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Electrostatic fields

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A single isolated proton is fixed on a surface. Where must another proton be located in relation to the first in order that the electrostatic force of repulsion would just support its weight?

    2)Two identical objects have charges from +6.0*10^-6 and -2.0*10^-6, respectively. When placed a distance d apart, their force of attraction is 2,0N. If the objects are touched together, then they moved to a distance of separation of 2d, what will be the new force between them?


    2. Relevant equations

    F=kq1q2/d^2

    3. The attempt at a solution
    1) the electrostatic force would be zero? And then how would u solve for the distance??
    2) I thought the inverse square law would mean that the new force would be 1/4 of its original (2.0) however the ans is .17
     
  2. jcsd
  3. Dec 6, 2013 #2

    Simon Bridge

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    Start by drawing a free body diagram for the object being levitated ... identify the forces (there's two) then write out the equations of those forces.

    What happens when the objects are touched together?
     
  4. Dec 6, 2013 #3

    Is there a gravitational force?
    I tried that (putting the electrostatic force equal to gravitational force)put won't the r and d cancel out?

    Also can u give me a hint of what happens when they touch?? Do they become neutral?
    If they become neutral how does that help in solving it?
     
  5. Dec 6, 2013 #4

    collinsmark

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    I'm guessing you are to assume that the proton is near the surface of the Earth. In other words, assume the force of gravity is Fg = mg. That's the way I understood the problem.

    It seems that this problem makes the assumption that the objects are conductors. In other words, if the objects touch the charge can flow between them. How much total charge is on the objects altogether?

    It's also important that the objects are identical (this was stated in the problem statement). That means that after the objects touch, whatever charge is left is split equally between the two of them.
     
  6. Dec 6, 2013 #5
    The ans for the first one is .12m and if I made mg= electrostatic force the d would not be equal.
    Also what do u mean regarding the conductors?
     
  7. Dec 6, 2013 #6

    collinsmark

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    That's roughly the answer that I got. :smile:

    :uhh: Assuming you don't stray too far from the surface of the Earth, the gravitation force is constant; it doesn't depend on d.

    If you really wanted to use Newton's universal law of universal gravitation, you would use [itex] F_g = G\frac{mM_{Earth}}{r_{Earth}^2} [/itex], where [itex] M_{Earth} [/itex] is the mass of the Earth, and [itex] r_{Earth} [/itex] is the radius of the Earth -- not d, the distance between the hovering proton and the proton on the surface. The latter is a much shorter distance.

    It's just much easier to approximate the gravitational force as [itex] F_g = mg [/itex]. Equate that to [itex] F = k \frac{q q}{d^2} [/itex].
    If the objects were insulators, then touching them together wouldn't necessarily have much of an effect. The charge is stuck to the surface of the objects and stays stuck there (for the most part) even if the objects come in contact with each other. On the other hand if they are conductors, the charge can flow between them when they touch.
     
  8. Dec 6, 2013 #7

    Simon Bridge

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    Technically the problem should specify that the objects are conductors - but it may be implicit to this stage in the course that charge moves about between objects that touch each other. It will be in the earlier course notes.

    For the initial situation you know that $$F=\frac{kq_1q_2}{r^2}=\frac{-12(\mu\text{C})^2k}{d^2(\text{m})}=-2(\text{N})\implies 2d^2=-12k$$ ... can you write the equivalent relation giving F (the unknown force) for the new charges and distance ##r=2d##
     
    Last edited: Dec 6, 2013
  9. Dec 6, 2013 #8

    I got the first one!! Thank you
    As for the 2nd you mean I add up the two charges to give me 4.0*10^-6 C?? Then what? I still don't really understand
     
  10. Dec 6, 2013 #9
    Where is the -12 coming from?
     
  11. Dec 6, 2013 #10

    collinsmark

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    Good job! Very nice. :smile:

    Yes, eventually. This part comes after the objects touch each other. Before that though, you'll need to find an expression for d in terms of other known variables. See Simon's post for help on that.

    Later though, after the objects touch you can assume that the objects are equally charged. To find the total amount of charge you add up the +6.0 μC and -2.0 μC to get the total charge of +4.0 μC. (Conservation of charge.)

    Then you can assume that the objects are equally charged. If the total charge is 4.0 μC, and half of that charge goes on one object, and the other half goes on the other object, how much charge does each object have?

    q1q2

    (Keep in mind this is before the objects touch.)
     
  12. Dec 6, 2013 #11
    I got it! Thank you Collin and Simon. Can you explain to me why the force would be repulsive?
     
  13. Dec 6, 2013 #12

    Simon Bridge

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    You seriously need to reread the course notes from the very start of the section on electrostatics: like charges repel, opposite charges attract.

    You can also see it from the sign of the force in Coulomb's Law.

    Because like charges repel, surplus charge will always try to distribute itself so each tiny bit of charge is as far away from all the other bits as possible ... this means that two identical conductors in contact with each other will have the same net charge.

    My equation had a typo - the units in the numerator should have been micro-Coulombs-squared ... If I read the problem statement corrcetly: q1=-2μC and q2=6μC so q1q2=-12(μC)2.

    (I also left out an equals sign - all corrected in the original post now)

    But you seem to have got it anyway.
     
  14. Dec 7, 2013 #13
    So just to clarified , since one was negative and the other object was positive at the beginning, when they touched some of the negative charges flowed to the positive, thus making both of them negative and since they now have like charges, they repel?
     
  15. Dec 7, 2013 #14
    Or when they touch they will both have a positive charge and the positive charge will repel?
     
  16. Dec 7, 2013 #15

    collinsmark

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    There ya go. :smile:
     
  17. Dec 7, 2013 #16

    Simon Bridge

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    When the spheres touch, they share all the charge equally between them.
    Thus, each gets 1μC of negative charge and 3μC of positive charge - making each sphere positive overall. So they are both positive.

    If they were free to roll on a flat surface, you'd expect the spheres to roll towards each other, touch, then roll apart.

    I'm serious about you needing to revise the basics.

    Checking - I notice you've had trouble with electrostatics before.
    • Help with electric fields (in November) - involving the force on an alpha-particle between two charged "objects".
    • Coulomb Law (just last week) - involving the force between point charges.
    ... you seem to be going backwards - which can happen when you don't understand the basics.

    I urge you to review the basics of electrostatics.
    ... go through lesson 1 in the link, from the beginning.
    If you don't get this, you won't understand the rest.
     
    Last edited: Dec 7, 2013
  18. Dec 14, 2013 #17
    I understand it after doing more practice problems.
     
  19. Dec 14, 2013 #18

    Simon Bridge

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    Well done.
    If you don't practice you do tend to go backwards ;)
     
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