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It's far beyond anything I have experience with. I've solved derivatives in the past, but wasn't familiar with second derivatives. To be honest, I'm in way over my head on these subjects. But even so I'd like to keep working on trying to understand them. Thank you so much for your help.

Thank you for the response! I still have some questions about the variables. I know how to use Figure 6 to find the values for maximum interfacial field Ew/v or Et/w and water thickness rw/rt, but I'm still not sure what to do with ∈(E), d, or how to solve for ϕw or ϕv. In the end, all I'm...

I recently read a paper on using an electric field to drive water autoionizaton. I'm trying to figure out how to use the Laplace equation on pg 9; 4th paragraph; to solve for voltage. I'm also interested in how this equation would change if I replaced the hemispherical tip with a parallel plate...

I'm not sure if this will help, but this article, "Dynamics Behaviors of Droplet on Hydrophobic Surfaces Driven by Electric Field" (2019), describes in section 3. Results and Discussion that a water droplet on a substrate within an electric field is "mainly affected by the inertial force...

I've been researching water bridges and electrowetting to learn the effects of electric fields on water molecules but something continues to confuse me: if polar molecules can only rotate in an electric field, how is the water moving? Anyone familiar with this phenomenon? Any help is greatly...

I've been researching pem electrolyzers, but still don't understand how to mathematically express how platinum reduces the activation energy necessary to dissociate H2O. I've seen the Arrhenius equation solved before, but didn't understand how to get the values for it. Here are the operating...

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...

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.

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...

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

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...

The turns were adjusted during the 26kV sim, but I've made the necessary changes to the transistor connections. Hope they're correct. New simulation just under 20kV.

Thank you so much for your help on this. Other than the issue with the number of turns, I figured out that the schematic connections for the transistor as I understood them (which is probably wrong since the transformers don't look the same) was incorrect. After connecting the bottom of the top...

I'm very new to transistors and transformers, so I'm not sure how to calculate the inductance for this circuit. I'm working on it now.

I couldn't find any information on the turn ratio of a computer flyback transformer, so I used the turn ratio of an ignition coil: primary at 135 and the secondary at 11000.

Thank you for quick the response! That definitely got the circuit running, but the max voltage and current are much lower than I thought they'd be (1.26kv, 1.65mA). Can't figure out what else I'm doing wrong.

I found this schematic while watching the following video on building a DC 20kv power supply: I wanted to test the circuit in multisim, but I'm having trouble finishing it. None of the transformers available look like the one used in this schematic. Below are the transformers available on...

Thank you for the clarification! I was worried the usual capacitor math wouldn't be as exact.

Recently I've been researching parallel plate capacitors and was wondering what effects the material had on the charge capacity of the plate. I found one source for measuring the capacity based on its material, but haven't seen any textbook evidence to support it yet. Any feedback on the...

Thank you for that correction! Let's see if I know how to convert this... 3.5ml O2 = 0.0035L/22.4L = 0.00015625 O2(STP) 0.0003125 H2O(l) →0.0003125 H2(g) + 0.00015625O2(g) 0.0003125M H2O *18g = 0.005625g H2O

Thank you for that confirmation! Now I just need to figure out how to balance the equation. Cathode: 0.000625 H+(aq) + 0.000625e−→0.0003125 H2(g) Anode: 0.0003125 H2O(l) → 0.00015625O2(g) + 0.000625 H+(aq) + 0.000625e− 0.0003125M H2O *18g = 0.005625g H2O Did I get any of this correct?

Did I calculate the moles of H2 and O2 STP correctly? 7ml H2 =0.007L/22.4L = 0.0003125 H2(STP)

Thank for your correction! I think I may understand how to do this now... If 7ml of hydrogen and 3.5 ml of oxygen are produced at 1A per minute during electrolysis, how much water is being electrolyzed per minute? 7ml H2 =0.007L/22.4L = 0.0003125 H2(STP) 3.5ml O2 = 0.0035L/22.4L = 0.00015625...

Apologies. I've felt very very lost about how to approach this solution. Let's see if I get this correct: 7ml H2 =0.007L/22.4L = 0.00031M(STP) H2 3.5ml O2 = 0.0035L/22.4L = 0.00016M(STP) O2 0.00047M H20 * 18g = 0.00846g H20

Ok, let's see if I got it right this time. 2 moles of H2O undergoes electrolysis at 1 amp for 1 min. What is the mass of H2 and O2 gas produced? Cathode: 4 H+(aq) + 4e− → 2H2(g) Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− m = ZQ = mass produced Q = quantity of electricity in coulombs; I*t Z =...

Apologies. I think this is the full equation: https://study.com/academy/lesson/faradays-laws-of-electrolysis-definition-equation.html m = ZQ m = mass produced Z = proportionality constant in g/C Q = quantity of electricity in coulombs

https://www.ausetute.com.au/faradayl.html First Law of Electrolysis : The mass of a substance produced by electrolysis is proportional to the quantity of electricity used. Q/m Q = quantity of electricity in coulombs m = mass produced

I used these values when I first started learning about Faraday's Laws of Electrolysis: (1 * 60 * 4) / (4*96,500) = 0.0006ml (1 * 60 * 32) / (4*96,500) = 0.0049ml If this is correct, I thought I could algebraically substitute the values for H2 and O2 with 7ml and 3.5ml respectively and solve...

Apologies. I added the units to the numbers.

I've been doing a little independent research into PEM fuel cells and came across this Horizon mim fuel cell. I read the following specs, but didn't see anything about the volume of water being electrolyzed...

I've been researching dielectric breakdown for a while and came across this interesting experiment: https://tore.tuhh.de/bitstream/11420/1160/1/Size_dependence_of_the_dielectric_breakdown_strength_from_nano_to_millimeter_scale_TUB_Doc_version.pdf 0.3mm of BaTiO3 was placed between two...

I've learned a lot in a short amount of time, but thanks to both of your safety concerns I definitely think I should learn more about circuits and working with mains voltage before moving forward. I found one site to be especially useful, but please feel free to add any other resources. Thanks...

Thank you for the quick response! I'm using this variac model with fuse protection. If it pops all current will drop immediately.

I apologize for my inexperience. I've designed circuits with lower voltage, but I don't have any experience rectifying voltage yet. I'm sure this is something simple, but I thought I'd ask for advice first. The warning is very appreciated!

I built a high voltage ac power supply using a YouTube tutorial so I could use it in my parallel plate capacitor experiments. But since discovering that I actually need direct current to charge the plates correctly, the circuit needs to be rectified. I found a tutorial on building a full bridge...

I've been interested in the effects of electric fields on water for a while and came across this impressive demonstration of the water bridge experiment: The thing that fascinates me the most is how much the electricity arcs out of the water. For example at timestamp 8:43 the arc clearly goes...

You're absolutely right. I submitted 22.531μF as the solution to the capacitance equation. But the capacitance is 2.25307x10-11F (0.0000225307μF). $$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{-12})\times1\times0.0254469}{0.01}=2.25307\times10^{-11}F=0.0000225307μF$$ I used this online...

Ok, I was definitely confused earlier. I found a more detailed tutorial on this, so I hope I have a better understanding of time constants and charging time now: https://www.electronics-tutorials.ws/rc/rc_1.html 1t = 0.019Ω*0.00002253uF = 4.280833e-7 secs 5t = 5X4.280833e-7 = 0.00000214041 secs

I apologize this is my first time doing this kind of equation. I used this video to learn about time constants: In this equation I still have the -t value over the RC value: After the RC value was solved, I should've written these equations as: 6V*(1-2.71828-4.280833e-7/4.280833e-7)...

$$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{-12})\times1\times 0.0254469}{0.01}=0.000022531F=22.531μF$$ 6V*(1-2.71828-t/0.019Ω*0.0000225307uF) time constants 6*(1-2.71828-1) = 6*(0.632) = 3.79 sec 6*(1-2.71828-2) = 6*(0.865) = 5.18 sec 6*(1-2.71828-3) = 6*(0.950) = 5.70 sec 6*(0.982)...

No it's ok. This isn't really homework just a scenario I've been experimenting with mathematically. I guess it's more of an approximation. Here's a related post I made back in March. I just changed a few of the specs...

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

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

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

I found a physicsforums thread that I might be able to use: https://www.physicsforums.com/threads/layering-different-dielectric-materials-to-increase-breakdown-voltage.902113/ If I add the two max voltages and divide by the total thickness, I should get the dielectric strength of the...

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

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 :frown:. 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...

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

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...

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.