Parallel Plate Capacitors - Online Tutor, Practice Problems & Exam Prep

Capacitance, defined as C = qv, is crucial in understanding parallel plate capacitors. These capacitors consist of two charged plates separated by distance d, with capacitance given by C = εAd. The electric field E = vd is uniform between the plates, while outside it approaches zero. Understanding these principles is essential for applications in electronics and electrical engineering.

Parallel Plate Capacitors

Alright, guys. So we're a little bit more familiar with capacitance and capacitors. In this video, we're going to check out the most common type of capacitor you'll see called the parallel plate capacitor. Let's go ahead and check it out.

We already have the relationship between the capacitance for any given capacitor. It's just C=QV. But the most common type of capacitor you'll see is called the parallel plate capacitor, where you have one surface of charge that's going to be positive Q and another surface of some negative charge that's going to be minus Q. These are separated by some distance, which we'll call d. Now, these don't necessarily have to be rectangles. They can be circles or some odd shape, but they will have an area. That area A is going to be the same for both of them.

So, if you have this kind of situation where you have a parallel plate capacitor, which you have some area and some distance, then we actually have another formula for finding out what the capacitance is. It's ε₀×A/d, that vacuum permittivity constant, times the area divided by the distance between them.

These capacitors, these parallel plates, produce an electric field in between them because the electric field lines want to go from the positive charge over to the negative charge. You should remember the electric field between the plates is uniform. In other words, it's the same always everywhere. No matter where you look inside this parallel plate capacitor, the electric field is going to have the same magnitude. Whereas, if you go outside, if you look anywhere outside of the parallel plates, the electric field very quickly approaches zero. The equation for the magnitude of the electric field is E=V/d, where V just represents the voltage.

The electric field between a capacitor has nothing to do with the capacitance and the voltage. All it depends on is the charge and the area involved. This is actually an interesting consequence of Gauss's Law. The electric field and the equipotential surfaces have to be perpendicular, which means the electric field always points in the direction of decreasing potential.

Let's go ahead and check out this example right here of a parallel plate capacitor, and solve some problems. We're told what the area of these plates is in centimeters squared, the separation distance, and what the voltage is. So our first step is just going to draw a quick little sketch of what's going on. So, we have these plates and separation distance, in other words, d=0.01 meters, we have to change everything into meters, so we have the right amount of units.

Now, let's figure out what the charge on the capacitor is using Q=C×V. The capacitance for a parallel plate capacitor is C=ε₀A/d. Plugging in our values, we get Q=4.43×10−11 Coulombs.

For part b, we can figure out what the magnitude of the electric field between the plates is both ways. The electric field is equal to Q/ε₀A which works out to be 10000 Newtons per Coulomb. Alternatively, E=V/d which gives the same answer. So we just got the same answer using two different approaches.

Let me know if you guys have any questions with this, and I'll see you guys in the next one.

Two circular plates of radius 2cm are brought together so their separation is 5mm. What is the capacitance of these plates?

• A. 2.23×10⁻⁸
• B. 2.23×10⁻¹⁰
• C. 2.23×10⁻¹¹
• D. 2.23×10⁻¹²

Point Charge Inside Capacitor

Okay, guys. So this problem, this example problem has a bit of parts to it. So we're going to work it out together. So we've got these 2 parallel plates, and I'm going to draw them out like that. Now, I know what the charge is on either of those plates. Both of these plates are charged to 30 nano coulombs, and that's \( 30 \times 10^{-9} \) coulombs. I also know that the plate separation between these two positive and negatively charged plates is equal to 10 millimeters. But I need that in the right unit, so I have to convert that. That's going to be \( 0.01 \) meters. Okay? The first thing I'm supposed to do is figure out what the voltage is or the potential difference between the plates. So for part a, we're going to be looking for what variable is potential difference in voltage, that's \( V \). So we can use \( V \) is equal to let's see. We have a couple of different ways we can use this. The easiest approach is going to be is going to be using \( Q = C V \). Right? So this is our approach. So that means that \( V \) is equal to the charge divided by the capacitance. Now we have the charge that's accumulated in both of the plates. The one thing we need to find is the capacitance. So we can use that the other relationship between capacitance in a parallel plate capacitor is \( \epsilon_0 \times \frac{A}{d} \). Now, what happens is this capacitance right here is in the denominator. So what happens is when you divide everything over, that means that these two things in the numerator are actually going to go in the bottom, \( \epsilon_0 A \), and then this \( Q \) is going to be multiplied by this \( d \) right here. So you're going to have \( \frac{Q \times d}{\epsilon_0 \times A} \). Right? So it's just because this thing is a denominator. You can work it out yourself. So you've got the potential is equal to the charge, which is \( 30 \times 10^{-9} \). Now the plate separation is \( 0.01 \), which by the way is just this distance right here between the plates. And now we've got the \( \epsilon_0 \). So we've got that \( \epsilon_0 \) constant, \( 8.85 \times 10^{-12} \). Now, we have to figure out what the area is. Now, the area of these 2 1 centimeter by 1 centimeter plates is just going to be multiplying those two distances together, but we need to put them in the right units first. One centimeter is going to be \( 0.1 \times 0.1 \). So that means it's going to be \( 1 \times 10^{-4} \). That's going to be in meter squared. So that's what we actually have to plug into our calculators. So \( 1 \times 10^{-4} \), and then you plug all of that good stuff into your calculator, and you should get a voltage of \( 3.39 \times 10^5 \), and that's in volts. Alright? So that's the answer to the first part.

The second part now says, what is the electric field in between the plates? Now, again, we have a couple of different formulas for that, but the easiest way to figure out the electric field is by relating it back to the voltage. We know that \( E \) is equal to the voltage divided by the distance. And we actually have what both of those things are now. If you wanted to use the other equation, so if you wanted to actually use this, \( \frac{Q}{\epsilon_0 \times A} \), that also would have been perfectly valid to do that. So you're more than welcome to do that on your own. So, just go ahead and plug in whatever formula you want to use. So I decided to go with the voltage divided by the distance. So in other words, I have the electric field is the voltage \( 3.39 \times 10^5 \) divided by the distance \( 0.1 \), and you should get \( 3.39 \times 10^7 \), and that's going to be Newtons per Coulomb. So that's the answer to part b.

Now this third part here is the interesting one. How much energy will it take to move a charge, a negative 5 coulomb charge, from the positive plate over to the negative plate? So we're just going to go ahead and assume that this plate is positive. That's negative \( Q \) or positive \( Q \). This is going to be the negatively charged plate, and we know that the potential difference, \( \Delta V \), is going to be final minus initial. So what happens if we were to take a negative charge and move it across this distance right here? How much energy would it take to do that? Well, let's think about it. We're taking a charge, which is \( Q \), and removing it across a potential difference. So what formula is going to relate that to the energy? Well, that's going to be the work formula. So the work is going to be equal to the negative \( Q \times \Delta V \). We just have to figure out what our final minus initial is. Now if we're going from the positive to the negative plate, then this \( \Delta V \) is actually going to be minus the potential difference, which we know is \( 3.39 \times 10^5 \). So this is actually what our potential difference is. Just in this specific case, we're removing a charge from here to here. If we were told it was going the negative to the positive, then it would have been a positive potential difference. Okay? So make sure that you have the sign correctly on that. So we're just going to plug that formula down into here, and we're going to get the work that's done. Alright. So we've got the work is equal to negative. Now, I've got negative 5 times \( 10^{-9} \), and that's going to be nano coulombs. Right? Yeah. It's \( 5 \times 10^{-9} \). And now times the potential difference, negative \( 3.39 \times 10^5 \). So now what happens is 2 of the negatives will cancel out, but one of them will still remain, which means that the work that's done is equal to negative \( 1.7 \times 10^{-3} \), and that's in joules. So the fact that this number right here is negative means that we're taking a charge which is on a positive plate and removing it to where it doesn't want to go. We're moving a negative charge to a lower potential, which means that we actually have to give this system some energy. So that means that this work is work that's done on the system, and this work is done on the system by you. So you actually have to do the work in order to move this charge from the positive plate over to the negative plate because you're going against what it naturally would want to do. Okay? So these are the answers to the questions. Let me know if you guys have any questions or concerns, and I'll see you guys in the next one.

A 3 F capacitor is given a potential difference across its plates of 10 V. What is the charge built up on its plates? If the source of the potential difference across the plates is removed, but the plates maintain their charge, what is the new potential difference across the capacitor if the distance between the plates is doubled?

• A. V_new = 5 V
• B. V_new = 10 V (no change)
• C. V_new = 15 V
• D. V_new = 20 V
• E. V_new = 30 V