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Thread: Capacitors

  1. #1 Capacitors 
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    I had 2 questions about capacitors :

    1) If you put a slab of metal between the plates of a capacitor without touching either plate, does the capacitance increase, decrease or remain the same? why?

    2) A capacitor has a capacitance of 7.2 ┬ÁC. How much charge 50 V must remove to decrease the potential difference of its plates?

    In the second question, I don't really understand what he wants. The only law I have in mind is q = CV. But that gives the amount of charge on the plates, right?

    Any help is appreaciated, thanks.


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  3. #2  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    1) The capacitance will increase. How much depends on the thickness of the metal between the plates or more importantly the separation between the plates of the capacitor and metal inserted between. However, even though the capacitance increases the maximum voltage will also decrease. Take a look at.

    http://hyperphysics.phy-astr.gsu.edu...ic/pplate.html

    and use the formula for 2 capacitors in series.

    2) I cannot understand "How much charge 50 V must remove to decrease the potential difference of its plates?" It would make sense to ask how much charge must be removed to decrease the potential difference by 50 V. In this case you just use the formula you stated.


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  4. #3  
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    Thanks.

    But if we use the formula for 2 capacitors in series, wouldn't the capacitance decrease?

    In series, 1/Ceq = 1/C1+ 1/C2

    Thus, Ceq will be smaller then that of 2 capacitors.

    or maybe I misunderstood...?
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  5. #4  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Yes that is correct if you replace one capacitor with two of the same capacitor in series you get a smaller capacitance, half in fact.

    But if you replace it with two capacitors of twice the capacitance in series you get the same capacitance.

    And if you replace it with two capacitors of more than twice the capacitance in series you get a higher capacitance.
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  6. #5  
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    Ah! I understand. And naturally the 2 new capacitors will have a higher capacitance than the original one because the distance decreased and C α 1/d.

    Ok, but one last question. Why did the metal bar behave like a capacitor? the original 2 bars were connected to a current, so they became charged. But what about this bar in the air between them?
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  7. #6  
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    Well I checked my book and I think I figured it out. It was a rather stupid question but that's frequent

    Thanks a lot for your help.
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  8. #7  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Draw a picture of two capacitors in series, and you will notice that the wire between the capacitors and the plates of the two capacitors which that wire connects to are actually not connected to anything else. If you connect those same two plates with more wires, then since these two plates are already connected by a wire nothing about the circuit changes. You can keep adding wires until you basically have just a solid block of metal between but not touching two plates. But that is exactly what we have when we inserted a block of metal between the plates of a capacitor.
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  9. #8  
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    Yes I understand. The only trouble I have is what charges to put on the plates : in the series model there are 4 plates so 2 +ve and 2 -ve. For the new one only 3 plates. I put a -ve charge near the -ve pole. The plate opposing it (the one in the middle) will be +ve and the last one -ve. But that leaves the last plate -ve although it is connected to the +ve pole of the battery (We use DC not AC in this chapter). So, is that a problem?

    I know I'm asking many questions, but I'm probably gonna teach this to my friends, so I want to be prepared to any question.

    Thanks again.
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  10. #9  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Well I assume the metal in between has some thickness in which case the two sides of this metal acts like two different plates, positive charges going to one side and negative charges to the other. It would be an interesting scientitific question as to what would happen if you inserted only a sigle atoms thickness of metal between the plates of the capacitor. I have an idea however, but it comes with an explanation of an alternate way of looking at the whole problem.

    Remember that website reference I gave you? The formula showed not only a dependence on the distance between the plates but also on the permittivity of the material between the plates. Well another way to do the whole problem is to calculate (or measure) the effective permittivity after the metal has been inserted, while keeping everything else the same (harder to do of course). Well a one atom thickness of metal inserted between the plates of a capacitor would probably better handled by this second approach. The atom thickness of metal would have to be imbedded in something and then the permittivity of that whole object could then be measured.
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    I thought about permittivity in the begining. But the problem is that I couldn't find if the permittivity of metals are less or greater than air.

    The only thing I noticed is that permittivity is α 1/permeability. So I searched for permeability of metals and found copper less than vacuum (so permittivity increase) but alluminium more than vacuum (so permittivity decrease).

    I don't know though. What is permittivity? "The quality of a material that allows it to store electrical charge". What does that give me?
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  12. #11  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Well the permitivity is how the material responds to an electric field by allowing the charges in it to separate. This reduces the electric field in the material itself. For a conductor the permitivity is extremely high allowing a complete separation of charges and reducing the electric field inside to zero. When adding different layers of different materials together I think it adds just like capacitance in series, that is
    1/e = 1/e1 + 1/e2.
    Or with different thicknesses
    d/e = d1/e1 + d2/e2.

    So that when you put in a conductor in a capacitor (like in your problem) the permitivity changes from e0 (vacuum permittivity) to e where
    d/e = d1/e0 + d2/(very large number) + d3/e0
    (and d2 is the thickness of the conductor)
    Since the d2/(very large number) is a very small number you basically get just
    d/e = d1/e0 and d3/e0
    which is why it is equivalent to two smaller capacitors with separations of d1 and d3.
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  13. #12  
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    Ok thanks for this information. They just gave us a rule with permittivity without explaning what it is actually.

    Well I think I asked enough questions to cover that problem.

    Thanks and see you later.
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  14. #13  
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    Another neat experiment with a capacitor is if you have a decent di-electric seperating the + and - and then charge it with a high enough voltage, a propulsive force is exerted in the + direction. The force has a direct relation to the amount of voltage used and the quality of the di-electric.
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