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Homework Help: Charged spheres

  1. Aug 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Three small spheres are placed at fixed points along the x-axis, whose positive direction points towards the right.

    Sphere A is at x = 46.0 cm, with a charge of –6.00 μC.
    Sphere B is at x = 55.0 cm, with a charge of 6.00 μC.
    Sphere C is at x = 59.0 cm, with a charge of –7.00 μC.

    (a) Calculate the magnitude of the electrostatic force on sphere B.

    (b) What would be the magnitude of the electric field at the point where sphere B was located?

    2. Relevant equations

    Coulomb's law [tex]\vec{F_{1,2}}= k_e \frac{q_1q_2}{r^2}[/tex]

    electric field: [tex]\vec{E}=\frac{\vec{F_e}}{q_0}[/tex]

    [tex]\vec{E}= k_e \frac{q}{r^2}[/tex]

    3. The attempt at a solution

    For part (a), I have used Coulomb's law as follows:

    [tex]\vec{F_{A,B}}= (8.9 \times 10^9) \frac{(6 \times 10^{-6} C) (6 \times 10^{-6} C)}{9^2}[/tex]

    = 3.95 x 10-3 N

    [tex]\vec{F_{C,B}}= (8.9 \times 10^9) \frac{(7 \times 10^{-6} C) (6 \times 10^{-6} C)}{4^2}[/tex]

    = 0.02336 N

    Taking FA,B as negative and FC,B to be positive:

    0.02336 - 3.95 x 10-3 N = 0.0194 N

    But the answer must be 196 N. What's wrong?

    (b) I tried the 3rd equation equation above but it didn't work...

    [tex](8.9 \times 10^9) \frac{196}{9^2} = 2.15 \times 10^{10}[/tex]

    Again this is wrong. It must be 32700000 N/C. Any help with this problem is appreciated.
  2. jcsd
  3. Aug 7, 2010 #2
    You need to convert your cm to m :smile:

    For (b) you need to sum up the electric fields of A and of C at point B since electric field is a vector quantity.
    Last edited: Aug 7, 2010
  4. Aug 8, 2010 #3
    Thank you VERY much! I got the right answer for part (a) & (b). But there is one more question here...

    Sphere B is still missing. Give the x-coordinate of the point on the x-axis where the field due to spheres A and C is zero.

    How do I need to approach this problem? :confused:
  5. Aug 8, 2010 #4
    Last edited by a moderator: May 4, 2017
  6. Aug 8, 2010 #5
    can you do an example on this question?
  7. Aug 8, 2010 #6
    Sorry for the noobish question, but why is [tex]\vec{F_{A,B}}[/tex] taken to be negative, while [tex]\vec{F_{C,B}}[/tex] is positive?
  8. Aug 8, 2010 #7
    Draw the force vector out and you'll realize they point in opposite directions. We arbitrarily designate the one pointing towards the right as positive and the one pointing towards the left as negative.

    Do an example for the whole question? No we only help with parts of the question not the entire question.
  9. Aug 9, 2010 #8
    [tex]\frac{-6 \times 10^{-6}}{(46-x)^2} + \frac{6 \times 10^{-6}}{x^2} = 0[/tex]

    I tried solving this

    [tex](-6 \times 10^{-6}) x^2 = - (6 \times 10^{-6})((46)^2-92x+x^2)[/tex]

    And I just grouped the tems and then solved them. I even used the equation solver on my calculator, but the answer doesn't come out! Is there something wrong with what I did in the first step?
    Last edited by a moderator: May 4, 2017
  10. Aug 9, 2010 #9
    You should write this instead:

    [tex]\frac{-6 \times 10^{-6}}{(x)^2} + \frac{7 \times 10^{-6}}{(13-x)^2} = 0[/tex]

    Note that you should use the charges of A and C, instead of those of A and B (B is non-existent in this part of the question!). 13 is the distance between A and C.

    After you find x, add x to 46 to get the x-coordinate.
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