Ch 25: Capacitors & DielectricsWorksheetSee all chapters
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Ch 02: 1D Motion / Kinematics
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Ch 21: The First Law of Thermodynamics
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Ch 25: Capacitors & Dielectrics
Ch 26: Resistors & DC Circuits
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Example #1: Capacitance of Spherical Capacitor

Transcript

Hey guys. Let's do an example. What is the capacitance of two 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 two 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 k, q over r squared r hat. So, our integral is going to look like the integral from A to B of k, q over r squared, r hat 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 k, q over r squared, d, r, okay? And it's a really easy integral, right? It's 1 over r squared, which is negative 1 over r, so this is going to be positive k, Q, 1 over r from A to B, this is going to be k, Q, 1 over B minus 1 over a. Alright, I'm going to 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 going to write it in a different way because most books include it in a different way k, Q I'm going to find the least common denominator which is a, b, so this is going to a over a, b minus b over a, b, that's going to be k, Q times a minus b over a, b, okay? And I wrote it like that because this is how most books are 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 k, Q, a minus b over a, b and you see that those Q's cancel, this is 1 over k a, b over a minus b and if you remember that k is 1 over 4pi epsilon naught, just remember that relationship, this is 4pi epsilon naught a, b over a minus b, that is the capacitance of these concentric spherical shells. Alright guys, that's it thanks for watching.

Example #2: Capacitance of Cylindrical Capacitor

Transcript

Hey guys. Let's do an example. What is the capacitance per unit length of two 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 dot 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, in that electric field is going to be k, sorry, 2 k lambda over r, r hat. So, our integral is going to look like from a to b, 2k lambda over r, r hat dotted into d, r, r hat as, we're doing the radial direction. So, this whole thing is going to look like the negative integral from a to b of 2k lambda over r, d, r, okay? And once again, this is also very easy integral 1 over r is just the log, so it becomes negative 2k lambda, ln of r from a to b, this whole thing is going to be negative 2k lambda, ln of b minus ln of a, we can combine these two by saying, that's b divided by a. So, it's a negative 2k lambda ln of b divided by a, 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 2k, lambda ln, a divided by b, but we're not done, we still need to find the capacitance, the capacitance is going to be Q over V, let me get myself just a little more space, it's Q over V, which is going to be Q over 2k lambda ln, a over b. Now, what is lambda? Lambda is the charge per unit length, right? So, this is going to be Q over 2k, Q over L, ln a over b. So, those Q's cancel, the L comes into the numerator and this is going to be L, 2k, ln, 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 the 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 1 over 4pi epsilon naught, then 1 over 2k becomes 2pi epsilon-naught and this is over ln, a over b, and that is the capacitance per unit length of infinitely long concentric cylindrical shells. Alright guys, thanks for watching.