Three layers of a capacitor - Voltage divider + field strength.

• Mutaja
In summary: Cair=##8.85*10^-12## * εr * \frac{A}{d} = ##8.85*10^-12## * 1... = 2.95*10-10.Ctotal=1.1615*10-9.Net charge on the equivalent total (Q?):Q = V * C = 100V * 1.1615*10-9 = 1.1615*10-7Using this Q across all capacitors is wrong - obviously. I get voltages of 86.28 and 115.04, which is still wrong, but closer - and at least I've understood
Mutaja

Homework Statement

In a plate capacitor, the plates has a area of ##100cm^2## and a distance of 3mm. The insulation between the plates is a 1mm glass plate (εr = 10), a 0.5mm thick mica plate (εr = 5) and the rest is air. The insulation layers is parallel with the capacitor plates. The capacitor is connected to a 100V voltage supplier.

a) find the voltage across the layers.

b) How big is the field strength in each layer?

The Attempt at a Solution

For a.

I'm using the current divider rule.

Vglass = $\frac{\frac{10*0.01m^2}{0.004mm}}{\frac{10*0.01m^2}{0.004mm}+\frac{5*0.01m^2}{0.0035mm}+\frac{0.01m^2}{0.003mm}}$ * 100V = 56.88V.

This answer seems about 10x larger than it should be - which is why I'm pretty sure that I'm on the right track here. Well, that's only based on the fact that the random chance that I've come up with a completely wrong formula and get an answer that is almost exactly 10 times larger than what it should be, is fairly low.

What am I missing here?

As always, and feedback is very much appreciated.

- Mutaja.

The layers are stacked between the plates and have thicknesses of 1 mm, 0.5 mm, and 1.5 mm. So that models as three capacitors in series. Being in series they must all carry the same current when any current flows. So current division isn't the way to go.

You have the information required to calculate the individual capacitor values. Why not start there?

1 person
You don't need to compute capacitances and the area of the plates is immaterial.

You can write an equation for the voltage drops across each region with E, the field in the air gap, as the unknown. You know what the relative E fields are in each region and you also know that the sum of the voltage drops has to equal 100V. You're right, your computed value for Vglass is close to 10x what it should be.

1 person
gneill said:
The layers are stacked between the plates and have thicknesses of 1 mm, 0.5 mm, and 1.5 mm. So that models as three capacitors in series. Being in series they must all carry the same current when any current flows. So current division isn't the way to go.

I realize that, but voltage divider makes sense, which is what I thought I attempted at least.
gneill said:
You have the information required to calculate the individual capacitor values. Why not start there?

Ok. This is my new attempt:

Cglass=##8.85*10^-12## * εr * $\frac{A}{d}$ = ##8.85*10^-12## * 10 * $\frac{0.01m^2}{0.001m}$ = 8,85*10-10.

Assuming this is correct, I now know c. If I want to use this to find the voltage, I need Q. Knowing myself, though, I've probably overlooked a useful formula.

Or are we now back at voltage dividing since I know more of the size of these capacitors related to each other (assuming, of course, I compute C for mica and air as well).

rude man said:
You don't need to compute capacitances and the area of the plates is immaterial.

You can write an equation for the voltage drops across each region with E, the field in the air gap, as the unknown. You know what the relative E fields are in each region and you also know that the sum of the voltage drops has to equal 100V.

Hmm. This has always been a challenge for me. In my initial attempt, I've written an equation for the voltage across the "glass capacitor" - wrongly. How would your method compare to that? "the field in the air gap" doesn't make much sense to me. Is it the electrical field strength (is that what the E is for?).

Sorry for the confusion, and thanks a lot for your help.

rude man is correct that you can write an equation for the voltage drop across each region if you know how the relative permittivity affects the net field strength (relative to vacuum) within the region. It's a good approach and quicker to achieve than what I was suggesting.

What I was going for is a variation of the voltage divider method that takes advantage of the way charge is distributed in serially connected capacitors. The usual voltage divider formulas get messier when there are more than two capacitors in series.

By computing each of the capacitor values and then the total equivalent capacitance, a known total voltage (100 V) applied gives you the net charge on the equivalent capacitor. For series capacitors the charges are all the same so that it will be the same charge on each of the capacitor "layers", and hence you know the potential across each layer via V = Q/C. With potential difference and thickness you then have your field strengths for each layer.

gneill said:
By computing each of the capacitor values and then the total equivalent capacitance, a known total voltage (100 V) applied gives you the net charge on the equivalent capacitor. For series capacitors the charges are all the same so that it will be the same charge on each of the capacitor "layers", and hence you know the potential across each layer via V = Q/C. With potential difference and thickness you then have your field strengths for each layer.

Computing each of the capacitor values:

Cglass=##8.85*10^-12## * εr * $\frac{A}{d}$ = ##8.85*10^-12## * 10 * $\frac{0.01m^2}{0.001m}$ = 8.85*10-10.

Cmica=##8.85*10^-12## * εr * $\frac{A}{d}$ = ##8.85*10^-12## * 5 * $\frac{0.01m^2}{0.0005m}$ = 8.85*10-10.

Cair=##8.85*10^-12## * εr * $\frac{A}{d}$ = ##8.85*10^-12## * 1 * $\frac{0.01m^2}{0.0015m}$ = 5.9*10-10.

Ctotal=2.36*10-9

Net charge on the equivalent total (Q?):

Q = V * C = 100V * 2.36*10-9 = 2.36*10-7

Using this Q across all capacitors is wrong - obviously. I get voltages of 266 and 400 using V = Q/C. Being able to see that I've done a silly mistake somewhere and not being able to spot it is unfortunate, at best.

Hopefully it's not all wrong and I've made at least *some* progress.

Thanks once again for helping me out.

Mutaja said:
Computing each of the capacitor values:

Cglass=##8.85*10^-12## * εr * $\frac{A}{d}$ = ##8.85*10^-12## * 10 * $\frac{0.01m^2}{0.001m}$ = 8.85*10-10.

Cmica=##8.85*10^-12## * εr * $\frac{A}{d}$ = ##8.85*10^-12## * 5 * $\frac{0.01m^2}{0.0005m}$ = 8.85*10-10.

Cair=##8.85*10^-12## * εr * $\frac{A}{d}$ = ##8.85*10^-12## * 1 * $\frac{0.01m^2}{0.0015m}$ = 5.9*10-10. <--- check the power of ten.

Ctotal=2.36*10-9
How do capacitors in series add? (as a check: The total should end up smaller than any of the individual capacitors).

Net charge on the equivalent total (Q?):

Q = V * C = 100V * 2.36*10-9 = 2.36*10-7

Using this Q across all capacitors is wrong - obviously. I get voltages of 266 and 400 using V = Q/C. Being able to see that I've done a silly mistake somewhere and not being able to spot it is unfortunate, at best.

Hopefully it's not all wrong and I've made at least *some* progress.
You're making progress. Fix up your capacitor values and net capacitance.

1 person
Thank you! I figured out the voltage across each layer. My power of 10 mistake should be - 11, and capacitors in series "equals" resistors in parallel. Well, the method of computing the values at least. My C total is ##5.2*10^-11##. V=Q/C is then correct.

I will give part b a shot in a few hours. I'm currently on the road so my apologies for that. But I guess the forum isn't going anywhere by the time I'm by my computer again heh.

Thank you so much for all your help so far.

Across each layer 1, 2 or 3, voltage V = distance d times parallel component of electric field E.

So: E1 d1 + E2 d2 + E3 d3 = V.
And then use the fact that E = Evacuor.

When you get your answer I'll let you know if it's right.

EDIT: NM, I see you went the other way already.

rude man said:
Across each layer 1, 2 or 3, voltage V = distance d times parallel component of electric field E.

So: E1 d1 + E2 d2 + E3 d3 = V.
And then use the fact that E = Evacuor.

When you get your answer I'll let you know if it's right.

EDIT: NM, I see you went the other way already.

Yes, I understand enough to attempt it, but seeing as I already have my answers, I'll leave this behind for now. When I get ahead of my schedule, I will look into it though. Any practice won't harm me, to say the least.

For my 2nd problem, it was a lot easier, obviously, when I now have the voltage across each capacitor. I simply used E = V/d (which I learned about in a previous thread on this forum), and I got answers that made very much sense.

Thanks a lot for your help, both of you.

1. What are the three layers of a capacitor?

The three layers of a capacitor are the two conducting plates and the dielectric material in between them.

2. How does a capacitor act as a voltage divider?

A capacitor acts as a voltage divider by storing and releasing electrical charge. When a voltage is applied to the capacitor, the charge is stored on the plates and the voltage across the capacitor is equal to the applied voltage. When the capacitor is connected to a circuit with a different voltage, the stored charge is released and the voltage across the capacitor changes accordingly.

3. What is the role of the dielectric material in a capacitor?

The dielectric material in a capacitor acts as an insulator between the two conducting plates. It has a high dielectric strength, which allows it to withstand high voltages without breaking down. It also stores energy in the form of an electric field, which increases the capacitance of the capacitor.

4. How does the field strength affect the performance of a capacitor?

The field strength, or the strength of the electric field, is directly proportional to the voltage across the capacitor and inversely proportional to the distance between the plates. This means that a stronger electric field will result in a higher capacitance and a lower voltage drop across the capacitor.

5. What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by several factors, including the area of the plates, the distance between the plates, and the type of dielectric material used. Generally, a larger area and a smaller distance between the plates will result in a higher capacitance, while using a higher dielectric constant material will also increase the capacitance.

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