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Electromagnatism - Find charge deos this sphere have

  1. Jun 7, 2007 #1
    1. The problem statement, all variables and given/known data
    The centre of a small sphere carrying a positive charge of 1.0 x 10^-8 C is fixed at 3cm above the centre of a small conducting ball, which had a negative charge and a mass of 0.05g. If the ball remains suspended, what charge does it have?

    Answer is [ 4.9 x 10^-9]

    but i can't seem to get that answer


    2. Relevant equations



    3. The attempt at a solution

    Because the ball is suspended i take it that the force between the two charged bodies must be equal but opposite to the force due to gravity so:

    F = mg

    F = 0.05 x 9.8 = 0.49

    F = (1/(4 x pi x epsilon 0)) x ((Q1 Q2) / r^2)

    (1/(4 x pi x epsilon 0)) = 9 x 10^9

    F = 9 x 10^9 x (Q1 Q2 / r^2)

    Q2 = 9 x 10^9 x (Q1 / (r^2 x F)

    Q2 = 9 x 10^9 x (10^-8 / (3^2 x 0.49)

    Q2 = 20.4



    please help :confused::confused::confused:
     
  2. jcsd
  3. Jun 7, 2007 #2

    Doc Al

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    Staff: Mentor

    Check your units.

    OK.

    Not sure what you did here.

    Solve the problem algebraically first. Set your two equations for F equal to each other and solve for Q. The last step is to plug in numbers.
     
  4. Jun 7, 2007 #3
    i see what i've done, it should be in Kg not g

    so

    F = 0.00005 x 9.8 = 4.91 x 10^-4


    ok well:

    F = x ((Q1 Q2) / r^2)

    so if i rearrange to get Q2

    Q2 = (1/(4 x pi x epsilon 0)) x (Q1 / (r^2 x F)

    yes???
     
  5. Jun 7, 2007 #4

    Doc Al

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    Staff: Mentor

    No. Redo that algebra. How did you get Q2 on the left side? (Looks like you divided by Q2 on the right, but multiplied by Q2 on the left. You have to do the same operation to both sides of an equation.)
     
  6. Jun 7, 2007 #5
    ah, stupid me!

    so

    Q2 = (1/(4 x pi x epsilon 0)) x ((Q1 x f) / r^2)

    Q2 = (9 x 10^-9) x (((1.0 x 10^-8) x (4.9 x 10^-4)) / 3^2)

    Q2 = 4.9 x 10^-3

    i, getting closer but, the answer should be 4.9 x 10^-9

    i thought it might be the units but they seem ok now
     
  7. Jun 7, 2007 #6

    Doc Al

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    Staff: Mentor

    algebra error

    This is not true. You are taking an equation like this:
    F = A*B

    And transforming it into this:
    A = B/F (NO GOOD!)

    (Imagine A & B are your Q1 & Q2)

    But you should get:
    1/A = B/F

    Which can be inverted to be:
    A = F/B

    Try this. Redo your algebra as I suggested earlier.
     
  8. Jun 7, 2007 #7
    ok, ignoring (1/(4 x pi x epsilon 0)) for the minute

    F = ( Q1 Q2 )/ r^2

    F / Q2 = Q1 / r^2

    1 / Q2 = Q1 / (F x r^2)

    Q2 = (F x r^2) / Q1
     
  9. Jun 8, 2007 #8

    Doc Al

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    Staff: Mentor

    You are back on track now.
     
  10. Jun 8, 2007 #9
    just plug in the values with appropriate units now
     
  11. Jun 8, 2007 #10
    Thanks for everyones help =)

    but i still cant get the corect answer:

    Q2 = ((F x r^2) / Q1) x (1/(4 x pi x epsilon 0))

    Q2 = ((4.91x10^-4 x 3^2) / 10^-8) x (9x10^9) = 3.97x10^15

    Above the question it is stated that
    (1/(4 x pi x epsilon 0)) = 9x10^9

    i still cant see how im not getting [ 4.9 x 10^-9 C ]
     
  12. Jun 8, 2007 #11

    Doc Al

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    Staff: Mentor

    Your "(1/(4 x pi x epsilon 0))" is in the wrong place. It should be right next to Q1.

    If you let k = (1/(4 x pi x epsilon 0)), then you have:
    mg = k Q1 Q2/r^2

    Solve for Q2 again (so you see where the k goes) and then plug in the numbers.
     
  13. Jun 8, 2007 #12
    try to do it on paper if you are not already
    because writing things out clearly with appropriate symbols is much simpler
    than saying (1/4 x pi x epsilon)) x (((F x r^2) / Q1))...........and so on
     
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