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@eitancahlon 18 сағат бұрын
Where is this question taken from?
@padraiggluck2980 Күн бұрын
👍
@FidiasLautaro Күн бұрын
in 10:13 shouldn't the Z component be 2vx cos(lamda)-g
@JNVbudgam-p8t 2 күн бұрын
thank you sooo much sir ❤❤
@DrBenYelverton Күн бұрын
Thanks for watching, I'm glad it was helpful!
@7177YT 3 күн бұрын
Very cool! Thx!
@ChaosSower99 3 күн бұрын
Thanks doc, you're a life saver. I was trying to find a mistake in my calculation for hours and your video poped up after the search, solving it
@DrBenYelverton 3 күн бұрын
Excellent, happy to help!
@hayanoor-xn5vh 3 күн бұрын
What is the free charge here?, Where it will be,1) inside the dielectric, if yes then it will be a conductor not a dielectric or 2) the outside charges that produce es the polarization in the dielectric, but why the name free?
@DrBenYelverton 3 күн бұрын
Your option 2) is correct. The term can potentially be a bit misleading as the charges are not necessarily free to move, but that's the standard terminology!
@ChiruQuestZone 5 күн бұрын
OMG!!! Such an amazing explanation!!! Have'nt found a video that better explains this concept in such an easy way! Thanks a lot!!
@DrBenYelverton 3 күн бұрын
Happy to hear that, I'm glad it was helpful!
@kamilwilk2417 6 күн бұрын
Electronics & Telecomunication student here, while explanation is pretty clear and i understand where does it come from, im kinda confused. We had to made a model of cable with two parallel wires and find value of capacitance. The model is maded in FEMM and using values from program i get different value of capacitance and now im curious which value is closer to be real. Is there any way to find out? Using values from femm, im just dividing charge over voltage difference between two wires, is that good way or not?
@DrBenYelverton 3 күн бұрын
My first thought is that maybe the approximation d >> a may not be valid in your model. Have you compared your value against the exact theoretical value involving cosh⁻¹? See: kzbin.info/www/bejne/jmmXfomCmpicppI
@erwinmanalastas5827 6 күн бұрын
My own understanding H = 1.028334744 meters .
@umylten4142 6 күн бұрын
Nice one! As you said it's doable using Newton's 2nd law, but tbh I prefer lagrangian mechanics for these type of problems (easier to manage the fact that the FOR relative to the slope is non-inertial), and it shows just how powerful it is here. Same stuff with coupled oscillators (springs, pendulums... we often had those in exams, I think once there was even a charged pendulum or smtg lol)
@DrBenYelverton 3 күн бұрын
I agree! Thanks for watching.
@anamariag8049 7 күн бұрын
11:00
@josephanthonycohenbrownsto8378 7 күн бұрын
I don't follow why you multiply |P| by delta x, delta y and delta z. Shouldn't it only be P times separation which is delta y. So shouldn't it only be |P|*delta y ?
@DrBenYelverton 3 күн бұрын
The polarisation P is defined as dipole moment per unit volume, so we need to multiply P by the volume of the cuboid to get the dipole moment.
@MissPiggyM976 8 күн бұрын
Well done, many thanks!
@es4617-f6m 8 күн бұрын
is it coincidence that the plane of potential zero is a sphere? If I had for example a grounded cube, how would that translate to the problem?
@DrBenYelverton 8 күн бұрын
Well, not a total coincidence as we've chosen the charge and distance to be just right to make the potential zero on that sphere. But I suppose we are lucky in the sense that a spherical equipotential exists in the first place. It would be impossible to apply the method of images in the case of a cube - I can't imagine any arrangement of charges that would give a cube-shaped equipotential!
@alexanderson8701 8 күн бұрын
I'm retired, and I'm using ( - wasting ?) my time going over my ol' University-Physics lecture-notes. ... I was stuck for *days* on a page in my notes about the Capacitance of Parallel-Wires, and the books didn't help. I tried the Internet, but *nothing* worked/helped me until I watched *your* presentation. ... *Many* thanks. - Now I can *finally* turn-that-page -----> [ - I'm *now* a Subscriber.]
@DrBenYelverton 8 күн бұрын
Sounds to me like a great way to spend your retirement! I'm glad the video helped. Feel free to let me know if there's anything else you get stuck with as I'm always looking for new video ideas.
@murtazavepari8937 8 күн бұрын
What if the m mass was performing rolling motion?
@DrBenYelverton 8 күн бұрын
You'd just need to modify the kinetic energy term to include rotational KE.
@aman6782 9 күн бұрын
Muchas gracias señor por la video, por el son mí ppt topic en mi universidad, gracias por la video
@DrBenYelverton 9 күн бұрын
I'm glad the video was helpful!
@edcoad4930 9 күн бұрын
Glorious video! Took me a moment to understand the problem with a circular mass on the slope rather than a block because all I was thinking was "rolling". Lagrangian mechanics and Feynman's trick are so very satisfying indeed.
@DrBenYelverton 9 күн бұрын
Thanks, I'm glad you enjoyed this!
@drscott1 9 күн бұрын
👍🏼
@monaipsm 9 күн бұрын
absolutely a wonderful way to solve a seemingly complex question
@ibragimshamsiyev_ 9 күн бұрын
Great video. What's this book?
@DrBenYelverton 9 күн бұрын
Thanks! I've seen various versions of this problem over the years, not sure where it first appeared.
@atila8623 10 күн бұрын
Nice
@queenabeskuma7443 11 күн бұрын
Notes: 1) Divergence of D is zero is analogous to divergence of B is zero, so there are no sources or sinks and lines must be in close loops. 2) curl of D is curl of P, in magnetostatics ( which means there is no time dependance on E), curl of B is miu_0 J. But J is not curl of something. So this means the D loop is going to circulating around "the curl of P", as B loop is circulating around J vector. 3)In terms of the relations B is more like E, as D is more like H. Where B and E are considering the effect of everything, D and H are only about free charges and free current.
@queenabeskuma7443 11 күн бұрын
Thank you for making this video. It is so helpful! I have one small questions. You said that when a vector field has zero divergence and zero curl then it is zero, does that have other constraint? For example, (2𝑥,−2𝑦) has zero divergence and curl.
@Googleeknowledge1111 13 күн бұрын
Thank you for giving us free information
@DrBenYelverton 12 күн бұрын
I'm glad it's helpful!
@mr.gyaneshwaran5992 14 күн бұрын
brilliant explanation! thanks a ton! it can be used to explain levitating liquids.
@DrBenYelverton 12 күн бұрын
Indeed, an interesting application!
@how-about-no_56 18 күн бұрын
Hey, can you take up leaky capacitor in your next video? Thanks!
@DrBenYelverton 16 күн бұрын
Maybe! What would you like to learn about them specifically?
@how-about-no_56 Күн бұрын
@DrBenYelverton Just why they leak and some numerical problems, i you could.
@ÖzgürCan-f7f 20 күн бұрын
I have tried to understand for 1 month. You just save me. Millions time thank you Sir. 🙏🙏🙏
@DrBenYelverton 16 күн бұрын
Excellent, that's great to hear!
@BharathKumar-ox4fx 20 күн бұрын
Maxwell's assumption or logic must have been explained in the beginning.
@juniorcyans2988 22 күн бұрын
I really enjoy such kind of problems, because I can learn a lot by just solving one single problem. Thank you very much!
@DrBenYelverton 16 күн бұрын
I'm glad this was helpful! Thanks for watching.
@Math1729_ 23 күн бұрын
What if the particle was on a hemisphere instead such that the hemisphere is kept on a frictionless surface so the hemisphere would also move, so how can we approach that problem?
@DrBenYelverton 17 күн бұрын
You'd need to modify the kinetic energy term in the Lagrangian to account for the motion of the hemisphere. Will try to cover this in a future video!
@nushratmahjabeen4305 24 күн бұрын
Thank you for the splendid explanation. Well, where would I make changes if the ball rolls AND slides too?
@DrBenYelverton 17 күн бұрын
There are two major differences in this case - firstly, the relation v = rω no longer applies, and secondly, you need to account for energy losses due to sliding friction. If you know the coefficient of friction you can solve the problem numerically, but I'm not sure if it can be done analytically!
@rummyrummyrum76 24 күн бұрын
Can you conserve angular momentum with the force of gravity acting on the ball's center of mass a distance R away from the step edge axis of rotation, would that not count as torque? and void the conservation of angular momentum about that axis?
@DrBenYelverton 16 күн бұрын
There's a torque due to the weight, but it produces a negligible angular impulse during the collision since the time interval is small, so we can equate the angular momenta immediately before and after the impact.
@geon79 24 күн бұрын
Very interesting video, but there is a catch: I don't agree with one approximation you are making. It is true that ω is small, but it also true that the radius of Earth is much larger than the height of the tower! It means that the centrifugal acceleration isn't negligeable at least at mid latitudes. Let's calculate the centrifugal force in function of the latitude, considering that the height of the tower is negligable compared to the radius of Earth (Re) and taking Earth as a perfect sphere. The distance from the axis of Earth |r|=Re·cos(λ), so the vector r is [0,−Re·sin(λ)·cos(λ),Re·cos²(λ)]. Plugging the vectors r and ω into the expression of the centrifugal acceleration we obtain a(centr)= −ω x (ω x r)= [0,−Re·ω² ·sin(λ)·cos(λ),Re·ω² ·cos²(λ)] Let's plug in some numbers: |ω|=7.292E-5 rad/s, Re≈6.37E6 m and let's consider the case of λ=45°, then a(centr)=[0,−0.0169, 0.0169] m/s². The x component is always 0 and the z component is much smaller than g, so it can be ignored. We are left with a negative y component, though, that accelerates the particle towards the equator. How does it compare to Coriolis acceleration? For a tower 100 m high, vz at the end of the fall is −44.3 m/s, so the x component Coriolis acceleration reaches aₓ(Cor)=−2·ω· vz · cos(λ) =0.00457 m/s², which is actually ~3.7 time smaller than the y component of the centrifugal force! If my calculations are correct, at mid latitudes, a falling particle is more deflected towards the equator by the centrifugal force than it is to the East by the Coriolis force. Notice, however, that the y component of the centrifugal force tends to 0 as we get closer to the equator, while the Corriolis force does not, so at low latitudes the latter prevails and becomes the only source of deflection at the equator.
@omarazami7377 24 күн бұрын
Terrific video!! Griffiths only allow us to visualize uniform polarization (and therefore only surface bound charges). It makes perfect sense that if the polarization is changing (I.e. has a derivative) then charges “don’t cancel out” internally. I’m going to show this to a student I’m tutoring. Thank you again!!
@DrBenYelverton 17 күн бұрын
Excellent, I'm glad it was helpful!
@ivarszickus4570 25 күн бұрын
I understand the case where the sphere is grounded. There you can set V=0 and you have clear boundary conditions for your Poisson equation. However I completely fail to understand the case where the sphere isn't grounded. What are the boundary conditions here? I thought the point was to take some image charges that yield the same boundary conditions for the Poisson equation but these conditions are nowhere mentioned. How would we know what the potential on the surface of the non-grounded conductor is in advance? Isn't this essential information that we need before we can even start using the method of images? I've breen trying to understand this for days and it's really driving me mad.
@barnabasperenyi1785 16 күн бұрын
looking for the same answer here
@ivarszickus4570 16 күн бұрын
@@barnabasperenyi1785 I think I understand it better now. So we start with an uncharged spherical conductor. We bring a charge Q near it. The charge on the sphere redistributes itself and the surface is now at a constant potential V. We don't know the value of this potential yet, but it doesn't matter I think. In our images setup, we set the first charge q as before and then add a charge q' at the center that will make sure the spherical shell is at this potential V, which we still don't know exactly but whatever it is, there is some carge that gives it. With this setup, the potential and therefore the electric field must be equal in the original setup and the images setup, as long as we're outside the shell. This follows from the unicity of the solution to the Poisson equation with boundary conditions. Now we can use Gauss's Law. Take a Gaussian surface that encloses the conductor but not the charge Q. The flux through this surface must be zero because the sphere has no net charge. Now take the same Gaussian surface in the images setup with the three point charges. There is no conductor here but still the electric field is the same when we are outside the region. Since E is the same everywhere on the surface, the flux through this surface is the same and must also be zero. Since the flux through a closed surface depends only on the enclosed charge, the total charge inside the surface must be zero. Hence q'=-q. From this we would also be able to calculate the potential on the conducting sphere's surface.
@paxshild4924 26 күн бұрын
Hello Dr. Ben, an interesting hypothetical question for you, if we were to teleport a chunk of neutron star mass the size of a golf ball here on earth , would it sink down to the core of the earth, creating earthquakes as its gravitational tidal forces rip through the crust , or will it explode? If it were to explode, it would be interesting for you to make a video on it. In a sense, how do you calculate the energy of that explosion? Its a question for fun. I think the audience might enjoy it.
@DrBenYelverton 17 күн бұрын
I like the idea! But what would cause it to explode?
@paxshild4924 15 күн бұрын
@DrBenYelverton I asked Dr. Christopher S Barid from Texas for a better answer and he had to say it in this way. Neutron star materials are extreme material indeed , masses being packed at ridiculous volumes staked up will have very large amount of gravitational potential energy, if we suddenly teleport a golf ball sized material what weighs roughly about an astroid 6 mile wide wouldn't have enough gravity of it's own to hold that potential energy and will rapidly expand resulting in perhaps a megaton yield explosion
@shrivatsa8604 15 күн бұрын
@@DrBenYelverton I guess it'd be more dramatic if we were to suddenly teleport the center most part of it as the energy density will be maximum as it holds the weight of all the layers. When suddenly teleported to earth the pressure difference will be astronomical resulting in an explosion. The energy can be calculated using mere classical physical as the kinetic energy / neutron makes it move less than 1% the speed of light , also the escape velocity is approximately 300m/s<<< C at its surface (golf ball).
@shrivatsa8604 27 күн бұрын
Hello, sir. I have a question related to the concepts in your recent video. If we take an electron-positron pair as point particles, without a definite radius, their collision doesn’t happen in a classical way, and calculating the annihilation time becomes challenging. Using Coulomb’s law and calculus, we could derive a time-to-distance relation as they accelerate towards each other. However, as they approach close distances, the velocity derived from potential approaches the speed of light, and near the Planck length, Coulomb's law breaks down due to high uncertainty, making exact calculations difficult. Could you explain how to determine the time taken for such a particle-antiparticle annihilation, addressing these relativistic and quantum aspects?
@DrBenYelverton 16 күн бұрын
Both very difficult issues to account for! To handle the quantum mechanics side of things properly, you'd likely need some advanced QED which is outside of my expertise. I'd imagine it's not possible to calculate one specific time though as QM is inherently probabilistic. For the special relativistic correction, this is also not straightforward as you'd need to take into account both electric and magnetic effects. It can probably be done but I'd need to spend a while thinking about it!
@shrivatsa8604 16 күн бұрын
@@DrBenYelverton I appreciate it, If possible, I’d love to know more about your specific area of expertise within physics. I have a range of questions in various fields, and understanding your focus areas would help me tailor my questions to topics where your insights would be most applicable.
@sinecurve9999 28 күн бұрын
So the particle absorbs a photon then emits a virtual photon that then decays into a particle anti-particle pair? What is meant by "in the presence"? These equations are independent of distance between the pair production and particle X. The particle could just as well be in the same atom or halfway across the observable universe, but as long as it somehow receives an impulse from the pair production, conservation of momentum is satisfied. This seems suspect because you would think that these events should be causally connected. The restrictions on what X can be is extremely loose in this framework. I could substitute X with "Milky Way Galaxy" and the math would still be the same. Could X be another photon? It seems all you have shown here is that pair production cannot occur in a universe only containing a single photon if conservation of energy and momentum are to be satisfied.
@DrBenYelverton 27 күн бұрын
The photon decays into a particle-antiparticle pair, then either the particle or antiparticle exchanges a virtual photon with what I called particle X. See e.g. Wikipedia's article on pair production for a Feynman diagram illustrating this more clearly. You're correct that there is lots of physics missing in this analysis! It's a quantum-mechanical interaction between particles, and for a full treatment we'd need to use QED. In your example of a galaxy, the photon would really be interacting with a specific particle somewhere in the galaxy, not the entire galaxy at once. About the distance between the photon and particle X - there is always a probability for the interaction to happen, but of course the probability decreases with separation and eventually becomes negligible. In the case of a nucleus, classically we'd say that the electric field it produces has infinite extent and influences particles at arbitrarily large distances, but in practice the field is so weak at large distances that it's negligible. So, the phrase "in the presence of" does indeed mean that there must be another particle somewhere in the universe, it's just that the process is far more likely to happen the closer that particle is to the photon. Finally, pair production could indeed happen with a second photon instead of a nucleus, provided that it has enough energy. The Feynman diagram is similar, except there is now no particle X and the second photon is not virtual.
@Nxck2440 28 күн бұрын
Woah, I always wondered about this. Never thought I'd see an explanation on exactly this, thanks! Also, do you happen to have a video about how magnetic fields are kind of like electric fields but in a different reference frame? I heard there is a link based on special relativity but it's hard to understand.
@r2k314 28 күн бұрын
search Y.T.: How Induction Helped Einstein Discover Relativity! Physics - problems and solutions
@DrBenYelverton 27 күн бұрын
Excellent, this is certainly a very specific topic and I'm glad it was helpful! I haven't covered the relationship between electric and magnetic fields in relativity yet, but it's something I've been meaning to do for a while.
@NerdZEY 29 күн бұрын
"Why are you learning this?" A robotics enthusiast: Nothing really important (CiWS)
@mausamkalita9375 29 күн бұрын
Oh man, what you did!
@Anselm-wb8we Ай бұрын
thank you
@darwinvironomy3538 Ай бұрын
can you do about geothermal model of the earth probably? using heat equation on steady state with a core that generates heat and a mantle. is it reasonable?
@DrBenYelverton 28 күн бұрын
I suppose the complication is that you probably need to account for convection in the mantle!
@darwinvironomy3538 27 күн бұрын
@@DrBenYelverton i see!
@mrcooper3139 Ай бұрын
How would the transmission coefficient, which shows the probability of penetration for nuclei, can be used to solve this problem?
@mohsenrezaei5965 Ай бұрын
Can you tell what force keeps the bead on the top point of equilibrium? theta=pi. thank you for this solution
@shrivatsa8604 Ай бұрын
Hello sir, I have read your website and found the thesis of yours on the topics the influence of planetary and stellar companion and debris disks. Quite interesting. So, could you please tell me an overview on what is this about?
@DrBenYelverton Ай бұрын
So, in other planetary systems it is often much easier to detect debris discs than it is to detect the planets themselves. Essentially this is because the asteroids that make up the discs gradually grind themselves down into small dust particles through collisions, and the total cross-sectional area of the many dust particles is much greater than the cross-sectional area of a typical planet. By studying the structures of the discs (e.g. gaps and asymmetries) we can infer things about what sort of otherwise-undetectable planets might be present. I spent a lot of time during my PhD thinking in particular about how multiple-planet systems can open gaps in a disc through a mechanism called secular resonance, which occurs when the orbits of the asteroids and planets precess at the same frequency and the asteroids' orbits become very eccentric as a result.
@adityk8138 Ай бұрын
Why we use this relation in band brake and rope pulley problem both while in band brake, pure slipping is there but in case of rope-pulley pure rolling is there between rope and pulley. This result is better suits for band brake(because pure slipping is there i.e maximum value of friction u×R is generated). Please help through video. Please. louuu from india.
@phyarth8082 Ай бұрын
Nice property of wave-particle duality de Broglie equation to reduce approx. d=1 fm result. I calculated static distance exactly for hydrogen atom using Casimir effect and this distance for hydrogen atom is d=0.583 fm.