Capacitance Using Calculus - Online Tutor, Practice Problems & Exam Prep

To find the capacitance of two concentric spherical shells with radii a and b (where a < b), consider the charge on each sphere as ±q. The capacitance (C) is defined as the charge (q) divided by the potential difference (V). The electric field (E) between the shells is given by Gauss's law as E = kq/r². The potential difference is calculated using the integral of the electric field, resulting in V = kq(1/b - 1/a). Thus, the capacitance is expressed as $=\frac{4\pi \epsilon (ab)}{(a-b)}$.

Capacitance of Spherical Capacitor

Hey, guys. Let's do an example. What is the capacitance of 2 concentric spherical shells? One of radius a and one of radius b with a less than b. Consider the charge on each sphere to be plus or minus q. Alright. 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 going to 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 r direction, the radial direction. Okay? What is the electric field going to be between these 2 spherical shells? Well, if I look at an arbitrary point between them, it's only going to be due to the inner sphere. That's what Gauss's law tells us. That is going to be kq over r squared r hat. So our integral is going to look like the integral from dotted into dr r hat. Just in the radial direction. That's what we're integrating. Alright. So this whole thing looks like the negative integral of a to b, kq or r squared, d r. Okay? And it's a really easy integral. Right? It's 1 over r squared which is negative one over r. So this is gonna be positive kq over r from a to b. This is gonna be kq over b minus 1 over a. Alright. 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 a different way because most books include it in a different way. Kq, I'm gonna find the least common denominator which is ab. So this is gonna be a over ab minus b over ab. That's gonna be kq times a minus b over ab. Okay? And I wrote it like that because this is how most books are gonna 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 kq(a-b) over ab. And you see that those q's cancel. This is 1 over k(ab) over (a-b), and if you remember that k is 1 over 4πε₀, just remember that relationship, this is 4πε₀ab over (a-b). That is the capacitance of these concentric spherical shells. Alright guys, that's it. Thanks for watching.

Capacitance of Cylindrical Capacitor

Hey, guys. Let's do an example. What is the capacitance per unit length of 2 concentric infinitely long cylindrical shells? One of radius a and one of radius b with a less than b. 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 b. Okay? 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 going to be the charge per unit voltage. So what we have to do is find the voltage between these two cylinders, right between this distance right here, and then divide the charge by that. Okay? That potential difference, that voltage, is going to be negative integral of e.dx. Okay? In whatever direction. Once again, we're working in the radial direction. Now between these two, the electric field is only going to depend upon the inner cylinder. That's what Gauss's law says and that electric field is going to be k, sorry, 2kλ over r &hat;{r}. So our integral is going to look like from a to b, 2kλr &hat;{r} dotted into d&hat;{r} because we're doing the radial direction. So this whole thing is going to look like the negative integral from a to b of 2kλdr. Okay and once again this is also a very easy integral, one over r is just the log. So this becomes -2kλln(r) from a to b. This whole thing is going to be -2kλ(ln(b)-ln(a)). We can combine these two by saying that's b divided by a, so it's -2kλlnba. Okay. Let me give myself some breathing room here. Another trick that we can do to simplify this because everyone's going to do it is we can bring this negative inside the log and that's just going to reciprocate the division. So this is going to be 2kλlnab. But we're not done. We still need to find the capacitance. The capacitance is going to 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 2kλlnab. Now what is λ? λ is the charge per unit length. Right? So this is going to be q over 2kqoverlnab. So those q's cancel, the l comes into the numerator and this is going to be l over 2klnab. 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 that this is an infinitely long cylinder which means that l is infinity which means that the capacitance is also infinity. But the capacitance per unit length is not infinity. If you remember that k is 14πε_0, then 12k becomes 2πε_0 and this is over lnab. And that is the capacitance per unit length of infinitely long concentric cylindrical shells. Alright guys, thanks for watching.