Hey, guys. So in this video, we're going to be talking about these things called capacitors. Alright, we're going to be working with capacitors much more in-depth later on in the course. So let's go ahead and get familiar from right now. All right, let's get started. So if you have two parallel plates and they have equal and opposite charge, this is actually the important part. They're going to produce an electric field between them. We know that charges produce electric fields. So in this specific arrangement, if you have two plates and they have equal but opposite charges, they produce a uniform electric field. That uniform just means that it's the same at all locations. Always has the same magnitude everywhere. Always. It's just the same. Okay. And to see why that is, we just have to remember that electric fields always point from positive charges. Whoops. They point from positive charges or from the positive plate towards the negative plate. And the reason for that, right? Because basically, you can think of this plate of positive charges as just a whole bunch of tiny little mini positive charges. We know these many positive charges have electric fields that point outwards. So what ends up happening is that this electric field that points in this direction right here these guys are going to point downward over here and then these guys right here, they're right next to each other. These electric fields there, pointing in this direction, are going to group up and basically cancel each other out and point in this direction as well. Now we have the negative charges on the opposite side of the plate that want to produce an electric field that points inwards like this. And just like in this case, where basically they symmetrically cancel out the symmetric, the symmetric electric field lines from these negative charges is going to result in an electric field that points downwards. They're just going to cancel each other out. Just is just like it for the positive charges. In other words, what we end up getting is we just end up getting this uniform electric field that is always in this direction from the positive plate towards the negative plates. So this is E. And the deal is that this at any point that you decide to measure the electric field E is always going to be the same so not. No matter if I chose here or here or here anywhere like that. These these are always going to be the same. So I'm going to write equal signs. So that is going to be the same. Is this E and this is going to be the same. So it's always the same at all of these parts. Got it. And so if you have these two parallel plates that have equal and opposite charge, that's what we call a capacitor. So that is the definition of a capacitor: equal and opposite charge, and they're separated by some distance. And basically all the capacitors we're going to talk about them a lot more later on in the course are, there are these things that store charge and this energy between them. All right, we'll talk about that later. And so we talked about the direction of the electric field in this diagram right here. Now the equation for that electric field is just Q divided by epsilon. Not so. Let me write that a little bit neater. So this Greek letter right here that kind of looks like an E, but it's called Epsilon—the times A, which is the area of each plate. So this new Epsilon not right here is called the vacuum permittivity, which isn't a name you probably don't need to know. And the value for that constant is \(8.85 \times 10^{-12}\) and it's got some units associated with you don't really need to know that now. This absolute not is actually related to our other constant that we know called Coulomb's constant. And the way that these things are related is that K, that \(8.99 \times 10^{9}\) is actually equal to \( \frac{1}{4 \pi} \times \epsilon\). Now, this is something we're going to talk about much later on. We're going to talk about it in greater detail. But it's just a good thing to know for now that these two constants are related and we have the A is just the area of each plate, whereas Q is the charge on each plate. Now, remember, these things are equal and opposite. So let's say one is going to be negative. Two coulombs one is going to be positive. Two coulombs. The Q's is going to be two in that case, and then A is the area of each plate. That depends on the geometry. Sometimes they could be square. Sometimes they could be cylinders, things like that. We'll get into it later. But this is the equation for inside the plates. So, in other words, the space right between the plates Here, this is going to be \( \frac{Q}{\epsilon \cdot A} \) but outside of the plates. What if you were actually to go at some point over here? What does the electric field in this area? Well, outside of the plates, it's just going to be zero. So anywhere outside of the plates, the electric field is zero anywhere inside of the plates. It's this equation right here. Got it. That's all we really need to know. For now, let's go ahead and start dealing with some problems. So we've got the electric field in between two parallel plates; we're given the magnitude of the electric Field 1000 Newtons per coulomb, and we're told with the area of these electric plates are so we just need to figure out the charge on each plate. In other words, we're looking for Q, so let's go ahead and start out with our equation. So the electric field between two parallel plates in a capacitor is going to be \(\frac{Q}{\epsilon \cdot A}\). So we have what the electric field is, and we have what this area is. This epsilon not is just a constant. So let's go ahead and figure out what this Q is. All right, so let's go ahead and move these things over to the other side. We've got Epsilon, not A we're going to bring over. So that means that Q is equal to \(\epsilon \cdot A \cdot E\). So we've got \(8.85 \times 10^{-12}\). You don't really need to know the units for that. You've got the area which is five centimeters squared. So I want to be careful here. Remember that five centimeters squared is not 0.5 m squared. You actually have to do the conversion twice because there's an exponent here of two. Alright, so just don't make this mistake. It's a very, very common mistake. So do not do this. It's not this. Instead, five centimeters squared is equal to \(5 \times 10^{-4} m^2\) because again, we have to apply that conversion twice. All right, Just just want to sort of refresh your memory. So we've got \(5 \times 10^{-4}\), and then you got the electric field, which is just 1000. So if you go ahead and work this out, you're going to get a Q that's equal to \(4.43 \times 10^{-9}\) that's going to be in Coulombs. All right, so that's the charge that's built up on each one of these plates. Man, my highlighter is not working. Alright, guys, that's it for this video. Let's go ahead and take a look at some examples.

# Electric Fields in Capacitors - Online Tutor, Practice Problems & Exam Prep

### Intro to Capacitors

#### Video transcript

### Example 1

#### Video transcript

Hey, guys. Let's work this one out together. So we have a capacitor that produces an electric field. We've got these 2 plates with 2 charges, and we're going to have to double the charge half of the area between the plates. And we need to figure out what happens to the electric field. This is going to be one of those proportion reasoning problems. We have to figure out what happens to E. So in other words, E_{new} as we change some variables is going to be what of E_{old}. Right? So this is how we handle these kinds of problems. So let's go ahead and set up what the electric field is between 2 plates. Right? So this electric field is going to be the charge divided by epsilon_{0}, some constant divided by the area as well or times the area, I guess. Just make sure this is in the denominator. Okay. So then what happens is that this charge, right, this Q_{new} now has to become 2 times the original charge. And then you're not going to actually double the both charges together because remember, this Q just represents the charge on each plate. So, in other words, you don't have to, like, multiply it by 4 or anything like that. Right? So it just becomes 2Q. And then this area, so this A_{new} , is just going to become one half of A_{new}. So now let's go ahead and using these two new variables, let's set up our new electric field. So this new electric field, which I'm going to call E_{new}, is just going to be Q_{new} divided by epsilon_{0} times A_{new}. And now all we do is we just substitute these expressions. And oh, I'm sorry. I didn't mean to write A_{new}. That should just be A. So we just have to substitute these expressions in for our new equation right here for E_{new}. So this is just going to be 2Q divided by epsilon_{0} times and then this is going to be 1/2 of A. So now what we do is now we've substituted those expressions in, we just have to see if we can pull all of these factors and constants out to the outsides, and then we're basically just going to extract the original expression. So what happens is this is going to be 2 divided by 1/2. So when you divide by a half, you're going to multiply it by 2. So you're going to have to multiply it by 2 again, and then you're going to get Q divided by epsilon_{0} times A. Now notice that this is just the original expression for E. Right? This is just the original electric field. So in other words, what happens is that we've gotten 4 times the original electric field, and that's our answer. So, if you double the charge and then half of the area between the two plates, then your new electric field is going to be 4 times stronger than the old electric field. And that's basically the answer. That's what happens to the electric field. It multiplies by 4. Alright? Let me know if you guys have any questions about this stuff.

An electron moves into a capacitor at an initial speed of 150 m/s. If the electron enters exactly halfway between the plates, how far will the electron move horizontally before it strikes one of the plates? Which plate will it strike?

^{−6}m

^{−5}m

^{−5}m

^{−4}m

## Do you want more practice?

More sets### Your Physics tutor

- INT FIGURE EX23.25 shows a 1.5 g ball hanging from a string inside a parallel-plate capacitor made with 12 cm×...
- INT Electrostatic cleaners remove small dust particles and pollen grains from air by first ionizing them, then...
- One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting el...
- CP A proton is traveling horizontally to the right at 4.50 * 10^6 m/s. (a) Find the magnitude and direction of...
- CP A proton is traveling horizontally to the right at 4.50 * 10^6 m/s. (b) How much time does it take the prot...
- CP A proton is traveling horizontally to the right at 4.50 * 10^6 m/s. (c) What minimum field (magnitude and d...
- a. Starting from rest, how long does it take an electron to move 1.0 cm in a steady electric field of magnitud...
- INT A proton is fired horizontally into a 1.0×10^5 N/C vertical electric field. It rises 1.0 cm vertically aft...
- A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet han...
- CP A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proto...
- CP A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proto...
- Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. ...
- An electron is released from rest at the center of a parallel-plate capacitor that has a 1.0 mm spacing. The e...