two smaller spheres, each with 46 protons and a radius of 6.9 10-15 m. What is the repulsive force pushing these two spheres apart?
    I know that F=k(q1q2/r^2) I'm not sure what the charges are. Would it just be 46x46 orr...

    [​IMG]
    2.) Three positive point charges of q1 = 4.5 nC, q2 = 6.0 nC, and q3 = 2.0 nC, respectively, are arranged in a triangle, as shown. Find the magnitude and direction of the electric force on the 6.0 nC charge.
    I did change the charges from nC to C, nC is a billionith of a columb correct? Do I have to find to F1,3 F2,3 F1,3? Just to find q2? Also since I have no angles how would i break it into x & y components?
     
  2. jcsd
  3. Jan 13, 2006 #2

    mrjeffy321

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    The charge on a single proton is 1.602177 E-19 C, so 46 protons would have 46 times the charge.

    Yes, the nano prefix means 1 billionth, or 1 E-9.
    I would find the forces that q1 and q3 are imparting on q2. You should have two forces, each point a different direction with a different magnitude. Add the two forces together.
    You do have angles; they just aren’t given to you. From the diagram, it would see that the three charges are each located 1 m from a common center. You will need to use this to determine each charge's distance from each other.
    So for example,
    q1 and q2 are at the corners of a right triangle, the distance between them is the hypotenuse, which is equal to the square root of (1^2 + 1^2). The force between them will be a long this hypotenuse.
    I don’t know how well I explained that, does that make any sense?
     
  4. Jan 13, 2006 #3

    Doc Al

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    Hint: The charge on a proton = +e (where e is the fundamental unit of charge); the charge on an electron = -e. (Look it up.)

    Correct.
    You need to find the force on q2, which is the vector sum of the forces F1,2 and F3,2. (F1,3 doesn't matter.)
    Since the dimensions of the triangle are given, you can figure out any angle you might need.
     
  5. Jan 13, 2006 #4
    for the first problem I'm going to do 46x1.60e-19 then divide that by (6.9e-15)^2 then multiply it by 8.99e9 correct? I'm just using coloumbs law or is it more because I keep getting the wrong answer 1.39e21 & 1.52e21
     
  6. Jan 14, 2006 #5

    Doc Al

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    1) Realize that there are two charges interacting: q1 = q2 = 46x1.60e-19 C.

    2) You are given the radius of each sphere; what you need is the distance between their centers.
     
    Last edited: Jan 14, 2006
  7. Jan 14, 2006 #6
    so would it be 2(r)^2
     
  8. Jan 14, 2006 #7

    Doc Al

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    Imagine two spheres, side by side, each with a radius = r. What's the distance between their centers?
     
  9. Jan 14, 2006 #8
    2r right? would i still square it though
     
  10. Jan 14, 2006 #9

    Doc Al

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    Yes, the distance will equal twice the radius. That distance will get squared when applying Coulomb's law.
     
  11. Jan 14, 2006 #10
    okay thank-you
     
  12. Jan 14, 2006 #11
    I can't seem to get the correct answer can someone check my work?
    Its going to be F= (8.99e9)[(46x1.6e-19)/(2x6.9e-15²)] When I calculated everything I made sure I did it separately and I did it about 10 times the answer i got which i think is most correct is 6.95e20
     
  13. Jan 14, 2006 #12

    Doc Al

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    Looks to me like you're making the same two errors I pointed out earlier:

    1) I only see the charge appearing once in your calculation, but it appears twice in Coulomb's law: as [itex]q_1 q_2[/itex] (or in this case, [itex]q^2[/itex], since the charges are the same).

    2) The distance must be squared. Note that D = 2R, but [itex]D^2 \ne 2R^2[/itex]. Instead, [itex]D^2 = (2R)^2 = 4 R^2[/itex].
     
  14. Jan 14, 2006 #13
    oo okay i understand so its the q1q2 is goin to be (46x1.6e-19)(46x1.6e-19)
     
  15. Jan 14, 2006 #14

    Doc Al

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    That's correct.
     
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