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Battery connected to capacitor

  1. May 1, 2014 #1
    In connecting a battery to a capactor, doubling the spacing halve the capacitance while keeping voltage fixed.
  2. jcsd
  3. May 1, 2014 #2


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    Why not? What is your understanding of how capacitors work?
  4. May 1, 2014 #3
  5. May 1, 2014 #4
    I have managed to start with a derivation with V=Ed to show that if d=2d, for a battery connected to a capacitor, then c is halved.
    But I do not know how should I show what happens when distance is doubled as capacitor is unconnected to battery.
  6. May 1, 2014 #5
    It states nothing....wouldn't really be asking to obvious other:frown:se

    I want the derivation...
  7. May 1, 2014 #6
    Fairly poor. I at least do know that it is a device used to store charges?
  8. May 1, 2014 #7


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    Yes, it stores a charge and the voltage on the capacitor after it reaches steady state is whatever voltage you put on it. If you hook it to a 12V battery, it will hold 12volts. This is independent of the value of the capacitor, but the value does tell you how much charge it stores, which is not the same thing as voltage.

    If you hook a capacitor DIRECTLY to a battery it will most likely explode. Any idea why?

    EDIT: You edited your post while I was entering this one. Going back and editing your entries makes a thread confusing. Better to just do a new post (in the thread) with the update, OR ... put EDIT: below your original post and then enter the new thing. Hm ... I now see 2 posts in a row. Maybe I just missed the first one.
  9. May 1, 2014 #8
    Interesting input! It never occurred to me the practicality of a capacitor. In analogy, a capacitor stores as we know and we can view it in similarity to a tank of a certain capacity. The batter that is connected to a capacitor serves as a water tap. The capacity of the capacitor is independent of the rate at which water flows through the water tap.

    Edit: It will explode because the capacitor has a finite capacitance.
    Last edited: May 1, 2014
  10. May 1, 2014 #9


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    At a simple level...

    You understand about the inverse square law, right? The electric field strength from a point charge is inversely proportional to the square of the distance from that charge.

    If the charge is spread out along an infinite wire (with a fixed amount of charge per unit length), do you understand that the relationship changes so that the electric field strength is inversely proportional to the distance from the wire?

    If the charge is spread out along an infinite plane (with a fixed amount of charge per unit area), do you understand that the relationship changes again so that electric field strength is a constant independent of distance from the plane?

    A capacitor is approximately a pair of planes with a fixed charge per unit area. The charge on the two planes is equal and opposite. As a result, the electric field strength is constant between the two planes and is zero in their exterior. Electric potential (aka voltage) is given by the integral of electric field strength on (any) path running from the arbitrarily chosen zero point to the point where the potential is measured. If you double the distance between the two plates and keep the field strength constant, you've doubled the resulting integral. That means you've doubled the voltage difference between the two plates.

    Twice the voltage holding the same charge means you've halved the capacitance. In order to hold the voltage fixed (because that's what batteries do), that means that the capacitor must now only be holding half the charge.
  11. May 1, 2014 #10
    Yes it drops off like 1/r2

    For a battery connected to a capacitor and where d = 2d:

    I begin with:

    EA = Q/ε0

    E = Q/Aε0

    But V = E.d
    ∴V = Qd/Aε0

    Expressing the above as Q:

    Q = VAε0/d

    But suppose d dobules. So, d= 2d
    ∴C =Aε0/2d

    The formula for capacitance is C = Q/V:
    From this,

    Q = [VAε0/2d]/V which simplifies to Aε0/2d
    ∴C = Aε0/2d

    for C = Aε0/2d to be true, it must be the case that C/2

    therefore, if d doubles, then C must be halved because if it were not then, we get an expression not equivalent to C/2 via reductio ad asurdum. (on a side note, what does this tells me about the voltage and charge?)

    I really shouldn't be spending all the time doing such derivation because it's really jeopardizing my grades and it isn't really what my course focuses at the moment but it's harder to just regurgitate.

    Really, a derivation of what happens when d=2d in the event a battery is disconnected would be great.
  12. May 1, 2014 #11


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    Having done this a number of times (and never experienced a resulting explosion), I disagree.
  13. May 1, 2014 #12


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    So you've derived that C must be halved if d is doubled. But you already know the relationship between voltage and charge. voltage = charge/capacitance.

    With a battery connected, voltage is held contant. An ideal battery will source (or sink) whatever current is required to maintain a fixed voltage difference between its two terminals.

    That means that as you separate the capacitor plates (which, if the charge on the plates were fixed would normally increase the voltage difference between them) the voltage increases only slightly. The increased voltage leads to a current that drains into the battery (against the grain, charging the battery). This drains the stored charge on the capacitor until equilibrium is regained.

    Already provided. If the battery is disconnected, the charge on the plates is fixed and the voltage doubles.
    Last edited: May 1, 2014
  14. May 1, 2014 #13
    He said 'most likely' so universal violation does not applies here.
  15. May 1, 2014 #14


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    That answer is relevant but not really the reason. I suppose if a capacitor had infinite capacitance, it could handle the infinite current that is really the problem.

    Now, in a real-world battery, you can't GET infinite current you just get however much the battery can provide so it may not explode but the point is that the capacitor acts like a short circuit when first connected ... THAT is the key point to understand ... and short circuits draw all the current you can feed them and too much current all at once can make violent things happen inside a physical device. It can also depend on the construction. There are LOTS of ways to make capacitors. Some have liquids inside, some don't and there are different dielectrics and so forth.
  16. May 1, 2014 #15
    I see it. I overlooked the relationship C = Q/V.
    Rearranging, V = Q/C and where C = C/2, then V = Q/(C/2) = 2 Q/C.
    This means with a batter connected, distance is halved and V remains constant then Q must double. Am I right?
  17. May 1, 2014 #16
    I'm still confused about the part where the battery is disconnected.
  18. May 1, 2014 #17


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    Not quite. You halved the denominator. The quotient is fixed. What has to happen to the numerator?
    Last edited: May 1, 2014
  19. May 1, 2014 #18


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    With the battery disconnected, the charge on the plates is fixed but the voltage is no longer fixed.

    Decrease capacitance and keep charge the same. What has to happen to voltage?
  20. May 1, 2014 #19
    It has to be halved.

    Edit: wait, what equation are we referring to? V = Q/C?
  21. May 1, 2014 #20
    Voltage has to increase.
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