A brief introduction to capacitors and capacitance for students in introductory calculus-based physics courses such as AP Physics C. For more information, please visit APlusPhysics.com
It disappeared! (OK, I think I'm funny, you already knew that). I should have specified in the video, but because we're focusing only on the magnitude of the potential difference, knowing V would be used in the capacitance equation, I neglected to follow it through. Negative potential going one way across the parallel plates would be a positive potential going the opposite direction, or start at the oppositely charged plate, you still get the same answer. Apologies for not explaining better!
hi this is an awesome video! Thank you for making it. Im just wondering why you integrate from RA to RB instead of from RB to RA for cylindrical and spherical capacitors.
Practically we're talking about the same thing. V is typically used for "potential difference," which is the same as Delta-V, change in electric potential.
At 12:23 when you make the substitution because you know that Ra is less than Rb, it is possible to leave the equation the way it is (1/Rb - 1/Ra) instead of simplifying?
In the last example, when solving for electrical field, why do you not use (charge density symbol) * (volume), similar to your previous examples, and instead keep q in the equation?
yerbah1 the integral of E in respect to L from positive infinity to L can be written as the integral of (KQ) / (R^2) in respect to d from positive infinity to D. And the integral of it is -(KQ) / R assuming potential infinite distance away is 0. which cancels the first negative.
Thanks for the video, but something seemed to go wrong around 5:30. You said lambda=Q/L, then wrote sigma*L=Q, then said "sigma" while you wrote lambda. And shouldn't it be sigma since the cylinder has a two dimensional surface?
You are correct, I should have stuck with lambda (linear charge density) throughout. And for our purposes of deriving the electric field with Gauss's Law, we can treat the inner wire as a line charge, since we're only worried about the electric field between the two. Good catch!
At 2:53, where does the negative sign go when you replace E?When we are calculating electric field, should we use formula k•Q1/d2 for point charge and density/ E0 for plate charge? Or we can use both of them.
Hi pengrun. I was focused on the magnitude so I didn't worry about it in this problem. density/e0 comes from a derivation of kq1/r, so they're mathematically equivalent.
How come we use the linear charge density for the conducting rod but not the surface charge density for the conducting sphere when calculating the electric field?
Because it is a lot easier to just use Gauss's Law to find the electric field due to the spherical capacitor, already assuming the charge. In short -- it wasn't required. :-)
Hi dan love the videos i have been binge watching today lol. how come in the problem with the cylindrical capacitors, u treat E as a constant and pull it out of the integral in the first step, but then substitute in lambda over yada yada yada in the second step when you are trying to solve for V.... If you could respond tonight that would be amazing... Thanks!!
Ben Eckardt Hi Ben. We used Gauss's Law to solve for the Electric Field of the cylindrical capacitor. Because we defined a Gaussian surface with appropriate symmetry, the electric field at all points on that surface will be the same. The video on Gauss's Law goes into further detail on the point if you'd like more background info. Thanks, and good luck!
So Mr. Fullerton, I got for the question 2*Pi*L*E0 / ln(Rb/Ra) when I did the integration in the opposite direction (different limits of integration). How and when would I know that I need to multiply the limits by a negative sign, like in here to flip the fraction in the natural log. Thank you, and have a great day!
Imagine the cylinder is a soda can. If we want its area, we could pretend to cut it down its length to make a rectangle. The length of one side of the rectangle is l, the other side is the circumference of the circle, or 2*pi*r
you just saved my day!
you're killing it with that leather jacket
Is that good or bad?
Dan Fullerton It's excellent.
oh, i see. it is nice of you to explain it so explicitly. You are the best instructor i have ever seen, thank you again. happy every day
Thanks Alex -- glad the video helped clarify things!
Glad you're enjoying the videos. And yes, I have heard that before, but I especially appreciate the "young" part! ;-)
I love the videos thank you for posting them
It disappeared! (OK, I think I'm funny, you already knew that). I should have specified in the video, but because we're focusing only on the magnitude of the potential difference, knowing V would be used in the capacitance equation, I neglected to follow it through. Negative potential going one way across the parallel plates would be a positive potential going the opposite direction, or start at the oppositely charged plate, you still get the same answer. Apologies for not explaining better!
can you shed more light on the limits of integration am lost checking my notes they used them in opposite as rb/ra and a lost and confused
Absolutely.
hi this is an awesome video! Thank you for making it. Im just wondering why you integrate from RA to RB instead of from RB to RA for cylindrical and spherical capacitors.
I think that he has them swapped.
It should be ln(RB/RA) as well.
to the second example( 6.02) why 2pai r l comes from dA??thank you
Practically we're talking about the same thing. V is typically used for "potential difference," which is the same as Delta-V, change in electric potential.
Hello Dan,
would it make a difference if we said that capacitance = Q / ΔV instead of Q/V?
At 12:23 when you make the substitution because you know that Ra is less than Rb, it is possible to leave the equation the way it is (1/Rb - 1/Ra) instead of simplifying?
In the last example, when solving for electrical field, why do you not use (charge density symbol) * (volume), similar to your previous examples, and instead keep q in the equation?
Just showing alternate ways / versions of solving the problems.
Near 2:55 you drop the negative off of E, why is that?
Thank you very much
What is the equation of electric field using sigma? and epsilon? I never saw that in my life before
where did the negative go at 3:00 ?
yerbah1 the integral of E in respect to L from positive infinity to L can be written as the integral of (KQ) / (R^2) in respect to d from positive infinity to D. And the integral of it is -(KQ) / R assuming potential infinite distance away is 0. which cancels the first negative.
Thanks for the video, but something seemed to go wrong around 5:30. You said lambda=Q/L, then wrote sigma*L=Q, then said "sigma" while you wrote lambda. And shouldn't it be sigma since the cylinder has a two dimensional surface?
You are correct, I should have stuck with lambda (linear charge density) throughout. And for our purposes of deriving the electric field with Gauss's Law, we can treat the inner wire as a line charge, since we're only worried about the electric field between the two. Good catch!
At 2:53, where does the negative sign go when you replace E?When we are calculating electric field, should we use formula k•Q1/d2 for point charge and density/ E0 for plate charge? Or we can use both of them.
Hi pengrun. I was focused on the magnitude so I didn't worry about it in this problem. density/e0 comes from a derivation of kq1/r, so they're mathematically equivalent.
okay, thank you.
is there a missing negative sign at 3:22?
How come we use the linear charge density for the conducting rod but not the surface charge density for the conducting sphere when calculating the electric field?
Because it is a lot easier to just use Gauss's Law to find the electric field due to the spherical capacitor, already assuming the charge. In short -- it wasn't required. :-)
Hi dan love the videos i have been binge watching today lol. how come in the problem with the cylindrical capacitors, u treat E as a constant and pull it out of the integral in the first step, but then substitute in lambda over yada yada yada in the second step when you are trying to solve for V....
If you could respond tonight that would be amazing... Thanks!!
Ben Eckardt Hi Ben. We used Gauss's Law to solve for the Electric Field of the cylindrical capacitor. Because we defined a Gaussian surface with appropriate symmetry, the electric field at all points on that surface will be the same. The video on Gauss's Law goes into further detail on the point if you'd like more background info. Thanks, and good luck!
Thanks Dan! you're the man... wish u were my teacher!
What he says at 13:48 is gold
Hi Zhixuan. No particular reason to go from Ra to Rb instead of the other way around... you could go the other direction if you wanted.
Wouldn't it make a difference? If we flip the limits of integration that would require us to multiply the whole integral by -1...?
littledivergirl1105 Yes. Assuming you're focusing on the magnitude, however, not a major concern.
Thank you! I had the same question myself.
So Mr. Fullerton, I got for the question 2*Pi*L*E0 / ln(Rb/Ra) when I did the integration in the opposite direction (different limits of integration). How and when would I know that I need to multiply the limits by a negative sign, like in here to flip the fraction in the natural log.
Thank you, and have a great day!
If you get a negative number for the answer, take the absolute value so you have a positive capacitance.
Why do you use the integral negative E.dl to find voltage. Why not keep it positive?
Because that's the formula and the physics (can't change the sign just to keep things positive, have to follow the rules...)
Dan Fullerton Hello Mr. Fullerton, in my textbook this comes up as -integral of E*dr, is dl the same thing?
i mean why there is a L, i know the derivative of the area of that circle is 2 pai r, thank you for replying
Imagine the cylinder is a soda can. If we want its area, we could pretend to cut it down its length to make a rectangle. The length of one side of the rectangle is l, the other side is the circumference of the circle, or 2*pi*r
Goat even a decade later