Intro To Dielectrics - Online Tutor, Practice Problems & Exam Prep

Dielectrics enhance the capacitance of capacitors, described by the equation $C=kC₀$, where C is the new capacitance, k is the dielectric constant (always > 1), and C₀ is the original capacitance. In scenarios with constant charge, inserting a dielectric decreases voltage and potential energy. Conversely, with constant voltage, charge increases, raising potential energy. Understanding these relationships is crucial for solving capacitor problems effectively.

Intro To Dielectrics

All right, guys. So, for this video, we're going to take a look at these objects called dielectrics and how they play a role in capacitors and capacitor circuits. Let's check it out. Basically, what it does is it always increases the capacitance of that capacitor. Now the equation we're going to use to relate those is c=k⁢cc0, where c is going to be effectively the new capacitance, and k is going to be a constant. The c0 is going to be what 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 dielectric constant. That dielectric constant is a number with no units and it's always greater than 1. It basically serves to weaken the electric field inside a capacitor. So that means if you stick a dielectric, it'll always weaken what 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 1, then that means the electric field is always going to be less. Now, a lot of the time in these problems, you won't know which equations to use. There are 2 basic ways that we can connect dielectrics 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 dielectric material is inserted between the capacitor. So that's given by this green little material right here. Now we know that this dielectric material is going to increase the capacitance of this capacitor. But what happens is if 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 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 going to increase. We know that the charge is going to remain the same. The only way that happens is if the capacitance increases and the voltage decreases. So that means that the voltage has to decrease between the plates to maintain that constant charge. Now we also know that the charge 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, then it means the potential energy also decreases. Now what ends up happening is that the potential energy ends up doing some work in compressing the capacitor together, so that that energy actually goes somewhere. Now what about the energy density? Well, we see that the energy density, the electric field is always going to be weakened by the presence of a dielectric material. So that means that this is always going to 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, in this case, we have the battery that is still connected when the capacitor when the dielectric material is inserted. So now you have a battery, and then you still have these positive charges that build up on the plates, the negative charges build up on these plates over here. And now what happens is, rather than the battery being disconnected, you have the dielectric 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, then 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.

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 μFb) 192 V
• B. a) 1.50 μFb) 21.3 V
• C. a) 1.50 μFb) 64 V
• D. a) 2.0 μFb) 21.3V

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

Partial Dielectrics

Hey, guys. Let's do an example. What is the new capacitance of the 2 capacitors that are partially filled with dielectrics shown in the following figure? We have an a and b situation going on here, so let's just address them each distinctly.

For part a, we have a dielectric that fills the top half of this capacitor. What this is going to be like is going to be like 2 capacitors in series. Imagine if this was plus q, then we'd have a minus q here on the inner surface of this dielectric, a plus q on the outer surface of the dielectric, 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 dielectric and the lower capacitor does not. We'll call those c1 and c2. So c1 is going to be κε_0 A, but the distance is d over 2. So this is 2 κε_0 A over d once I rearrange it. C2 is not going to have any dielectrics, so no κ. It's ε_0. The area is still A, and the distance is still d over 2. So this is 2 ε_0 A over d. Now since these are in series, then I would say that the equivalent capacitance, so the total capacitance of this physical capacitor, is 1 over C equals 1 over 2 κε_0 A over d plus 1 over 2ε_0 A over d.

The least common denominator is 2 κε_0 A over d, so I need a κ over a κ. So this is going to be 1 plus κ over 2 κε_0 A over d. C equals 2 κ over 1+κε_0 A over d. This coefficient right here looks kind of like an effective dielectric constant. It looks almost like its own dielectric constant, just a number times the capacitance.

Now let's do part b. This time, we put a dielectric halfway through on the left side. The same here is going to be the potential difference or the voltage. That's just going to be whatever it is across the plates whether or not the dielectric is there. These look like they are in parallel, right? They both have the same voltage. So let's find, I'll call this one C1, I'll call this one C2. Let's find those capacitances. C1 is going to be κε_0. The area, though, has half of the capacitor, so it's A over 2, but the distance is the same. So this is κ over 2, ε_0 A over d. C2 is just going to be ε_0, no κ, right, because it's in a vacuum. The area, once again, is half. It has half the capacitor. So this is A over 2d. This is just one half ı ε_0 A over d. Now these are in parallel, so I can just add them. Once I've added them, we get this one half, this κ over 2 plus this one over 2. They have the same denominator, so I can just say it's κ+1 over 2, and notice this also looks like a normal capacitor but with some sort of weird dielectric constant. But either way, this is the capacitance for part b.

Alright, guys. Thanks for watching.

Here’s what students ask on this topic:

What is the role of a dielectric in a capacitor?

A dielectric in a capacitor serves to increase the capacitance of the capacitor. This is described by the equation $C=k{C}_0$, where $C$ is the new capacitance, $k$ is the dielectric constant (always greater than 1), and $C_0$ is the original capacitance. The dielectric constant weakens the electric field inside the capacitor, which allows the capacitor to store more charge for a given voltage.

How does inserting a dielectric affect the voltage and potential energy in a capacitor with constant charge?

When a dielectric is inserted into a capacitor with constant charge, the capacitance increases. Since the charge $Q$ remains the same, the voltage $V$ across the capacitor decreases, as given by $V=Q/C$. The potential energy $U$ also decreases because it is given by $U=\frac{Q^2}{/}$. The energy density, which depends on the electric field, also decreases.

What happens to the charge and potential energy when a dielectric is inserted into a capacitor with constant voltage?

When a dielectric is inserted into a capacitor with constant voltage, the capacitance increases. Since the voltage $V$ remains constant, the charge $Q$ on the capacitor increases, as given by $Q=CV$. The potential energy $U$ also increases because it is given by $U=\frac{Q^2}{/}$. The energy density decreases due to the weakening of the electric field by the dielectric.

What is the dielectric constant and how does it affect a capacitor?

The dielectric constant, denoted as $k$ or $\kappa$, is a dimensionless number that quantifies the ability of a dielectric material to increase the capacitance of a capacitor. It is always greater than 1. The capacitance of a capacitor with a dielectric is given by $C=k{C}_0$, where $C$ is the new capacitance and $C_0$ is the original capacitance. A higher dielectric constant means a greater increase in capacitance.

How do you determine which equations to use when solving problems involving dielectrics in capacitors?

To determine which equations to use when solving problems involving dielectrics in capacitors, you need to identify whether the capacitor is connected to a battery (constant voltage) or disconnected (constant charge). For constant voltage scenarios, use $Q=CV$ and $U=\frac{Q^2}{/}$. For constant charge scenarios, use $V=\frac{Q}{/}$ and $U=\frac{Q^2}{/}$. Understanding the context of the problem will guide you to the correct equations.