Capacitance and Charge

Discussion in 'MCAT Study Question Q&A' started by MrNeuro, 03.31.12.

1. MrNeuroGold Donor

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According to the figures, decreasing which of the following would create the greatest increase in charge stored per unit voltage on an axon membrane in its rest state?
A. Leakage channel resistance
B. Na+ channel resistance
C. Area of the membrane surfaces
D. Thickness of the membrane

Explanation:
D. "Charge stored per unit voltage," Q / V, is the definition of capacitance, C. The equation for the capacitance of a parallel-plate capacitor is C = k&#1013;0A / d, where k is the dielectric constant, &#1013;0 is a universal constant (the permittivity of free space), A is the area of each plate, and d is the distance between the plates. Of the choices given, only D, decreasing the thickness of the membrane (that is, decreasing d), would increase the capacitance, C.

ummmm

how do you work through this without knowing C = k&#1013;0A / d

i used V=Ed and C=Q/V

if you decrease d V decreases but alos according to C=Q/V decreasing v increases C

so how would you work through this without knowing C = k&#1013;0A / d

Na channel has a smaller resistance than leakage channels
also just want to double check is the voltage across the capacitor equal to 1/leakage + 1/Na = 1/R?? if so wouldn't decreasing the Na resistance have the greatest impact on increasing the voltage? im not that great w/ ratios and proportions but wouldn't the same magnitude of change have a greater impact on overall Na if it was applied to Na vs leakage channels?
Last edited: 03.31.12
2. MedPR

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You should know that A and B are immediately out since resistance doesn't affect the amount of charged stored on a capacitor.

Without knowing that equation, you should still know that the bigger the plates are the more charge can be stored, which rules out answer C. Think about it, can you fit more electrons on a 1meter plate? Or a 50 meter plate?

You should know that equation though.

Edit: Ok you added more stuff.

Yes, you added the resistors correctly, but you also typed out C=Q/V. Where in that equation do you see R? Remember, voltage drop across parallel circuit elements is equivalent, so the voltage drop across the capacitor. Even if the voltage across it was less (due to a resistor in series) the charged stored would not decrease. It would simply take longer for the capacitor to charge.
3. pfaction

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The more you answer MedPR, the more I learn I don't know my material at all. Thanks for answering.

Just so you know, OP: Q=CV where C is constant in that equation. Q is proportional to V, C does not matter.
4. MrNeuroGold Donor

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so you're essentially saying the magnitude of Q is independent of V???

if so then huh? i thought Q = CV

huh????? medpr just said that Q is independent of V now your saying that Q and V are dependent upon each other?

EDIT

ahhh i think i may understand what you guys are saying now

are you saying that V exerts a change on Q indirectly via capacitance?

where a increase in V would decrease capacitance which in turn would decrease the charge?
waitttt nooo that sounds wrong medpr said that Q remains constant as well w/ any voltage...

IM SOOO CONFUSED someone help!
Last edited: 03.31.12
5. MedPR

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V in C=Q/V is not the voltage of the battery, it is the voltage between the capacitor plates. As you build up a potential difference between them, the capacitance goes down. This is why dielectrics are insulators, they reduce the potential difference between the plates of a capacitor.
6. MrNeuroGold Donor

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So are you saying that the V acquired from the the parallel resistors cannot be used to find the charge if you are given the capacitance?
7. pfaction

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The voltage of the battery and the voltage of the plates are the same if there's no dielectric involved in them. That's because you're charging something to the max of the battery. I may be spewing wrong information so I'll stop and let MedPR explain and back it up if needed.
8. MrNeuroGold Donor

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thanks for the help pfaction and medpr

im asking how is Q (stored) independent of V
9. MedPR

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Yes, I believe that is correct. The V in C=Q/V is the potential difference the capacitor builds up as it charges. Remember that capacitors are just like batteries in that they store energy. As you charge up a capacitor, it gets exponentially more difficult to continue to add more charges because a potential difference is building up in between the plates.
10. chiddler

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I may be misunderstanding the question, but I think that's incorrect. Voltage splits evenly across each parallel path. Current does not, sure. But voltage does.

So in a 3 way parallel circuit like the above with 2 R's and 1 C, if you have voltage drop of two of the parallel resistors, you can multiply that value by 1.5 to find the voltage across the capacitor.
11. MedPR

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But is that the same as the voltage that exists between the capacitor plates?
12. pfaction

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It would have to be split evenly between all three. The one thing I know is that in parallel, both resistors and capacitors have to have the same voltage. In series, capacitors have the same Q, resistors have the same I.
13. MrNeuroGold Donor

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so which results in a greater change in charge

decreasing the resistance of the leakage channels
decreasing the resistance of the Na channels

im asking because if you decrease the resistance in the circuit your increasing current (do you also decrease resistance?) so you should increase Voltage????

so decreasing which resistance would have a greater impact on overall Voltage?

EDIT
waaaaiiittttt i think i got it

the voltage doesn't change because the battery is in parallel w/ the capacitor right???
14. pfaction

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Are we talking about voltage in the circiut, or voltage in the capacitor? Can't change voltage in the resistor because it's the battery that provides the V, so unless you add a battery, battery V is constant.
15. MrNeuroGold Donor

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V of the capacitor isn't it the same as V of the resistors?
16. MedPR

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The resistors don't matter because eventually the capacitor is going to get hit by the battery. No resistors = it will charge faster, not more.
17. chiddler

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Should be, right? Where else does the voltage drop occur in a capacitor.
18. MedPR

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Doesn't the potential within the capacitor change as it charges? So it won't always be the same..? Idk, maybe I don't know what I'm talking about :/
19. pfaction

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The potential is getting closer and closer to the battery's potential before it stops shooting or absorbing electrons. That's as much as I know.

Good night y'all, starting TBR boot camp tomorrow. I don't know if it's a good idea with only 27 days before my exam, but I can't keep ****ing up.
20. MrNeuroGold Donor

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Yeah I think it does. I just had a flashback LOOOL. Potential difference increased as the capacitor games charge and stops once it's potential difference equals the voltage of the battery.... I think (LOOOL I love Siri she just wrote part of that message driving in the rain)
21. chiddler

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yes, it's changing. PE = qV. Q changes, but V can stay the same.

edit: oop qV doesn't apply to capacitors only point charges.

same thing though