Electric breakdown in capacitor

[Erjr2]

I cant understand where i am going wrong here. I appreciate any help.

[studiot]

This is a more suitable size of file 300k instead of 3000k.

Greyscale also helps if you don't need the colout.

And please put it up the right way.

I have just spent the time I could have been reading it doing this for you.

I will be back later when I have read it.

[Erjr2]

Okay sorry, will keep that in mind. Thanks.

[Erjr2]

It is supposed to say 10^3 not 10^-3 at the end there.

[studiot]

Ok I have read through your working and here are my thoughts.

1) It is nice and - a pleasure to read, unlike some.

2) You have correctly deduced that for dielectrics havinf the same cross sectional area in series the electric field strengths (or potential gradients) are inversely proportional to their relative permittivities. Well done.

3) You have correctly used this to establish that the air layer will be the first to breakdown and subsequent working is correct to calculate that voltage.

However

2) You need to be carful with terminology: permeability refers to magentic effects. The correct electric term is permittivity.

3) Apart from the silly sign in the power of 10 I agree with your arithmetic so far as it goes and you have found the sign error yourself.

4) But the question was What is the maximum voltage that can be impressed and you have not addressed what would happen after breakdown of the air. Would the capacitor fail?

In fact I suggest that the air layer would fail (ie become ionised and of low resistance) and the capacitor would continue to function with only the porcelain dielectric acting.
Thus you can recalculate the voltage where the porcelain would fail by considering the air a short circuit.

 

Does this give the answer they seek?

[Erjr2]

Thank you very much for a good answer. And thanks for clearing up permeability and permittivity, guess I just got the two mixed up.

Yes indeed, I did not take into account that the capacitor could keep functioning even though the air has become ionised. 

Thanks again for great help ! 

[studiot]

A learning point here.

Your analysis route was quite complicated.

When they teach series and parallel circuits, using resistors the fact that in a series circuit the same current flows in all series elementrs is highlighted.

Capacitors do not have conductive current, but capacitors in series have the same charge. This is often overlooked.

But it can be used to develop simple and useful relationships.

[Erjr2]

Yeah I agree, it felt like much work for so little. I'll be sure to keep that in mind the next time.

Thanks again for great answers.