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Homework Help: Physics Theory Questions. Electrostatic/magnetic

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A positively charged object is brought near to, but does not touch, one end of the a neutral metal rod on an insulated stand. The opposite end of the metal rod is grounded. The ground is removed, and then the positively charged object is removed.

    As a result of this procedue described above the metal rod will be :

    a) neutral
    b) have a net zero charge
    c) have a positive net charge
    d) have a negative net charge

    2. When white light from the sun strikes a flint glass beat, the white light is separated into its component colors.

    Which of the following statements contains a valid prediction of the relative indices of refraction for red and violet light and a justification of that prediction?
    a) Index of refraction of red light in flint glass is greater than that of violet light because red light refracts more inside the flint glass.
    b) Index of refraction of red light is less than violet because it refracts less.
    c) Index of refraction of violet light is greater than red because it refracts more
    d) Index of refraction of violet light is greater than red because it refracts less.
    2. Relevant equations

    3. The attempt at a solution

    1. if you brought a positive charge to it, it would become positive and ground removes electrons, but as the grounding is removed and the object would it become net zero, or neutral? Or remain positively charged

    2. I said b, because red refracts least right?
    Last edited: Jan 11, 2013
  2. jcsd
  3. Jan 11, 2013 #2

    Simon Bridge

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    It helps to draw a picture of each stage.

    You have worked out that the charges in a conductor are free to move, and that a neutral conductor contains equal amounts of positive and negative charge.
    You don't need to know what the charge is carried on to do the problem.

    1. bringing a positively charged rod close to the conductor.
    ... what sort of charges are attracted to the positive rod and what sort are repelled?

    2. ground the opposite side:
    ... what charge is the opposite side?
    ... what does this mean for the connection to ground?

    3. remove the ground then the rod
    ... the conductor started off with a balance of charges, will this balance remain (have charges been added or removed?)
  4. Jan 11, 2013 #3
    1. Well, actually if i remember snells law says that they repel right? or is that different, im sorry my electrostatics is blurry. But since its neutral i thought the positive charges would just transfer. Electrons would be attracted to the + and + would repel.

    2. negative right? and i have no idea

    3. Well thats kind of my question, i dont understand what happens, would the electrons flow back in, that were being grounded?
  5. Jan 11, 2013 #4
    When the positive charged object is brought close to the metal rod, it attracts the opposite charge. In this case, the negative charge would be attracted to the positive charge, right? Because opposite charges (positive and negative) attract and the same charges (negative + negative, positive + positive) repel. So if the negative charge is attracted to the side closer to the positive charged object, the other side of the rod would have to be what charge? This charge has to counter act the growth of negative charge on the other side of the rod. Grounding in electricity essentially means that the object being grounded neutralizes. In this case, the opposite end of the metal object is being grounded. So now you have a neutralized opposite end and a negative end being attracted still with the positive charge. When the positive charge is released, the negative charge does what? It leaves the concentrated area and disperses. So what would the final net charge be?
  6. Jan 11, 2013 #5
  7. Jan 11, 2013 #6
    yes, and sorry I don't remember optics too well.
  8. Jan 11, 2013 #7

    Simon Bridge

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    Snell's Law is the Law of Refraction in optics.

    - charge is transferred by contact (strictly: "electrical contact")
    - opposite charges attract

    ... these two bits of information are essential to completing this problem.
    Go through the questions again, bearing these two facts in mind.
    Like I said - it helps to draw a picture of each stage.

    At step one - what's the picture?
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