Prof. Anderson I have a question about the approximation process. Why does the s/2 in the denominator cancel but the s or s/2 in the numerator either stays as s or is simplified to s. If s is small enough to cancel in one part why not both parts? And also thank you so much for your videos! You are a beacon to of light on the shores of understanding to those who have capsized in an ocean of mediocre physics teachers. Cheers!
I realize this is probably late for you, but just in case somebody else comes across this video and has the same question: You don't want to get rid of the small terms too soon, otherwise you can get some nonsense results. For example look at about 7 minutes at the equation at the bottom of the screen: you can't just say "well s/2 is much smaller than y so we can ignore it", because then you get y^2 - y^2, which is a zero in the numerator and thus the whole dipole would be zero. You still need the s terms because they multiply y and thus have an effect on the overall equation. If you want to be more a bit more rigorous, think about it this way: if you multiply something small by something else that is small (like itself), you get something which is *really* small and you can neglect it. That's what happens when you square s in this video. You can't really do that for s*y, because y is big and can't get canceled out by s. If you look at about 10:10, the reason you are only left with y^2 there is that (s/2)^2 is neglected because it's really small, while the cross terms cancel out (which they don't at minute 7). This is probably why it's so confusing, Prof. Anderson doesn't really go into detail at that point and so you might not see it, but if you expand out the terms and only get rid of the really small ones, you get the correct equation. Don't know if this was clear enough, the short version is that you square the parentheses and only cross out the (s/2)^2 terms (because if you square something small you get something very small, thus negligible), while the s*y cross terms sometimes cancel, but sometimes don't. And that's how you get those formulas.
@@bogdancorobean9270 While looking at my lectures, I noticed that they did this a ton and never really had someone explain why it was happening so I was left to guessing. So thank you for taking your time to explain this out. Cheers.
Each direction has to have it's own electric field equation, Ex, Ey and Ez. Both Ex and Ez are = 0. There are not charges in the X direction, and there are no charges in the z direction.
But if y >> s then dont that mean the charge is the sum of the nearby charges that is essentially 0. So how are we allowed to carry on with the calculation without putting q as 0+, thereby stopping us from ariving at anything other than 0+. How do we know this is a legal assumption that we can use the original charge at such distance?
Kaplan’s mcat prep booklet says ““One very important equipotential line to be aware of is the plane that lies halfway between +q and -q, called the perpendicular bisector of the dipole. Because the angle between this plane and the dipole axis is 90° (and cos 90° = 0), the electrical potential at any point along this plane is 0” .. you said at the green point the x components will cancel, and this the electric potential will be zero. They seem perpendicular to the dipole axis. Is this saying the same thing or no? I just don’t know why the potential difference would be zero perpendicular to the dipole. When i look at the electric field lines of a dipole it seems like work can be done if you through a test charge on any point along the perpendicular axis. It would accelerate towards the negative side assuming test charge is positive. Anything will help. Thanks
Yes, potential is zero along the perpendicular bisector of the dipole. This, of course, does not mean the electric field is zero, since the electric field is the negative gradient of the potential. Cheers, Dr. A
In his example at ruclips.net/video/Yz_tYsH-5gY/видео.html he explains that the field on the x-axis will be in the negative x direction because the y-components will cancel and the x-components will essentially be double but in the negative x direction in relation to the coordinate plane. Hope that helped even tho 4 months after you probably already finished your class haha :)
Walter Lewin and Matt Anderson are the only physics teachers that showed me the beauty of the subject!
Thanks!
This was the best explanation I've ever seen !!!
Thank you professor Anderson!
Your follower from Iraq, well explained, Professor ❤
man, you are so good, bless you
Thank you so much!!! I finally got it🤩
wish my university have this kind of physics professor
Prof. Anderson I have a question about the approximation process. Why does the s/2 in the denominator cancel but the s or s/2 in the numerator either stays as s or is simplified to s. If s is small enough to cancel in one part why not both parts?
And also thank you so much for your videos! You are a beacon to of light on the shores of understanding to those who have capsized in an ocean of mediocre physics teachers. Cheers!
I realize this is probably late for you, but just in case somebody else comes across this video and has the same question:
You don't want to get rid of the small terms too soon, otherwise you can get some nonsense results. For example look at about 7 minutes at the equation at the bottom of the screen: you can't just say "well s/2 is much smaller than y so we can ignore it", because then you get y^2 - y^2, which is a zero in the numerator and thus the whole dipole would be zero. You still need the s terms because they multiply y and thus have an effect on the overall equation.
If you want to be more a bit more rigorous, think about it this way: if you multiply something small by something else that is small (like itself), you get something which is *really* small and you can neglect it. That's what happens when you square s in this video. You can't really do that for s*y, because y is big and can't get canceled out by s. If you look at about 10:10, the reason you are only left with y^2 there is that (s/2)^2 is neglected because it's really small, while the cross terms cancel out (which they don't at minute 7). This is probably why it's so confusing, Prof. Anderson doesn't really go into detail at that point and so you might not see it, but if you expand out the terms and only get rid of the really small ones, you get the correct equation.
Don't know if this was clear enough, the short version is that you square the parentheses and only cross out the (s/2)^2 terms (because if you square something small you get something very small, thus negligible), while the s*y cross terms sometimes cancel, but sometimes don't. And that's how you get those formulas.
@@bogdancorobean9270 While looking at my lectures, I noticed that they did this a ton and never really had someone explain why it was happening so I was left to guessing. So thank you for taking your time to explain this out. Cheers.
@@bogdancorobean9270 Thank u sm sir!
You are just the best.. THANK YOU SO SO SO MUCHHHHH
Lina Berkani,
You're very welcome. Glad you're enjoying the videos.
You might also like my new site: www.universityphysics.education
Cheers,
Dr. A
yoooo Doctor Strange
But what about 3 dimensions? How do we write this final equation using (x, y, z)-axis in a arbitrary P position? Awesome lecture by the way.
Each direction has to have it's own electric field equation, Ex, Ey and Ez. Both Ex and Ez are = 0. There are not charges in the X direction, and there are no charges in the z direction.
But if y >> s then dont that mean the charge is the sum of the nearby charges that is essentially 0. So how are we allowed to carry on with the calculation without putting q as 0+, thereby stopping us from ariving at anything other than 0+. How do we know this is a legal assumption that we can use the original charge at such distance?
Thank You sir......
Kaplan’s mcat prep booklet says ““One very important equipotential line to be aware of is the plane that lies halfway between +q and -q, called the perpendicular bisector of the dipole. Because the angle between this plane and the dipole axis is 90° (and cos 90° = 0), the electrical potential at any point along this plane is 0” .. you said at the green point the x components will cancel, and this the electric potential will be zero. They seem perpendicular to the dipole axis. Is this saying the same thing or no? I just don’t know why the potential difference would be zero perpendicular to the dipole. When i look at the electric field lines of a dipole it seems like work can be done if you through a test charge on any point along the perpendicular axis. It would accelerate towards the negative side assuming test charge is positive. Anything will help. Thanks
Yes, potential is zero along the perpendicular bisector of the dipole. This, of course, does not mean the electric field is zero, since the electric field is the negative gradient of the potential.
Cheers,
Dr. A
@@yoprofmatt Hey Dr. A. Thanks for the reply! I really appreciate.
15:28 why do we add
Im really impressed you can do that left handed and backwards
York,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
Hello Sir, I have a question. At 26.47 you told the E(x) is negative as it is pointing down. But why is it? Why is it negative?
In his example at ruclips.net/video/Yz_tYsH-5gY/видео.html he explains that the field on the x-axis will be in the negative x direction because the y-components will cancel and the x-components will essentially be double but in the negative x direction in relation to the coordinate plane. Hope that helped even tho 4 months after you probably already finished your class haha :)
I am still confused why s/ 2
Dr. Strange
This answer is wrong, it should be + sin(\theta) not -sin(\theta) in the final expression
ok
Cool beans.
This is wrong 🤦🏻♂️ godness... It should be opposite as answer. Eaxial = 2 * Eequatorial...everywhere it's written like that???
Thank you so much
You're most welcome
Cheers,
Dr. A