Intro To Dielectrics - Video Tutorials & Practice Problems

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1

concept

Intro To Dielectrics

Video duration:

6m

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Alright, guys. So for this video, we're going to take a look at these objects called Die Electrics and how they play a role in capacitors and capacitor circuits. Let's check it out. So basically what a die electric is Is this an insulate er or an insulating material that you'll stick in between charged plates? You'll stick in between a capacitor and what it does is it always increases the capacitance of that capacitor. Now, the equation we're gonna use to relate those is the C equals K times. See not where C is gonna be effectively, the new capacitance and K is gonna be a constant. The sea not is going to be with the capacitance is in a vacuum or otherwise. The old capacitance, if you will now this k constant or Kappa. But the Greek letter Kappa is called the dye electric constant that die electric constant is a number with no units, and it's always greater than one. And it basically serves toe weakened the electric field inside of capacitor. So that means if you stick a die electric, it'll always weaken with the electric field is and that's given by this equation right here where the new electric field is going to be the old electric field or the electric field in a vacuum divided by this K constant. And if it's always greater than one, that means that he's always gonna be less Now. A lot of times in these problems, you won't know which equations to use. There's two basic ways that we can connect. I electrics to capacitors. Let's go ahead and check them out. The first is where you'll have a battery that's initially hooked up to a capacitor, and you'll have these charges that build up on these plates right here. So you have positive charges and negative charges. Now what happens is the battery is disconnected. And then what happens is the DI electric material is inserted between the capacitor, so that's given by this green little material right here. Now we know that this capacity that this dialectic material is going to increase the capacitance of this capacitor. But what happens is that the battery is not connected. Then these charges have to remain the same. These charges have to remain conserved. So in this situation we have constant charge, and what happens is in the in these examples, we can relate the charge, the capacitance and the voltage to see how these variables will change. And we know that the capacitance is gonna increase. We know that the Q is going to remain the same. The only way that happens is if the capacity increases and the voltage decreases. So that means that V has to decrease between the plates in order to maintain that constant charge. Now, we know also that the Q is going to remain the same. So what does that mean for the potential energy? Well, we know that the capacitance is going to increase, and if this is in the denominator, that means the potential energy also decreases. Now, what ends up happening is that the potential energy ends up doing some work in compressing the capacity together. So that energy actually goes somewhere. Now, what about the energy density? What we see that the energy density, the electric field, is always going to be weakened by the presence of a dialectic material. So that means that this is always gonna go down. And that means that the energy density is going to decrease over here. And by the way, this equation also holds for our other scenario where we have a constant voltage. Let's go ahead and check out how that works. So now the in this case, we have the battery that is still connected with the capacity when the dialectic material is inserted. So now you have a battery and then you still have these positive charges that build upon the plates. The negative charges buildup on these plates over here and now what happens is rather than battery being disconnected, you have the dialect of material that's inserted while the battery is still connected. Now what happens is we know that the voltage wants to decrease. But if this thing is still connected to the battery than that, what happens is that these positive charges sort of get released from the batteries, and they basically build up on the plates to effectively increase the charge. Let's see how that works just by using the equation. So what happens is you get these extra charges that build up. We know that the capacitance of the material with the dye electric is going to increase. But in this case, if the battery is still connected, then that means that the voltage across the circuit has to be constant. And if the voltage is constant, but the capacitance increases, the only way that's possible is if the charge also increases. So you have to maintain the same exact ratio. So that means that Q has to increase for these situations. Now let's see what happens with the potential energy. We know that this voltage is gonna remain constant and the capacity is gonna increase. So in this situation, the potential energy also increases. And just how is in the last example? The energy density still depends on the electric field, which always gets weakened. So that means that the energy density always decreases. So this is how these variables and things like that how they change. You just have to make sure that when you're when you're doing your problems, you understand which scenario we're working with here. So the question to ask, Is the Di electric still connected or inserted when the battery still connected? Or is the di electric inserted after the fact And that's basically go to govern, which equations you're gonna use. All right, so let's see how all of this stuff works out. When do some practice problems. So this example says that a capacitor is connected to a battery. And then we're supposed to figure out what the charge on this capacitor is after the dye electric is inserted while it's still connected to the battery. So while it's still connected to the battery, this means that the voltage is going to be equal to a constant. So we know which equations we're gonna use. So let's check it out. We're supposed to be figuring what Q is. So we have a relationship between Q and See we know and Syracuse C and B. We know Q is equal to see times V. So what happens is when you stick this die electric material in between this capacitor right here, it's going to increase the capacitance. We know that that's going to be effectively increased. We know the new capacitance is going to be that k constant times the old capacitance, which is this three Ferried capacity right here. So that means that the new capacities of this thing after the dye electric is inserted is going to be K, which we know is to over here times the original capacities, which is the three ferrets. So what happens is this has to be six ferrets. So now we know what the six fared capacitor is now in order to maintain the same voltage, we know that that voltage has to remain, Constance. So what happens is that the voltage is going to be Q divided by C. Now you know that the voltage has to be nine. So let's go ahead and plug everything in right? We know that the charge is going to be the new capacitance, which is equal to six times the voltage, which is constant Times nine. In other words, we get a charge of 54 columns. That's basically how to solve these problems. First Step is gonna be figuring out which equations you're gonna use. Is the battery still connected? Is it not connected? Are using constant charge or constant voltage? All right, let me know if you guys have any questions, we're gonna do some more practice problems to get more familiar with this stuff. Alright, guys, let me know if you have any questions

2

Problem

Problem

A capacitor in a vacuum is charged to 64V between its plates, then disconnected. Initially, each plate has 32μC. An insulating slab of dielectric glass with k = 3 is placed between the plates. a) What is the capacitor's new capacitance? b) What is the new voltage across the capacitor?

A

a) 2.0 μF

b) 192 V

B

a) 1.50 μF

b) 21.3 V

C

a) 1.50 μF

b) 64 V

D

a) 2.0 μF

b) 21.3V

3

Problem

Problem

A parallel plate capacitor is formed by bringing two circular plates, of radius 0.5 cm, to a distance of 2 mm apart. The capacitor is made so that it has a dielectric of constant κ between the plates. When the charge on the capacitor is 3 nC, the voltage of the capacitor is 5000 V. What is the dielectric constant?

A

κ=0.172

B

κ=0.580

C

κ=1.72

D

κ=5.80

4

example

Partial Dielectrics

Video duration:

4m

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Hey, guys, let's do an example. What is the new capacitance of the two capacitors that air partially filled with di electrics shown in the following figure? Okay, we have an A and B situation going on here, so let's just address them each distinctly okay, for part A, we have a die electric that fills the top half off this capacitor. What this is going to be like is this is going to be like to capacitors in parallel. Sorry. In Siris. Okay, imagine if this was plus Q Then we'd have a minus que Here on the inner surface of this die electric Ah, plus Q on the outer surface of the die electric and a minus Q on this plate. So this looks like a single capacitor, and this looks like a single capacitor where the upper capacitor has a di electric and the lower capacitor does not will call those C one and C two. So see, one is gonna be Kappa. Absolutely not. That area is a but the distances d over to. So this is to Kappa Absolutely not a over D. Once I rearrange it, C two is not gonna have any dye electric. So no, Kappa, it's absolutely not. The area is still a and the distance is still deal over to. So this is to absolutely not a over T. Now, since these Aaron Siri's, then I would say that the equivalent capacitance. So the total capacitance of this physical capacitor is one. Oversee equals one over to Kappa Absolute. Not over d plus one over to absolutely not a over d. Okay, the least common denominator is to Kappa absolute. Not over d. So I need a Kappa over a Kappa. So this is gonna be one plus Kappa over to Kappa. Absolute. Not over D Okay, so the whole thing is going to be Let me give myself just a little bit of room here. C equals to Kappa over one plus Kappa. Absolutely not a over D. Now, this coefficient right here looks kind of like an effective die electric constant, right? Like if you put half of a die electric through, then this looks almost like its own die electric constant. Right? That's just a number times Epsilon. I still not over d Sorry. It's just a number of times, the capacitance. So it looks like it's effective dialect. Your constant. Okay, Now let's do part B. This time we put a dye electric halfway through on the left side. Now it's the same. Here is gonna be the potential difference or the voltage. That's just gonna be whatever it is across the plates, whether or not the dye electric is there. Okay, so these look like they're in parallel, right? They both have the same voltage. So let's find I'll call this one C one. I'll call this one C two. Let's find those capacitance is to see one is gonna be Kappa. Absolutely not. What's the area, though? Well, it has half of the capacitor, so I was half of the area, So this is a over to, but the distance is the same. So this is what we'll call it Kappa over two. Excellent. Not a over D. C. Two is just gonna be absolutely not. No Kappa. Right, Because it's in a vacuum. The area once again is half. It has half the capacitor. So this is a over to de This is just one half. Absolutely not. Over D Now, these air in parallel. Okay, so I could just add them. Okay, Okay, Once I've added them, we get this one hat in this case over two. Plus this one over to They have the same denominator. So I could just say it's K plus 1/2 and notice. This also looks like a normal capacitor, but with some sort of weird die electric constant. But either way, this is the capacitance for part B. All right, guys, Thanks for watching.

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