What is the breakdown voltage for this two dielectric capacitor (BaTiO3,LPDE)?

[HelloCthulhu]

Homework Statement

A copper parallel plate capacitor has an area of 0.049m. A 1mm thick low-density polyethylene (LDPE) film is placed in between two 3.5mm thick BaTiO3 plates and placed inside the parallel plate capacitor. Both dielectrics have an area of 0.049m. Find the maximum capacitance, voltage and charge for the capacitor. If this two dielectric capacitor were to exceed its maximum voltage, which dielectric will breakdown first?

Relevant Equations

Q=CV

V = Ed

electric constant = $ \epsilon_{0}=8.854\times 10^{-12}$

Capacitance:
$$C=K\epsilon_{0}\frac{A}{d}$$

Capacitance for 2 capacitors in series:
$$C_{T}=\frac{C_{1}\times C_{2}}{C_{1}+C_{2}}$$

Breakdown voltage the capacitor:
$$ V_{max}=\frac{V_{bd}}{\frac{C_{T}}{C_{1}}}$$

BaTiO3
thickness = 7mm = 0.007m
dielectric constant = K = 1260
dielectric breakdown = E = 97MV/m (97kv/mm)

(Structural, Dielectric and Ferroelectric Properties of Tungsten Substituted Barium Titanate Ceramics, 2009)

LDPE
thickness = d = 0.007m
dielectric constant = K = 2.3
dielectric breakdown = 3kV/m (300V/mm)

https://www.azom.com/article.aspx?ArticleID=428

In a two dielectric capacitor, if one dielectric breaks down the capacitor fails. To solve for the voltage across each capacitor in series, first find the capacitance:

Max voltage for BaTiO3:
(97e6 MV)× 0.007m = 679000 = 680kV

Capacitance:
$$C=\frac{(8.85\times 10^{-12})\times1200\times0.049}{0.007}=0.00000007434F=7.434\times10^{-10} F=74340pF$$

Charge for BaTiO3:
(680e3V)*(0.00000007434e-10F) = 5.05512e-12C = 5.05512e-6μC

Max voltage for Low Density Polyethylene - LDPE:
(27e6V)×0.001m = 27000V = 27kV

Capacitance for LDPE:
$$ C=\frac{(8.85\times 10^{-12})\times2.3\times0.049}{0.007}=0.000000000997395F=9.97395\times10^{-10}F=997.395 pF$$

Charge for LDPE:
(27e3V)×( 9.97395e-10F) = 2.6929665e-5C = 26.929665μC

Total capacitance:
$$C_{T}=\frac{(7.434\times10^{-10})\times(9.97395\times10^{-10})}{(7.434\times10^{-10})+(9.97395\times10^{-10})}=9.84190445\times10^{-11}F=984.190445pF$$

BaTiO3:
$$V_{max}=\frac{680\times10^{3}}{\frac{984.190445pF}{74,340pF}}=\frac{680\times10^{3}}{0.0132390428436911}=51363229.8V=51MV$$

LDPE:
$$V_{max}=\frac{460\times10^{3}V}{\frac{984.190445pF}{997.395 pF}}=\frac{460\times10^{3}V}{0.986760957293750}=466171.6491V=466kV$$

 

[kuruman]

I assume that the plate area is 0.049 m² not 0.049 m.
I am not sure why you say you have 2 capacitors in series when you have 3. You have a sandwich in which the "bread" is BaTiO3 and the "filling" is the plastic.
I disagree with your value of the total capacitance. For trouble shooting your solution, it would be easier if you found an algebraic expression for the equivalent capacitance starting from the well known expression $$\frac{1}{C_{eq}}=\frac{1}{C_B}+\frac{1}{C_L}+\frac{1}{C_B}$$ where subscript B stands for BaTiO3 and L for LDPE. Solve for $C_{eq}$ and then substitute the numbers.

 

[HelloCthulhu]

I've done a few these before, but now I'm worried I've been using the wrong equation. Could you take a look at the first problem I posted here?

https://www.physicsforums.com/threads/max-values-for-a-2-dielectric-capacitor.941359/#post-5955152

 

[kuruman]

I've done a few these before, but now I'm worried I've been using the wrong equation. Could you take a look at the first problem I posted here?

https://www.physicsforums.com/threads/max-values-for-a-2-dielectric-capacitor.941359/#post-5955152

I looked. That problem has two dielectrics. You can use that solution here if you consider double thickness for the LDPE dielectric. The equation in post #2 can be written as $$\frac{1}{C_{eq}}=\frac{1}{C_B}+\frac{1}{C_L}+\frac{1}{C_B}=\frac{1}{C_L}+\frac{2}{C_B}.$$

 

[HelloCthulhu]

Ok, let's see if I can get this right...

Total capacitance:

$$C_{eq}=\frac{2}{9.97395\times10^{-10}}+\frac{1}{7.434\times10^{-10}}=3350394444.19$$

$$C_{eq}=\frac{1}{3350394444.19}=0.0000000002984723F=298.4723pF$$

BaTiO3:

$$V_{max}=\frac{680\times10^{3}}{\frac{298.4723pF}{74340pF}}=\frac{680\times10^{3}}{0.0132390428436911}=169366470.3V=169MV$$

LDPE::

$$V_{max}=\frac{460\times10^{3}V}{\frac{298.4723pF}{997.395 pF}}=\frac{460\times10^{3}V}{0.299251851}=1537166.76V=1.54MV$$In a capacitor with a source voltage of 12.3kV and a plate separation of 0.008m, the electric field will be equal to 1.54MV (12320V/0.008m = 1540000V). Since the breakdown voltage of BaTiO3 is 169MV, LDPE will be the first dielectric to break down.

 

[kuruman]

Can you please fix your equations? Use two $ signs to bracket them instead of one. Also, your answer would be much more legible and easier to troubleshoot if you used symbols instead of numbers. Less stuff to write and you can always substitute at the very end. It's a good habit to acquire.

 

[HelloCthulhu]

Can you please fix your equations? Use two $ signs to bracket them instead of one. Also, your answer would be much more legible and easier to troubleshoot if you used symbols instead of numbers. Less stuff to write and you can always substitute at the very end. It's a good habit to acquire.

Thank you for the posting correction! Not sure how I forgot that. However, I'm still confused about what you mean by replacing numbers with symbols.

 

[kuruman]

Thank you for fixing the equations. Strictly speaking, your first equation is post #5 should have $\dfrac{1}{C_{eq}}=\dots$. You can fix that at some later time.

For BaTiO3 you have $$C=\frac{(8.85\times 10^{-12})\times1200\times0.049}{0.007}=0.00000007434F=7.434\times10^{-10} F$$ That should be $7.434\times10^{-8}$F. Check your other numbers.

By symbols I mean no numbers. I would write the equivalent capacitance as $C_{eq}=\dfrac{C_LC_B}{2C_L+C_B}$ and write the charge on the capacitor at a given voltage as $$Q=C_{eq}V=\frac{C_LC_B}{2C_L+C_B}V$$and the voltage across just the LDPE as $$V_L=\frac{Q}{C_L}=\frac{C_B}{2C_L+C_B}V.$$ Since $V_L+2V_B=V$ you can easily find $V_B$ and so on for whatever you do next.

Note that if you happen to make a mistake in calculating the value of one symbol (as you have in this case), it would be much easier to recalculate the final answer without having to redo the intermediate calculations.

 

[HelloCthulhu]

I'm still a little confused (still need practice with the derivations), but let's see how I do...

BaTiO3 capacitance:

$$C=K\epsilon_{0}\frac{A}{d}=7.434\times10^{-8} F$$

LDPE capacitance:

$$C=K\epsilon_{0}\frac{A}{d}=9.97395\times10^{-10}F$$

Total Capacitance:

$$C_{eq}=\dfrac{C_LC_B}{2C_L+C_B}=971.33 pF$$

BaTiO3 max voltage:

$$V_{max}=\frac{V_{bd}}{\frac{C_{T}}{C_{1}}}=52MV$$

LDPE max voltage:

$$V_{max}=\frac{V_{bd}}{\frac{C_{T}}{C_{1}}}=472kV$$

In a capacitor with a source voltage of 3776V and a plate separation of 0.008m, the electric field will be equal to 472kV (3776V/0.008m = 473kV). Since the breakdown voltage of BaTiO3 is 52MV, LDPE will be the first dielectric to break down.