# Max values for a 2-dielectric capacitor

• HelloCthulhu
In summary, a parallel plate capacitor with an area of 1 cm2 and a plate separation of 0.01m has a maximum capacitance of 0.635895pF when filled with water and a plastic bag with a thickness of 0.003m. The maximum voltage for the capacitor is 0.001392V and the maximum charge is 3.370243nC.
HelloCthulhu

## Homework Statement

A parallel plate capacitor has area A = 1 cm2 and a plate separation of d = 0.01m. Water at room temp (20°C) is poured into a ziplock bag and placed between the plates filling the volume of 1cm3. Find the maximum capacitance, voltage and charge for the capacitor.

## Homework Equations

Q=CV

ziplock:
bag thickness = 1.5mm x 2 (touches both sides of the capacitor = 3mm = 0.003m

dielectric constant = 2.3

dielectric breakdown = 25MV/m

water:
thickness = 0.007m

dielectric constant = 80.1

dielectric breakdown = 65MV/m

Capacitance for two dielectrics in series:
$$C=\frac{\varepsilon_0A}{\frac{d_1}{k_1}\frac{d_2}{k_2}}$$

Voltage for two dielectrics in series:
$${V_r} = {E_{r1}}{d_1} + {E_{r2}}{d_2}$$

## The Attempt at a Solution

Maximum capacitance:
$$\frac{(8.85\times10^{-12})\times0.0001}{\frac{0.003}{2.3}+\frac{0.007}{80.1}}=6.35895\times10^{-13}F=0.635895nF=$$Maximum voltage:
$$(65\times10^6V\times0.007m)+(25\times10^6V\times0.003m) =(45.5\times10^3V)+(75\times10^3V) = 455kV+75kV=530kV$$Maximum charge:
$$(6.35895\times10^{-13}F)\times(53\times10^{4}V)=(3.370243\times10^{-7}C)=337.0243nC$$

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Check your metric prefix value for your capacitor result. a nanofarad is ##10^{-9}\;F##, while a picofarad is ##10^{-12}\;F## .
Can you explain the reasoning behind your calculation of the maximum voltage for the capacitor? In practice, can the individual sections of the capacitor attain their maximum potential difference independently?

Yikes! Thank you so much for catching that! It was supposed to be picofarads.

Maximum capacitance:Maximum capacitance:
$$\frac{(8.85\times10^{-12})\times0.0001}{\frac{0.003}{2.3}+\frac{0.007}{80.1}}=6.35895\times10^{-13}F=0.635895pF=$$

Here's a link to a thread on maximum voltage for a capacitor with two dielectrics. I'm using the same formula, but I'm not positive if it's correct:

If you analyze the overall capacitor as two capacitors in series (combining the two plastic layers into one for convenience -- this won't affect the net capacitance or breakdown characteristics), then when some voltage is applied across the capacitor the two component capacitors will develop potential differences that depend upon their capacitance values. The smaller capacitance will take a larger share of the total potential difference.

The question then is, which of the two capacitors will break down first as the total potential difference is increased?

1.5mm strikes me as too thick but perhaps it was given in the problem statement.

Not that it really matter for this problem but I've just put a micrometer on a similar plastic bag and two layers were about 0.4mm thick in total.

HelloCthulhu
gneill said:
The question then is, which of the two capacitors will break down first as the total potential difference is increased?

Intuitively, I would think the dielectric closest to the charged plate would break down first. But I'm not sure how to prove this mathematically. Did you take a look at the thread I posted regarding two dielectric capacitor voltage? Derivations were included, but I'm not positive if they're correct.

HelloCthulhu said:
Intuitively, I would think the dielectric closest to the charged plate would break down first. But I'm not sure how to prove this mathematically.
The potential difference developed across the individual dielectrics is straightforward to calculate; it's just the potentials on the two capacitors of the equivalent model. You should have seen the capacitor potential divider calculations before.
Did you take a look at the thread I posted regarding two dielectric capacitor voltage? Derivations were included, but I'm not positive if they're correct.
Yes, but they weren't quite going for the same goal as you. They were looking for the maximum field strength for the second dielectric such that it would break down at the same time as the first dielectric. You're looking for the maximum potential (voltage) across the whole capacitor before one or both of its dielectrics fail.

HelloCthulhu
gneill said:
The potential difference developed across the individual dielectrics is straightforward to calculate; it's just the potentials on the two capacitors of the equivalent model. You should have seen the capacitor potential divider calculations before.
...You're looking for the maximum potential (voltage) across the whole capacitor before one or both of its dielectrics fail.

Thank you so much for helping me with this. I admit, I still don't fully understand how to use two capacitors in series as an equivalent model for the maximum voltage. But I just realized using Q=CV, I should be to able use the same equation I used to find the maximum capacitance to find the maximum voltage. At least I hope so:

Maximum capacitance:
$$\frac{(8.85\times10^{-12})\times0.0001}{\frac{0.003}{2.3}+\frac{0.007}{80.1}}=6.35895\times10^{-13}F=0.635895pF$$

Maximum voltage:
$$\frac{(8.85\times10^{-12})\times0.0001}{6.35895\times10^{-13}F}=(\frac{0.003}{2.3}+\frac{0.700}{80.1})=0.001392V$$

No, the answer should be at least in the kilovolt range.

When you have a dielectric in a capacitor, when a potential is applied across the outer plates it establishes an electric field between them, passing though the dielectric. The charges in the material of the dielectric will respond to this field by an induced charge separation to produce a field internally that opposes the external field (in an insulating dielectric the charges are constrained to move only by very tiny amounts, staying "pinned" to their atoms). The net effect is an effective charge separation that makes the dielectric "look like" a capacitor itself.

If you have several dielectrics stacked in a capacitor you can model the assembly as a set of capacitors connected in series.

Now, if you place a potential difference ##E## across the capacitor, that potential difference will be split across the two dielectrics according to the capacitor voltage divider principle.

You need to ensure that these two potentials, ##V_1## and ##V_2##, do not exceed the breakdown voltage of their respective dielectrics.

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I'm still pretty lost. Is 0.635895pF the maximum capacitance? I learned how to calculate it watching this video:

HelloCthulhu said:
I'm still pretty lost. Is 0.635895pF the maximum capacitance?
While the two dielectrics are intact (no breakdown occurs), there will be just one total capacitance value. The term "maximum capacitance" doesn't apply here. The value that you have calculated is the capacitance of the capacitor with the dielectrics in place.

You should round your result to an appropriate number of significant figures, since the given values are specified with much fewer than you're showing

HelloCthulhu
gneill said:
While the two dielectrics are intact (no breakdown occurs), there will be just one total capacitance value. The term "maximum capacitance" doesn't apply here. The value that you have calculated is the capacitance of the capacitor with the dielectrics in place.

You should round your result to an appropriate number of significant figures, since the given values are specified with much fewer than you're showing

Ok the part about the capacitance makes sense to me. I apologize for the sig figs. I'm still learning how to use those properly . But I'm still confused about how to solve for maximum voltage. Can you break down what it'll look like mathematically for two dielectrics? The only thing that makes sense to me is to add the two max voltages together, but you've already told me that's wrong. I'm hoping I at least know how to find max values for 1 dielectric. Your help is greatly appreciated because I'm completely lost.

Max voltage for plastic:

V = Ed

$$(25\times10^6V)\times0.003m=75\times10^3V=75kV$$

Max voltage for water:

$$(65\times10^6V)\times0.007m=455\times10^3V= 455kV$$

Capacitance for plastic:

$$C=K\varepsilon_0\frac{A}{d}$$

$$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{-12})\times2.3\times0.0001}{0.003}=6.785\times10^{-13}=0.679pF$$

Capacitance for water:

$$\frac{(8.85\times10^{-12})\times80.1\times0.0001}{0.007}=1.012693\times10^{-11}F=10.12693pF=10.127pF$$

Charge for plastic:

Q=CV

$$(75\times10^3V)\times(1.012693\times10^{-11}) =7.696197\times10^{-7}C=769.6197nC=769.620nC$$

Charge for water:

$$(455\times10^3V)\times(6.785\times10^{-13})=3.091725\times10^{-7}=309.1725nC=309.2nC$$

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Imagine that there is some potential difference ##E## impressed across the dielectric-filled capacitor. That total potential difference will be split across the two capacitors of the equivalent model (two series-connected capacitors). The split of the potential will depend upon the individual capacitance of the capacitors in the equivalent model. Each will have a breakdown voltage that depends upon the dielectric associated with it. If the potential across either capacitor exceeds the breakdown voltage of that capacitor, the overall device fails.

So you need to establish how the potential splits across the two capacitances, then find the maximum applied potential difference such that neither of the breakdown voltages of the two component capacitors are exceeded.

You could formulate the problem working from the maximum voltage of each capacitor (your 75 KV and 455 KV), determining what the corresponding potential difference would be across its compatriot component to make sure it's within its safe operating range, and if safe determine the overall potential difference across the device. (for a given potential difference across a capacitor there is a certain charge associated with it, and for series connected capacitors the charges are equal).

The bottom line is that you have effectively two capacitors in series, and if you place some potential difference across the pair the potential will be split across them according to the capacitor potential divider relationship. If either potential exceeds the breakdown potential for its capacitor, the device fails.

gneill said:
Imagine that there is some potential difference ##E## impressed across the dielectric-filled capacitor. That total potential difference will be split across the two capacitors of the equivalent model (two series-connected capacitors). The split of the potential will depend upon the individual capacitance of the capacitors in the equivalent model.

For the series capacitor equation, I'm following the example in this video:

$$\frac{1}{0.679pF}+\frac{1}{10.127pF}=10.806pF=0.092541pf=0.093pF$$

But I don't understand how this can help me solve for the maximum voltage.

Each will have a breakdown voltage that depends upon the dielectric associated with it. If the potential across either capacitor exceeds the breakdown voltage of that capacitor, the overall device fails.

Would that mean that the lowest breakdown voltage is the maximum voltage?

So you need to establish how the potential splits across the two capacitances, then find the maximum applied potential difference such that neither of the breakdown voltages of the two component capacitors are exceeded.

You could formulate the problem working from the maximum voltage of each capacitor (your 75 KV and 455 KV), determining what the corresponding potential difference would be across its compatriot component to make sure it's within its safe operating range, and if safe determine the overall potential difference across the device. (for a given potential difference across a capacitor there is a certain charge associated with it, and for series connected capacitors the charges are equal).

I still don't understand how to use 75kV and 455kV to solve this. Do I add them and divide by 2? Or subtract 75kV from 455kV. I'm probably missing something very simple. Could you show me what the equation is supposed to look like?

HelloCthulhu said:
Would that mean that the lowest breakdown voltage is the maximum voltage?
It means that the breakdown voltage that is exceeded first with increasing overall potential difference for the capacitor determines the maximum voltage, where the overall potential difference for that situation is then the maximum voltage for the device.. The overall potential difference is split across the component capacitors, and each one has a potential difference limit beyond which it will fail. As the overall potential difference is increased, eventually one of them will reach its breakdown limit.

HelloCthulhu said:
I still don't understand how to use 75kV and 455kV to solve this. Do I add them and divide by 2? Or subtract 75kV from 455kV. I'm probably missing something very simple. Could you show me what the equation is supposed to look like?
Those numbers are the limits of the individual capacitors making up the equivalent model of the device. What must be the potential across the device to exceed one or the other of those potential differences?

I think that you need to investigate the capacitor voltage divider and how the potential is split across the capacitors making up the divider.

I found a physicsforums thread that I might be able to use:

https://www.physicsforums.com/threa...terials-to-increase-breakdown-voltage.902113/

Baluncore said:
Given; Barium titanate, dielectric constant = 1200, dielectric strength = 1.2 Mv/m. Mica, dielectric constant = 3, dielectric strength = 118 Mv/m. We can hypothesise the layer capacitors as;
C1 = Barium titanate. A 1 mm layer gives a voltage rating of V1 = 1.2 kV, with say C1 = 1200 pF.
C2 = Mica. A 1 mm layer gives a voltage rating of V2 = 118 kV, with say relative C2 = 3 pF.
C3 = C1.

We know that voltage will be shared in inverse proportion to capacitance, so; V1 / V2 = C2 / C1. If we operate C2 at the breakdown voltage V2, the voltage across C1 will then be a trial value of V1;
V1 = V2 * C2 / C1. So; V1 = 118 kV * 3 / 1200 = 0.295 kV which is OK as it is below the maximum 1.2 kV calculated earlier.

To survive a single step turn-on transient, the answer to your question is that;
1. The breakdown voltage of the combined sandwich is 0.295 kV + 118 kV + 0.295 kV = 118.59 kV. But since it is now 3 mm thick, the effective dielectric strength has become 39.53 kV/mm = 39.53MV/m.
2. The capacitance of the series sandwich is Ct = 1/ (2/C1 + 1/C2 ) = 2.985 pF. But since it is now 3 mm thick, that makes the equivalent dielectric constant = 8.955

If I add the two max voltages and divide by the total thickness, I should get the dielectric strength of the capacitor:
$$\frac{(455kV+75kV)}{0.01m}=53kV$$

Combining the individual breakdown voltages in any way to find some net maximum voltage doesn't make sense. They are independent limits.

What you might do is find a relationship between the two potential differences on the capacitors when its in operation and see if when one just reaches its breakdown point the other is still safe or has already exceeded its own breakdown threshold. For two capacitors in series, how does the total potential divide between them?

Then use the established first breakdown point to determine the overall potential difference across the series connection (capacitor voltage divider equation).

HelloCthulhu
gneill said:
What you might do is find a relationship between the two potential differences on the capacitors when its in operation and see if when one just reaches its breakdown point the other is still safe or has already exceeded its own breakdown threshold. For two capacitors in series, how does the total potential divide between them?

Then use the established first breakdown point to determine the overall potential difference across the series connection (capacitor voltage divider equation).

Alright I think I'm finally understanding what you're referring to. I'm using a formula I found on this link to solve for the voltage drop across the plastic dielectric:
https://www.electronics-tutorials.ws/capacitor/cap_7.html

$$C_T\frac{C_1\times C_2}{C_1+C_2}$$

$$C_T=\frac{(6.785\times10^{-13}F)\times (1.012693\times10^{-11}F)}{(6.785\times10^{-13}F)+(1.012693\times10^{-11}F)}=\frac{0.000000000000000000000006871122005F}{0.00000000001080543F}=0.635895286pF=0.636pF$$

$$V_{C_{1}}=\frac{C_T}{C_1}\times V_T$$

I modified this equation to solve for the max voltage across the plastic's breakdown voltage(bd):
$$V_{bd}=\frac{C_T}{C_1}\times V{max}$$

$$V{max}=\frac{V_{bd}}{\frac{C_T}{C_1}}$$

plastic:
$$V_{c1}max=\frac{7.5\times10^{6}}{\frac{0.635895286pF }{0.678pF}}=\frac{7.5\times10^{6}}{0.9378986519174041}=79965.9961624562192461V=80kV$$

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Okay, that works

Another approach is to write KVL for the series connected capacitors that comprise the device. Suppose that some applied current has deposited a charge Q on the device, then:

##\frac{Q}{C_{cap}} = \frac{Q}{C_1} + \frac{Q}{C_2}##

because the the charge of series-connected capacitors is the same.

That's just a statement of the voltage across the assembled capacitor being the sum of the voltages across the series-connected components comprising it (Remember that for a capacitor, Q = CV so that V = Q/C).

Then you can examine one of capacitor terms on the RHS and find out what value of Q corresponds to its breakdown voltage which you calculated earlier. Check to see if that value of Q will cause the other term to exceed its breakdown threshold or not. That will tell you which capacitor will breakdown first, and tell you what the overall device's potential difference will be when it happens (since the potential difference across the device is the LHS of the expression, so that ##\frac{Q}{C_{cap}}##).

HelloCthulhu
I can't thank you enough for helping me out with this gneil! You're the best!

## 1. What is a 2-dielectric capacitor?

A 2-dielectric capacitor is a type of capacitor that consists of two layers of insulating material, or dielectrics, with a conductive material, or electrode, in between. This structure allows for the storage of electrical energy.

## 2. What is the purpose of a 2-dielectric capacitor?

The purpose of a 2-dielectric capacitor is to store and release electrical energy in a circuit. It can be used in a variety of applications, such as power supplies, filters, and timing circuits.

## 3. What are the factors that determine the max values for a 2-dielectric capacitor?

The max values for a 2-dielectric capacitor are determined by the dielectric materials used, the distance between the electrodes, and the surface area of the electrodes. Other factors such as temperature, voltage, and frequency can also affect the max values.

## 4. What are the advantages of using a 2-dielectric capacitor?

One advantage of using a 2-dielectric capacitor is that it can have a higher capacitance value compared to a single-dielectric capacitor. It also has a lower risk of breakdown due to the presence of two insulating layers.

## 5. How are the max values for a 2-dielectric capacitor calculated?

The max values for a 2-dielectric capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric materials, A is the surface area of the electrodes, and d is the distance between the electrodes. This formula can be applied to both parallel plate and cylindrical capacitors.

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