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Ch 27: Capacitors & DielectricsWorksheetSee all chapters
All Chapters
Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Kinematics
Ch 05: Projectile Motion
Ch 06: Intro to Forces (Dynamics)
Ch 07: Friction, Inclines, Systems
Ch 08: Centripetal Forces & Gravitation
Ch 09: Work & Energy
Ch 10: Conservation of Energy
Ch 11: Momentum & Impulse
Ch 12: Rotational Kinematics
Ch 13: Rotational Inertia & Energy
Ch 14: Torque & Rotational Dynamics
Ch 15: Rotational Equilibrium
Ch 16: Angular Momentum
Ch 17: Periodic Motion
Ch 19: Waves & Sound
Ch 20: Fluid Mechanics
Ch 21: Heat and Temperature
Ch 22: Kinetic Theory of Ideal Gases
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
Ch 25: Electric Force & Field; Gauss' Law
Ch 26: Electric Potential
Ch 27: Capacitors & Dielectrics
Ch 28: Resistors & DC Circuits
Ch 29: Magnetic Fields and Forces
Ch 30: Sources of Magnetic Field
Ch 31: Induction and Inductance
Ch 32: Alternating Current
Ch 33: Electromagnetic Waves
Ch 34: Geometric Optics
Ch 35: Wave Optics
Ch 37: Special Relativity
Ch 38: Particle-Wave Duality
Ch 39: Atomic Structure
Ch 40: Nuclear Physics
Ch 41: Quantum Mechanics
Sections
Capacitors & Capacitance
Parallel Plate Capacitors
Energy Stored by Capacitor
Capacitance Using Calculus
Combining Capacitors in Series & Parallel
Solving Capacitor Circuits
Intro To Dielectrics
How Dielectrics Work
Dielectric Breakdown

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.