Capacitance Using Calculus

Hey, guys, let's do an example. What is the capacitance of two concentric spherical shells? One of radius A and one of radius. Be with a less than be consider the charge on each sphere to be plus or minus que All right, remember that the capacitance mathematically is going to be the charge divided by the potential difference. Okay, for this arbitrary arrangement, what we need to do is find the potential difference between these two plates right here. In order to do that, we're gonna use our calculus equation that it's just the electric field dotted into our direction that we're looking at in this case, we're looking at the our direction, the radial direction. Okay, what is the electric fields gonna be between these two spherical shells? Well, if I look at an arbitrary point between them, it's on Lee going to be due to the inner sphere. That's what God's law tells us that is going to be que que over r squared our hat. So are integral is gonna look like the integral from a to B of cake over r squared our hat dotted into d. R r half just in the radial direction That's what we're integrating. All right, So this whole thing looks like the negative integral of a to be cake. Your r squared D r. Okay, that's a really easy integral right. It's one over r squared, which is negative. One over R. So this is gonna be positive. Que one over our from a to B. This is gonna be cake one over B minus one, over a. All right. I'm gonna give myself a little bit of room here. Now, you can leave this exactly like this. We're not done yet, But you can leave this answer like this. I'm just gonna write it in a different way because most books included in a different way Que que I'm gonna find the least common denominator, which is a B. So this is gonna be a over a B minus B over a B. That's gonna be cake times a minus, bi over a b. Okay. And I wrote it like that because this is how most books we're going to write this. Now. What we need to do is we need to find the capacitance, which is the charge per unit. Voltage. So this is Q over que que a minus bi over a B. You see that? Those accused cancel This is one over K A B over a minus B. And if you remember, that K is 1/4 pi. Absolutely not. Just remember that relationship. This is four pi s l A not a B over a minus bi. That is the capacitance off these concentric spherical shells. Alright, guys, that's it. Thanks for watching.

Hey, guys, let's do an example. What is the capacitance per unit? Length off. Two concentric, infinitely long cylindrical shells, one of radius A and one of radius. Be with a less than be Consider the charge on each cylinder to be plus or minus. Q. So what they're talking about is we have some cylindrical shell that's infinitely long with Radius A and some other concentric cylindrical shell that's also infinitely long of radius be and they have plus and minus Q. And this will form a capacitor. So what we want to do is find the capacitance. Now the capacitance is gonna be the charge per unit. Voltage. So what we have to do is find the voltage between these two cylinders between this distance right here and then divide the charge by that. Okay, that potential difference. That voltage is gonna be negative, integral of e dot dx. Okay, in whatever direction, once again, we're working in the radial direction. Now, between these two, the electric field is on Lee going to depend upon the inner cylinder. That's what gasses lost says. And that electric field is going to be K sorry to K lambda over our or hat So our integral is gonna look like from a to B two k lambda over our our hat dotted into D are are happy because we're doing the radial direction. So this whole thing is gonna look like the negative Integral from A to B of two K lambda over r D R. Okay, And once again, this is also a very easy integral one. Over R is just the log. So this becomes negative. Two k Lambda Ellen of our from A to B. This whole thing is gonna be negative. Two k lambda Ln of B minus, Ellen of A. We can combine these two by saying that's be divided by a So it's negative two K Lambda Elena be divided by a Okay, let me give myself some breathing room here. Another trick that we could do to simplify this because everyone's gonna do it is we can bring this negative inside the log, and that's just gonna reciprocate the division. So this is going to be two k Lambda Ellen, a divided by beef. But we're not done. We still need to find the capacitance. The capacitance is gonna be Q over V. Let me give myself just a little bit more space. It's Q over V, which is going to be Q over to. Okay, Lambda Ellen A over B. Now, what is Lambda Lambda is the charge per unit Length. Right. So this is gonna be Q over to K Q over L Ln a over b So those queues cancel the l comes into the numerator, and this is going to be L two k. Ellen A over B. So what I need to do is I need to divide this l over. I'm looking for the capacitance per unit length. The reason is is because this is an infinitely long cylinder, which means that Ellis infinity, which means that the capacitance is also infinity. But the capacitance per unit length is not infinity. If you remember, that K is 1/4 Pi epsilon. Not then 1/2 K becomes two pi epsilon. Not, And this is over, Ellen. A over B and that is the capacitance per unit. Length off infinitely long concentric cylindrical shells. All right, guys, Thanks for watching