Theoretical doubt regarding capacitance

  • #1
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I understand that the dielectric constant of a perfect isolated charged conductor is infinite when its placed in an electric field (generally uniform)... Now I also know that when a conductor is placed between a parallel plate capacitor such that the distance between two plates in those capacitors is completely filled by the conductor, the capacitance of that capacitor becomes infinite, because capacitance of a capacitor when a dielectric medium is placed between it is given by :

C=εoA/(d-t+t/K)

Where, εo is permittivity of free space, A is area of the plate of the capacitor, d is distance between two plates of the capacitor, t is thickness of the dielectric medium placed between the capacitor, K is dielectric constant of the medium.

Now if conductor fully occupies, K=∞ and t=d

Hence we get, C=∞.

Now here are my doubt:

Capacitance is defined as the tendency of conductor to "hold" the charge so that it can be accumulated. Now a conductor has no tendency to "hold" charge.(I am taking the case as above , a conductor occupies all space between capacitor. Hence there can be no induction.) It just passes over the charge across it. Hence logically its capacitance should be 0, but its infinite. Why ? :confused:

Same case here:

Suppose I connect a capacitor across a battery. If we observe the part of a wire which is resistance free and there is no other thing, what will be its capacitance across the parts ? Zero or infinite ? Also what will be the potential difference across the part of wire simply ?

Please help !!

Thanks in advance.... :smile:
 

Answers and Replies

  • #2
NascentOxygen
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If the plates of a capacitor are shorted together, you no longer have a capacitor.
 
  • #3
tiny-tim
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hi sankalpmittal! :smile:
Capacitance is defined as the tendency of conductor to "hold" the charge so that it can be accumulated. Now a conductor has no tendency to "hold" charge. (I am taking the case as above , a conductor occupies all space between capacitor. Hence there can be no induction.) It just passes over the charge across it. Hence logically its capacitance should be 0, but its infinite. Why ? :confused:
no, it should be Q/V, which is 0/0, which could be anything :wink:

also, Q will only be zero if you make the mistake of allowing the charge on the plate to move onto the conductor!

if you avoid this by placing charges ±Q on either side of a conductor, but so that the charges are "glued" so that they cannot move onto the conductor, then the charges on the conductor will rearrange themselves so as to cancel out the electric field, ie to make V = 0 …

so the capacitance of a conductor is Q/V = Q/0 = ∞ :smile:
 
  • #4
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If the plates of a capacitor are shorted together, you no longer have a capacitor.
Yes, but we can still define capacitance of a non-capacitor. For example: Consider one plate instead of two. It will be next to no capacitor, if we assume other plate to be infinitely far away, but we can still define its capacitance.

hi sankalpmittal! :smile:


no, it should be Q/V, which is 0/0, which could be anything :wink:
:confused:... From above, you mean that potential difference across a conductor is zero. How ? Ahh, E-field inside a conductor is zero and so is potential difference. But that is only the case in an external E-field. :confused: E-field means electric field. Also, when there is no potential difference across conductor, how do you expect the charge to flow across it ?? :confused:

Ahh..Ok !! So you mean that in this case it will not behave as a capacitor, and we judge potential difference across it as work done to move a unit charge across it, but here we do no work against the field ??? Right? But when p.d zero, there is no flow of charge.

also, Q will only be zero if you make the mistake of allowing the charge on the plate to move onto the conductor!
Means ??? :confused: What do you mean by "onto the conductor" ?

if you avoid this by placing charges ±Q on either side of a conductor, but so that the charges are "glued" so that they cannot move onto the conductor, then the charges on the conductor will rearrange themselves so as to cancel out the electric field, ie to make V = 0 …

so the capacitance of a conductor is Q/V = Q/0 = ∞ :smile:
Yes, that is Ok with me... But I am confused. First you said its 0/0, then 0 capacitance... :confused: Although this last case I do understand.
 
  • #5
tiny-tim
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:confused:... From above, you mean that potential difference across a conductor is zero. How ? Ahh, E-field inside a conductor is zero and so is potential difference. But that is only the case in an external E-field. :confused: E-field means electric field.
in equilibrium, the electric field inside a conductor is always zero :wink:
Also, when there is no potential difference across conductor, how do you expect the charge to flow across it ?? :confused:
i don't … when do we ever expect charge to flow across a capacitor? :wink:
Ahh..Ok !! So you mean that in this case it will not behave as a capacitor, and we judge potential difference across it as work done to move a unit charge across it, but here we do no work against the field ??? Right? But when p.d zero, there is no flow of charge.
no, i am treating it as a capacitor, so
i] i'm not letting the charge from the plate move onto the conductor (see below)
but
ii] i'm fixing Q and measuring V, instead of (as usual) fixing V and measuring Q
Means ??? :confused: What do you mean by "onto the conductor" ?
i mean that electrons are as usual free to move around the conductor, but that the electrons on the plate cannot move onto the conductor (or vice versa), they're stuck on the plate :smile:
Yes, that is Ok with me... But I am confused. First you said its 0/0, then 0 capacitance... :confused: Although this last case I do understand.
no, i said it's 0/0, which has no meaning

then i said it's ∞ (not 0) :wink:
 
  • #6
jtbell
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i mean that electrons are as usual free to move around the conductor, but that the electrons on the plate cannot move onto the conductor (or vice versa), they're stuck on the plate :smile:
Perhaps one might imagine infinitesimally thin insulators between the conductor and the capacitor plates. Or would the dielectric properties of the insulators mess up the analysis, even though they're infinitesimally thin?
 
  • #7
sophiecentaur
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Yes, but we can still define capacitance of a non-capacitor. For example: Consider one plate instead of two. It will be next to no capacitor, if we assume other plate to be infinitely far away, but we can still define its capacitance.



:confused:... From above, you mean that potential difference across a conductor is zero. How ? Ahh, E-field inside a conductor is zero and so is potential difference. But that is only the case in an external E-field. :confused: E-field means electric field. Also, when there is no potential difference across conductor, how do you expect the charge to flow across it ?? :confused:

Ahh..Ok !! So you mean that in this case it will not behave as a capacitor, and we judge potential difference across it as work done to move a unit charge across it, but here we do no work against the field ??? Right? But when p.d zero, there is no flow of charge.



Means ??? :confused: What do you mean by "onto the conductor" ?



Yes, that is Ok with me... But I am confused. First you said its 0/0, then 0 capacitance... :confused: Although this last case I do understand.
Are you confusing 'zero capacitance' with 'infinite capacitance' (- which is what you have between the slices in a conductor)?
 
  • #8
785
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Are you confusing 'zero capacitance' with 'infinite capacitance' (- which is what you have between the slices in a conductor)?
No thanks, I got it.

One more quick question:

Can I use the concept of Wheatstone Bridge in circuits involving capacitors ? Also, can I use Kirchoff's junction and voltage law in circuits only of capacitors ? I think I can but I am not sure. :(
 
  • #9
sophiecentaur
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The Wheatstone bridge (and others) works just as well for complex Impedances as with simple resistances. It was the only way to measure components when I was a lad.
 

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