potential difference of capacitor in circuitby iampaul Tags: capacitor, circuit, difference, potential 

#1
Mar1312, 05:03 AM

P: 78

I'm having a problem with this although I just finished the chapter about electric potential and potential energy in University physics by young and freedman and I need to understand this because the next chapter is about capacitors and dielectrics and there is an example just like this:
(((((((((((ll AllllB *The paretheses are just for spacing * I can't paste the pictures (((((((((((ll The three capacitors are in parallel and the potential difference between A and B is V, what is the potential difference across each capacitor. I know that the potential difference across each branch must be equal to V but i don't understand it clearly. I think i have read something about the potential difference being independent of the path. So my first question is why is the potential difference the same when in parallel. The second one is about the potential difference between the plates of the capacitor. Potential depends on distance. Lets say the potential is computed from a point charge q beyond A. The distance between A and B is different from the distance between the plates of the capacitor so the difference between potentials must be different??? What im trying to say is something like this: ((((((((((((((((((((((((((((((((((((((( ll ((((((((Q(((((((((((((((((((( AllllB (((((((((((((((((((((((((((((((((((((((((ll The point charge q causes a potential difference between A and B but that potential difference must be different from that between the plates of the capacitors. (This chapter is before resistance and current in the textbook im using ). I always get stuck at something in almost every chapter. Help please 



#2
Mar1312, 09:28 AM

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hi iampaul!
it's like height … if there are three roads between A and B on a hill, then the roads will have different lengths, but they will all go up by the same height similary, the potential difference (the voltage drop) between any two points A and B is the same for any route between them (in series, the voltage drops are different because they're measured between different points ) the field caused by two (or more) objects is the sum of the individual fields similarly, electric potential is additive … in this case, you add the potential difference caused by the charge q to the potential difference caused by the emf in the circuit 



#3
Mar1412, 02:53 AM

P: 78

My first question is now cleared up but the second one still confuses me. Shouldn't the potential difference between A and B be different from that experienced by the capacitor because they are measured in different points? The potential difference between the plates of the capacitor is measured between the plates and it must be different from that between A and B but in circuit problems we use the potential difference between A and B for the capacitor.




#4
Mar1412, 03:09 AM

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potential difference of capacitor in circuit
hi iampaul!
(if it wasn't, we'd just use wires with less resistance! ) so yes, the potential difference is different, but the difference is far less than the accuracy of the measurements we make 



#5
Mar1412, 03:34 AM

P: 78

What i really mean is something like this:
What if we have a wire with end points A and B and that the potential difference between A and B is caused only by a point charge q beyond A. The Capacitor is placed along the wire at a point between A and B. Because the potential on a point in the wire depends on its distance from q, the potential difference between A and B due to the charge q is different from that between the plates of the capacitor which are separated only by a very small distance. If the wire is short then the potential difference b/w A and B is just about equal to that b/w the plates but if it is long then there should be a big difference. Sorry for my bad english 



#6
Mar1412, 04:08 AM

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but i've never come across an exam question that combines an electric circuit with a outside charge, and i don't think it happens intentionally in real circuits have you come across it? 



#7
Mar1412, 06:06 PM

P: 1,351

there would be an electric field, and charges would immediately move to cancel the field, and with it the potential difference. 



#8
Mar1412, 06:16 PM

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#9
Mar1412, 06:54 PM

P: 1,351

In an actual circuit, not separated in two by capacitors, charge will distribute, so the external field will get canceled in the conductors, and the external charge will only produce a potential difference in the circuit for a very short time. 



#10
Mar1412, 08:26 PM

P: 78

Thanks for the replies.
So if the potential difference between A and B and the potential difference between the plates of the capacitor are not necessarily equal, why is it that we use the potential difference between A and B for V in C=Q/V when V should be the Potential difference between the plates??? In my book capacitance comes after the chapter about electric potential and, Current, resistance , EMF and circuits aren't yet discussed, that's why i need to understand it with concepts only involving electric potential. 



#11
Mar1412, 11:22 PM

P: 78

To clarify what my problem really is, i will give another example:
Suppose that we have a wire with endpoints A and B, and the potential difference of these points is Vab. Lets take two other points c and d between A and B. Because potential varies with the distance r ( im just a beginner so i understand potential only in terms of V=q/4ε_{o}r, i think V_{ab} is not equal to V_{cd}. Now if c and d happens to be the plates of a capacitor with a given capacitance and we wish to compute for its charge, we only know V_{ab} but not V_{cd}. But in examples that i have seen V_{ab} is used as V_{cd} and i can't understand why. I really need some help with this. 



#12
Mar1512, 02:23 AM

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it makes no practical difference! 



#13
Mar1512, 02:51 AM

P: 1,351

If you have a single wire without a capacitor, the potential difference will be 0 after the wire reaches equilibrium. You should have learned already that with static electricity all parts of a conductor are always at the same potential. If you use a capacitor in the wire there will be a potential difference across the capacitor, but it will be much lower than you would expect from V=q/4ε_{o}r[/QUOTE] (lower for a bigger capacitor) The precise value will be much too hard to calculate at this point, involving a partial differential equation for charge distribution in the wire, to make the electric field in the wire 0 everywhere except between the plates of the capacitor. 



#14
Mar1612, 05:38 AM

P: 78

i think i need to read the whole chapter again, i forgot a lot of concepts. I tried finishing the chapter too fast, i remembered nothing. Thanks for your replies :)




#15
Mar1612, 05:58 AM

P: 78

I'll try to get back here if im still confused. Thanks again




#16
Mar1712, 03:02 AM

P: 78

Ok i think i get it
V_{ab}= V_{ac}+ V_{cd}+ V_{db} V_{ac}= V_{db}=0 because they are in the conductor so: V_{ab}=V_{cd} 



#17
Mar1712, 05:25 AM

P: 125

I had this same confusion, and it drove me nuts. I still don't know if I fully understand it, so someone correct me if I am wrong.
With a capacitor fully charged and no current is moving, the electric field for all electrons is a net zero. This is not to say that electrons are not bouncing all over the place, because they definitely are, at high velocities. But this is just random motion, so while there are corresponding net electric fields causing the random acceleration of each electron, it is not a net acceleration in one direction creating a flow. Or simply, momentum is conserved, and the over all field is said to be zero. There is a potential difference over the plates though because there are no longer any electrons with their own repulsive force to cancel out the ones on the surface. It is the surface itself, or the material that is holding them back, and thus allowing for a potential difference. So the way I see it, the surface limit is what forces the electric field inside to be zero (electrons can't leave/pass the surface), but in doing so, there is charge built up on that surface. If you were to disconnect the charging device and connect the two places, charge would distribute itself until the potential yet again, over the now single conductor, is the same. 


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