Insight needed on Di-electric breakdown

[unam]

Hello,
Consider the following:
case 1: 2 parallel metal plates(sufficiently thick) separated by a di-electric and 1 plate is charged to a voltage exactly equal to break down of the di-electric. the other plate is tied to ground.
case 2: same as case 1 but instead of the plate tied to ground, it is tied to ground through a reasonably high resistance.
[let us not get into the soft-breakdown like topics and let us think of breakdown as when arc'ing happens]

1. Is there any difference in di-electric breakdown between case 1 and case2?
2. Does the di-electric breakdown even depend on the material properties of the plates? (ie higher R or lower R to ground matter even?) -- my thought is that the electrons due to high field start vibrating and acquire enough KE and you get enough electrons in the conduction band (you also have ionization etc, etc) so it is a di-electric property and not an electrical property of the plates (high or low resistance)?
3. if the voltage on plate is gradually increased in both case 1 and case 2 till you get di-electric breakdown, will the di-electric breakdown happen at the same instant in both cases? will case 2 need slightly higher voltage for break-down?

Thanks in advance

 

[NFuller]

Is there any difference in di-electric breakdown between case 1 and case2?

Practically speaking, no. As the capacitor in case 2 charges, the current through the resistor drops and thus the voltage change across the resistor drops. Eventually, the resistor will not shed any voltage and the capacitor is effectively tied to ground.

Does the di-electric breakdown even depend on the material properties of the plates?

No, it depends on the properties of the dielectric.

if the voltage on plate is gradually increased in both case 1 and case 2 till you get di-electric breakdown, will the di-electric breakdown happen at the same instant in both cases? will case 2 need slightly higher voltage for break-down?

It may take a little bit longer for case 2, since the resistor slows the rate at which the capacitor charges. In the steady state situation however, both capacitors will fail at the same voltage.

 

[unam]

Practically speaking, no. As the capacitor in case 2 charges, the current through the resistor drops and thus the voltage change across the resistor drops. Eventually, the resistor will not shed any voltage and the capacitor is effectively tied to ground.

No, it depends on the properties of the dielectric.

It may take a little bit longer for case 2, since the resistor slows the rate at which the capacitor charges. In the steady state situation however, both capacitors will fail at the same voltage.

Thanks NFuller for your answer. It does validate my thoughts (I wanted somebody better than I to tell me that I was not wrong).

Couple more points, See If you agree :

--- When I asked does the di-electric breakdown on the material properties of the metal plates? I was referring to Resistance of the plate in particular. Conceptually the external resistance to ground in case 2 is equivalent to the a metal plate (with inherently higher resistance) connected to ground. So , it may take a bit longer to breakdown due to the high Resistance but it will still breakdown at the same voltage.

---- Taking the above argument a bit further, in case 2 the Resistance in series with the metal plate approaches infinity (ie it is not connected to ground) but there is an air gap to ground (say). Then too, the breakdown voltage should be the same? or would it be different? (ie. have plate 1 tied to voltage source, plate 2 is not tied to anything but is disconnected from the ground ... you could have an air gap to ground say...). Then,

--- Arg1: you have a capacitive divider and so the field/voltage needed for di-electric breakdown would be larger in case 2
--- Arg2: whether you have the 2nd plate connected to ground or not, breakdown voltage is the same as case 1 as the breakdown voltage is solely dependent on the material properties of the di-electric.
---Arg3: since the 2nd plate is floating when the voltage on the 1st plate increases, the 2nd plate's potential also raises (concept of boot-strapping) such that the net potential difference is 0. In this case there is no di-electric breakdown?

Would be interested in how/what you think about this and how you would address the above arguments. Again, many thanks to you for the answer

 

[NFuller]

So , it may take a bit longer to breakdown due to the high Resistance but it will still breakdown at the same voltage.

Yes

Then too, the breakdown voltage should be the same? or would it be different? (ie. have plate 1 tied to voltage source, plate 2 is not tied to anything but is disconnected from the ground ... you could have an air gap to ground say...)

This is more subtle. You now have one side of the capacitor floating and the potential it feels will be determined by what's around it, i.e. a metal enclosure if its inside some device. If there is nothing around the capacitor such that it is in empty space, then the potential at the other plate is a linear function of the separation between the plates.

--- Arg1: you have a capacitive divider and so the field/voltage needed for di-electric breakdown would be larger in case 2
--- Arg2: whether you have the 2nd plate connected to ground or not, breakdown voltage is the same as case 1 as the breakdown voltage is solely dependent on the material properties of the di-electric.
---Arg3: since the 2nd plate is floating when the voltage on the 1st plate increases, the 2nd plate's potential also raises (concept of boot-strapping) such that the net potential difference is 0. In this case there is no di-electric breakdown?

The issue is that these are technically two different problems because of how the boundary conditions are being specified for the electric potential. For the case of a grounded plate the potential will go to zero at that plate while for the isolated floating plate the potential goes to zero at infinity.

 

[unam]

Yes

This is more subtle. You now have one side of the capacitor floating and the potential it feels will be determined by what's around it, i.e. a metal enclosure if its inside some device. If there is nothing around the capacitor such that it is in empty space, then the potential at the other plate is a linear function of the separation between the plates.

Thanks again NFuller-san. Pl. pardon my lack of understanding. I did not quite follow your comment that the potential at the other plate (interpreted as floating plate) a linear function of the separation between the plates. Perhaps I am mis-interpreting your comments? (in which case pl. forgive me). The following is the way I am thinking. (would appreciate if you can find the flaw in my argument).

Let us consider the case that there is no metal enclosure and the 2nd plate is some distance away from the ground (and hence floating). I think that that there is NO potential difference between the plates ie the floating plate follows the potential of the 1st plate being charged as no energy is spent or stored. This is similar to a spring where one end is being pulled. If the other end is fixed then work is done and energy is stored in the spring; if the other end of spring is free then no energy is spent and no energy is stored in the spring (neglect friction etc, etc, ). So if my statement is true, the 2nd plate is at the same potential as the 1st plate (potential difference between them is zero). In this case what is the breakdown voltage? is it the same or lower or higher than case 1 (where the 2nd plate was tied to ground)

The issue is that these are technically two different problems because of how the boundary conditions are being specified for the electric potential. For the case of a grounded plate the potential will go to zero at that plate while for the isolated floating plate the potential goes to zero at infinity.

 

[NFuller]

Let us consider the case that there is no metal enclosure and the 2nd plate is some distance away from the ground (and hence floating). I think that that there is NO potential difference between the plates ie the floating plate follows the potential of the 1st plate being charged as no energy is spent or stored. This is similar to a spring where one end is being pulled. If the other end is fixed then work is done and energy is stored in the spring; if the other end of spring is free then no energy is spent and no energy is stored in the spring (neglect friction etc, etc, ). So if my statement is true, the 2nd plate is at the same potential as the 1st plate (potential difference between them is zero). In this case what is the breakdown voltage? is it the same or lower or higher than case 1 (where the 2nd plate was tied to ground)

There is a potential difference because the potential is not constant throughout space. Their is no analogy here between a capacitor and a spring. It would take work to put charge on the plate even if the other plate did not exist. This thread is marked as an A level thread meaning that you have taken upper level courses on the subject. Are you familiar with solving boundary value problems?