# What is breakdown voltage?

In my introductory text (Resnick and Halliday, page number 732), it is written that

Another effect of the introduction of a dielectric is to limit the potential difference that can be applied between the plates to a certain value $$V_{{\rm max}}$$.

I am not able to understand what exactly the text is trying to say.

• But, what are your efforts ? The site is to ask questions which have to do something with concepts, not for "What does this mean".
– user363737
Jul 2, 2023 at 14:27
• "Certain"? With careful engineering and fabrication it can be made predictable... Jul 2, 2023 at 14:34
• Sir, do you mean that after inserting a specific dielectric, $V$ would surely decrease but there exists a special $V$ when it's value is maximum without breaking the dielectric, but still less than $V$ with air in between ? Jul 2, 2023 at 14:46
• In the context of real-world physics I would call the book wrong. In the theoretical context where you start with vacuum between the plates don't have significant field emission, the book is correct. This is rarely encountered in reality. Jul 2, 2023 at 15:22
• @JohnDoty Sir can you please write an answer to this question which consolidates this whole conversation so that i can note it down. Jul 2, 2023 at 15:47

## 2 Answers

There is no material you could use as a dielectric between the plates of a capacitor which will have a higher breakdown voltage than a vacuum. Therefore, the maximum voltage you can place across a capacitor will be the breakdown voltage of its dielectric and this "Vmax" will always be lower than the Vmax for the vacuum case. I think this is what they are referring to.

BTW I used an earlier edition of Halliday & Resnick as an undergrad 50 years ago. It was a pretty good textbook.

So, what is "breakdown voltage"? It is the voltage at which an insulator "breaks down" and instead becomes a conductor. What happens is this: in an insulator, the electrons that form the bonds between atoms are immobile: they are locked in place and are not free to move under the influence of an applied potential difference they way they can in the case of a conductor.

But there will be some (high enough) voltage where they are being pulled on hard enough that they start to move through the insulator anyway. This means there are now ions within the insulator which not only conduct current but also chemically attack and break up other molecules of the insulator.

Now you have a growing number of conductors inside the insulator, which is being degraded- and the high voltage you applied drives a sudden huge burst of current through the insulator at its most weakened spot. A spark results inside the insulator which either blows the capacitor apart or sets it on fire, or both.

• Have you ever worked with HV in vacuum? In practice, Vmax due to field emission may be lower than dielectric breakdown in the insulators. Jul 2, 2023 at 18:11
• no, I haven't. would not that characteristic depend sensitively on the surface condition of the plates? let me klnow if I should edit. -NN Jul 2, 2023 at 20:32

A material is made of a set of positive charges and negative charges bound together. Consequently, there are always dipoles, but the effective dipole moment can vanish because of the symmetry of the material. If an electric field $$\textbf{E}$$ is applied, the dipole moments $$\textbf{p}_i$$ will tend to be aligned for minimizing $$\Delta E = \sum_i \textbf{p}_i.\textbf{E}$$ depending on the material. If the field is large enough it is possible for electrons to move in the material, this situation is similar to a polarization catastrophe making the transition from an insulating state to a conductive state.

A breakdown voltage is the voltage generating an electric field large enough for a specific dielectric leading to a conductive state.