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Gauss' Law & charge inside sphere

  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data

    11gn3eq.jpg

    I need to find the total charge inside the small metal sphere, inside the big metal sphere aswell as outside the big metal sphere.


    2. Relevant equations

    What confuses me is the electric field vector. Since it's only poiting in one direction it can't originate from a single charge, or a group of charges. What could produce such a field?

    3. The attempt at a solution

    I know I'm supposed to use Gauss' law.

    For the small sphere. The amount and magnitude of electric field lines entering and leaving is the same. Which indicates that there is no charge inside the small sphere. As for the big sphere, inside and outside I honestly have no clue how to proceed. You should multiply the electric field with the perpedicular area. However if there is only one electrif field line, what area? Once again, so greateful for any idées/nods in the right direction!
     
  2. jcsd
  3. Mar 20, 2016 #2

    BvU

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    Hello Lamberto, benvenuto :welcome:
    That's not much of a relevant equation ! But let me help you a bit: the field between the spheres can only be spherically symmetric(can you explain why ?). They just drew one field vector, but they all point to the center. And at 8 cm from the center they are all 15 kN/C strong.
     
  4. Mar 21, 2016 #3
    Haha I'm sorry I was in a rush writing my question. So the field is spherically symmetric, due to the fact that the charges repell eachother and align at a equal distance from eachother, in such a pattern that they create a symmetric field? Could that be?

    However with your information I used Gauss' law to calculate the charge inside the small sphere accordingly:
    q = E x 4 π (8cm)2 x ε0

    And for the large sphere the total charge would be that of the small charge, in addition to the q= E x 4 π (17cm)2 x ε0

    (the surface area 4 π (8cm)2 and 4 π (17cm)2 because the electric field is 15 000 N/C at 8cm respective 17cm)

    Am I on the right track?
     
  5. Mar 21, 2016 #4

    BvU

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    Yes, I think you are on the right track. Don't rush, however, or you make unnecessary errors, like
    you of course mean "the total charge would be minus that of the small sphere"....

    yes, a charge on the small sphere separately (no big hollow sphere present) would give a spherically symmetric charge distribution and a spherically symmetric field, and the large sphere does not introduce any disturbances of that symmetry (because it is concentric -- if it would not be concentric you would have a different E-field and a different, non-uniform charge distribution on the inside of the large shell.)
     
  6. Mar 21, 2016 #5
    Yes of course electric field, it's a vector.

    Thank you for the help! I have a better understanding of Gauss' law now!

    This dose not concern the excerise but it's something I've been wondering about. Maybe you could help me or perhaps this is inappropriate for this forum. My knowledge about charges, conductors etc are lacking. In this case when the large sphere is positively charged (the electric field lines are pointing out of the sphere). Dose it mean that is has NO negativly charged particles (moving freely around) or that it has a greater amount of positive charges.
     
  7. Mar 21, 2016 #6

    BvU

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    Ah ! That is such a nice question that I am overjoyed you ask it !
    What was the total charge on this big hollow sphere again ? 48 nanoCoulomb ?

    Calculate with me:
    The mass is ##{4\pi\over 3}(0.15^3 - 0.1^3) * \rho \approx 80 ## kg ##\quad (\rho \approx 8000 {\rm \ kg/m^3}) \ ## Pretty heavy !.
    With an atomic weight of 55.85 kg/kmol that is 1.4 kmol.
    If neutral, each atom counts 26 electrons, so ## 1.4 * 26 * N_A \approx 2.2 \times 10^{25} \ ## electrons in the sphere, a whopping ## 2.2 \times 10^{25} \times q_e \approx 3.6\ ## MegaCoulomb.

    So if one in 74 billion electrons is missing you already get this electric field. There are plenty of electrons left over.

    I did cheat for effect: most of the Fe electrons are not moving about freely. Only the ones in the conduction band -- but that's a field of physics that I don't know much about. I suppose that on average it's one or two of the outer electrons that move pretty freely.

    Please correct my calculation errors -- what was this again about being in a rush :smile: ?
     
  8. Mar 21, 2016 #7
    Dude haha chill, but thank you for your input!

    EDIT: sorry not sure how that sounded but I am truly greatful :)
     
    Last edited: Mar 21, 2016
  9. Mar 21, 2016 #8
    But hey since you're this enthusiastic! Potential energy in a current. Positive charges loses potential energy as they move towards the negativ output whereas negativ charges gain potential energy as they move towards the positive output. Do the charged not interact with eachother as they move through the wire. And what happens with potential energy when it converts to heat or alike due to resistance? Like what happens to charges in a current as they run into a resistor?
     
  10. Mar 21, 2016 #9

    BvU

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    As we have seen, only a very small fraction of the charge that is present in a conductor has to be added or removed to generate a considerable electrical field.

    The way I learned it is that in general mobile charge carriers are electrons. But an electron that moves from A to B is completely equivalent to an equal but opposite charge moving from B to A. In transistors and diodes these "missing electrons" are called holes.

    The charges interact with each other: they still repel each other so -- as in your charged spheres -- they tend to 'travel' far apart.

    Resistance occurs if a material doesn't allow the charge carriers to move completely freely; a fraction of their energy is converted to motion of the consituents of the material (the atoms) and that is heat.
     
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