Hi professor, I really love your lectures. They give me dopamine. I have been suffering from mental illness for a long time. I'm lagged way behind than others. I hope that I have the opportunity to chase them. I don't want to be a loser. Thank you anyways.
Sir we can use formula kqx/(x²+r²)³/² as r is radius of ring and x is any distance on the axis of ring Due to -ve charge Electric is toward ring draw vectors. By puttting different values of x we get answers actually we have to take cases . Answer was too length that's why which concept i used i tell you Thank you sir for this amazing problem Love you from india
Dr. Lewin, PLEASE FORGIVE ME, BUT I AM TAKING ADVANTAGE OF THIS RECENT VIDEO BY YOU TO SAY SOMETHING OFF-SUBJECT, BUT OF MUCH SIGNIFICANCE IN YOU 802 LECTURE 20 ABOUT THE RL DEMO!!! You gave an outstanding lecture and demo of the L/R time constant with a huge 30H 4 Ohm inductor...you spent time talking about the slow charge AND DISCHARGE of the RL circuit, however, the big demo could only show the slow illumination of the light bulb and NOT the slow fade-out. There is an easy way to do this: Just put a fuse in series with the battery and close a switch which will blow the cheap fuse (or a circuit-breaker), taking the battery OUT of the circuit, but keep the current flowing...the bulb then will SLOWLY go dim!!! I would love to see a circuit powered by an inductor only!! To this day, I have yet to see it. I wish I was there!! BTW: You mentioned that later in the course you will show how such a huge inductor is made, bit I searched all of the lectures and never found it. I can't get a response from your UA-cam lecture series. You have really effected my life with your OCW videos!! THANKS MUCH!! --Dale
To find the electric field at a distance r from one ring we take a small piece, dq, and integrate: E = ∮ (K/r²) dq = ∮ [K/(x² + R²)] dq = KQ/(x² + R²) Since all vertical components cancel, we only get an E-field in the x-direction. If we draw a line from the location of dq to the x-axis then this line has the length r = √(x² + R²).
Between r and the x-axis we get an angle θ. The E-field from one ring in the x-direction becomes: E(x) = KQ/(x² + R²)·cos θ => E(x) = KQ/(x² + R²)·x/√(x² + R²) => E(x) = KQ·x/(x² + R²)³ᐟ² To find the total electric field at a point x, we use superposition of the fields from both rings. Let LR denote the left ring and RR the right ring. Angle θ₁ = θ for RR, and θ₂ = θ for LR. Locations are x₁ for RR, and x₂ for LR. Since none of the rings is centered in origo but LR is located at x₂ = -L/2 and RR at x₁ = +L/2 we must rewrite the formula for E(x): E(x) = KQ{ (x-x₁)/[(x-x₁)² + R²]³ᐟ²+ (x-x₂)/[(x-x₂)² + R²]³ᐟ²} Answer: (a) x = L x-x₁= L - L/2 = L/2, x-x₂ = L + L/2 = 3L/2 => E(L) = KQ{ (L/2)/[(L/2)² + R²]³ᐟ²+ (3L/2)/[(3L/2)² + R²]³ᐟ²} The E-field points in the negative x-direction. (b) x = 2L x-x₁= 2L - L/2 = 3L/2, x-x₂ = 2L + L/2 = 5L/2 => E(2L) = KQ{ (3L/2)/[(3L/2)² + R²]³ᐟ²+ (5L/2)/[(5L/2)² + R²]³ᐟ²} The E-field points in the negative x-direction. (c) x = -1.5L = -3L/2 x-x₁= -3L/2 - L/2 = -2L, x-x₂ = -3L/2 + L/2 = -L => E(-1.5L) = KQ{ (-2L)/[(-2L)² + R²]³ᐟ²+ (-L)/[(-L)² + R²]³ᐟ²} The E-field points in the positive x-direction. Note: Q is a negative charge for both rings.
I think field st hebpoint should be subtracted ie field of ring a - field of ring b as by the formula kq/r2 cos theta and we can calculate the net electric field as e1 vect- e2 vect
Im more interested in the general behaviour of electricity and electrostatics, and to quote you sir "Its really the electric forces that hold our world together" , we are electrical beings living inside a huge electrical field , and i suspect we could learn alot from the research of Royal Raymond Rife. But thats just me !!
It's good but not that great, it's a standard problem that students can do it they know the formulas, rather give them something interesting like There are two charges +q and -q both seperated by a distance x, an electric field line goes from +q making an angle A with the horizontal and reaches -q making an angle B. Find the relation between angle A and B.
Note that due to symmetry, the field on the x-axis will have zero y and z components, so the direction will be either in the plus x (when x = -1.5L) or minus x direction (when x = L or 2L). The x-direction force on the test charge will be proportional to two factors: the inverse square of the distance to the ring, and the sine of the angle made between the line from the ring to the test charge and the y-z plane. Let k = 1/ (4 * pi *epsilonzero) Consider only cases where x >= L/2 For the near ring, the field is E1 = -(Q/k)(1/[(x-(L/2))^2 + R^2])((x-(L/2))/([(x-(L/2))^2 + R^2]^(1/2)) or: E1 = -(Q/k)((x-(L/2))/([(x-(L/2))^2 + R^2]^(3/2)) For the far ring: E2 = -(Q/k)((x+(L/2))/([(x+(L/2))^2 + R^2]^(3/2)) And the total is E = E1 + E2 CASE A: x = L; E will be in the minus x direction E = -(Q/k) [((L/2)/([(L/2)^2 + R^2]^(3/2)) + ((3L/2)/([(3L/2)^2 + R^2]^(3/2))] CASE B: x = 2L; E will be in the minus x direction E = -(Q/k) [((3L/2)/([(3L/2)^2 + R^2]^(3/2)) + ((5L/2)/([(5L/2)^2 + R^2]^(3/2))] CASE C: x = -3L/2; E will be in the plus x direction E = +(Q/k) [(L/([L^2 + R^2]^(3/2)) + (2L/([(2L)^2 + R^2]^(3/2))] Sanity check: let R = 0 E1 = -(Q/k)((x-(L/2))/([(x-(L/2))^2]^(3/2)) = -(Q/k)(X-(L/2))/[((X-L/2)^2)^(3/2)] = -(Q/k)/([(X-(L/2)]^2)
Please tell me if I'm wrong, I have made a general formula for these charged rings in which I can calculate the magnitude of the electric field at any general point of their axis, then using the superposition principle I can calculate the magnitude of the electric field at x= L, 2L, -1.5L.
@@hiddenbros Try sanity checks with yours, like setting R to zero. Here's mine with R=0: For x=L have: E= (kQ/L²)(4+⁴/₉){←} For x=2L have: E= kQ(4/L²)(⅑+⅕){←} = (kQ/L²)(1+¹¹/₄₅){←} For x= -1.5L have: E= (kQ/L²)(1¼){→}
Re Moon landing televised walk: from APOD website: The signals received by Parkes telescope were sent to Sydney. From there the TV signal was split. One signal went to the Australian Broadcasting Commission, the other to Houston for the international telecast. The international signal had to travel halfway around the world from Sydney to Houston, adding a delay. So Australian audiences saw Neil Armstrong's historic first step 0.3 seconds before the rest of the world.
As you said this problem is a piece of cake for JEE students. Lets derive the general case at a distance x.... B(x) = B due to 1st ring+ B due to 2nd ring B(x)=[{mu°IR²/2(R²+x²)^3/2}+{mu°IR²/(R²+(L-x)²)^3/2}] On solving B(x)=(mu°IR²/2)[1/(R²+x²)³/²+1/{R²+(L-x)²}³/²]
My name is Ankit Agnihotri from India[Bharat] Uttar Pradesh. I am learning Physics and English from you. I love your way of teaching and want to become professor of physics like you. I'm highly obliged to you Sir
My doubt is specifically regarding two capillary tubes placed above one another. In case of insufficient length of one tube, will the water rising in the tube will cross the interface or would the meniscus adjust it radius on the interface for balance?
Sir, I am student from India , everyone in the class including me felt physics is boaring only because of my physics teacher i really understood the beauty of physics by seeing pw and your lecture videos . My teacher had made us to roat learn everything. She would just read every topics by standing in a place like a rock even though we have a smart board she wont use it she wont even use the board . You had broke the myth that physics is about roat learning. My teacher had made a beautiful subject boaring , i would consider her as a criminal as u said which is true , I really wanted to give a police complaint . Thank you for making the world to feel the beauty of physics . Waiting for your reply.
E(r) = 1/(4*pi*e0)*Q/r²*er and for the amount component in x-direction E = 1/(4*pi*e0)*[-Q/r1²*cos(a1)-Q/r2²*cos(a2)], with sin(a1) = R / r1 and sin(a2) = R / r2. 1) x = L , so r1² = L²/4+R² and r2² = 9/4*L²+R² Direction - X - Axis 2) x = 2*L, so r1² = 9/4*L² + R² and r2² = 25/4*L² + R² Direction - X - Axis 3) x = -1.5*L, so r1² = L² + R² and r2² = 4*L² + R² Direction + X - Axis
-kqy/(r^2+y^2)^3/2(not to conflict with x) individually though how do we do it add them? with respect to their positions calc at x=l for ex left ring for x=l put y=3/2l and right y=l/2
E-fields are vectors. x component is a cos or sin multiple depending on the angle you use. For points on the x axis, y & z components all cancel for each coil. Trig functions can be +ve or -ve. So can angles be ±, as can lengths.
If you want to be good in physics then you have just won 1st prize in the lottery of learning; you have found this website. It's all here for those who look.
Sir I don’t send you the answers personally but if you like I can post in the comments directly as I am a student of class 12 If not then can you please tell where can I send you the answers
E(x=L) = [kQ(l/2)/((l/2)^2 + (R)^2)^3/2 + kQ(3l/2)/((3l/2)^2+(R)^2)^3/2] (x direction) E(x=2L) = [kQ(3l/2)/((3l/2)^2 + (R)^2)^3/2 + kQ(5l/2)/((5l/2)^2+(R)^2)^3/2] (x direction) E(x=-1.5L) = [kQ(l)/((l)^2 + (R)^2)^3/2 + kQ(l/2)/((l/2)^2+(R)^2)^3/2] (-x direction) Professor Lewin, you don't have to put more difficult problems just because I'm watching your channel.
Take pictures of the problem with your phone, go to open AI chat GPT, paste the pictures and ask it to solve it, it will explain everything for you. Not intended to be a cheat, but a way to learn how to think about it.
Mind you ChatGPT is a language prompt, not a problem solver, it will only fool you that it has solved something but it will make some fundamental error in such an expert manner that it would be difficult to locate the error, not a good way to learn at all
I’m just beginning my physics journey and these reasons are still a big mystery to me. I’m really struggling to understand this. Could you please help me
BRO IT IS MORE STABLE WHEN IT IS MINIMUM JUST LIKE A STONE WANTS TO COME CLOSER TO EARTH TO BE IN A MORE STABLE STATE.........OR MOVING FROM HIGHER TO LOWER POTENTIAL(RIGHT SIR?CORRECT ME IF NOT) AND BRO DONT FORGET ENERGY DOESNT ACTUALLY TEND TO MINIMUM ITS THE POTENTIAL ENERGY THEY ARE REFERRING TO@@flatfootedlectures9335
General solutions: Total E-fields are: E=E1+E2 if x½l For x=L , 2L & -1.5L, all have E=E1+E2. For all below, let t=R/L such that t⁻ⁿ:=(t⁻¹)ⁿ:=(L/R)ⁿ=0 if L=0. For x=L have: E= kQ(4/L²)(1/√(4t²+1)³+3/√(4t²+9)³){←} OR: E= kQ(4L/R³)(1/√(4+t⁻²)³+3/√(4+9t⁻²)³){←} For x=2L have: E= kQ(4/L²)(3/√(4t²+9)³+5/√(t²+25)³){←} or E= kQ(4L/R³)(3/√(4+9t⁻²)³+5/√(1+25t⁻²)³){←} For x= -1.5L have: E= (kQ/L²)(1/√(t²+1)³+2/√(t²+4)³){→} OR: E= (kQL/R³)(1/√(1+1t⁻²)³+2/√(1+4t⁻²)³){→} ~~~~~~~~~~~~~~ ½,1½...1½,2½....1,2 Example solutions: Assuming that L=2(½l)~2R and k=1/4πεo~c²μo/4π~9*10⁹=9G and y=0 for all x values given: => E-field values are approximately: (-1.79kQ/L²)x^~{←}16.14G·Q/L², x=L(→) (-0.537kQ/L²)x^~{←}4.834G·Q/L², x= 2L (→) (-0.944kQ/L²)x^~{→}8.494G·Q/L²,x= -1.5L(←) [G=10⁹ here] Another example (zero sized rings): t=0 (R=0) => For x=L have: E= (kQ/L²)(4+⁴/₉){←} For x=2L have: E= kQ(4/L²)(⅑+⅕){←} = (kQ/L²)(1+¹¹/₄₅){←} For x= -1.5L have: E= (kQ/L²)(1¼){→} Example (L=0) leads to a quandary... 🤔 (Only if you don't multiply everything by 0 ) For x=0 (=L) have: E= kQ(4L/R³)(1/√(4+0)³+3/√(4+0)³){←} = (kQL/R³)(4²/√4³){←} = (2kQL/R³){←} =0 , as L=0. OR: E= (21½kQL/R³){←}...⁉️😕 =0 ✅☺️ (L=0) OR: E= (3kQL/R³){→}...⁉️😕 =0 ✅☺️(L=0) ...The quandary may get resolved if you use the fact that you are just overlaying the rings on top of each other, so there's now only 1 ring with double the charge, but all the fields cancel out in the middle via superposition, so the real answer is: E=0 , if L=0=x. (L is in all the above equations and it's 0) Explanation: Directions "-x^" are from the cosines (adj/r) of the angles that adjacents make with each ring and strengths depend on the inverse squares of the distances to each of the parts of the rings. All fields here have -Q charge and are proportional to adjacent side for each ring, so the E-fields are in directions (-1)*adj, so if adj is ←, -(←)=→ and if adj is →, it is ←. For each ring, you just add the cosine components of each part of the E-fields (-kδQ/r²) and the sin components add to zero. Then just add the resultants for each ring at the points (x, 0) for the given x values. (More details in reply comments below)
@@KeithandBridgetThat was the shortened version! Here's the solutions w/o assuming L=2R: Lhs: E1= -x^ kQ(x/r³), x^={←}. Rhs: E2= -x'^ kQ(x'/r'³), x'^={→}. Total E-fields are: E=E1+E2 if x½l E=E1-E2 if -½l
No mention of the ring thickness, so I will assume it is near zero. In this case the E-field components around a ring would cancel, leaving only components that are perpendicular to the ring. Coulombs law for an element of a ring of charge -Q would be: δE= k(-δQ/r²)cosθ for points along the x axis. r=√(R²+(x+½l)²), lhs (x0, x L~2R => E~ {→}(kQ/L²)(1/√(1¼)³+2/√(4¼)³)
Hi professor, I really love your lectures. They give me dopamine. I have been suffering from mental illness for a long time. I'm lagged way behind than others. I hope that I have the opportunity to chase them. I don't want to be a loser. Thank you anyways.
Hi sir
Very big fan of yours
Thank you for teaching us for these years
May you live a happy life
Namaste sir ! This is the problem I faced in my jee pYq series I thought like you but feared to do it so I skipped , now I realised what it was !
Sir we can use formula kqx/(x²+r²)³/² as r is radius of ring and x is any distance on the axis of ring
Due to -ve charge Electric is toward ring draw vectors. By puttting different values of x we get answers actually we have to take cases .
Answer was too length that's why which concept i used i tell you
Thank you sir for this amazing problem
Love you from india
Sir I love your videos!
Thanks for posting a good question Sir.
We can use the result : Electric field by a ring at a distance x on its axis E=KQ/(x^2+R^2)^3/2
multiplied by x
You missed a x in numerator :)
Dr. Lewin,
PLEASE FORGIVE ME, BUT I AM TAKING ADVANTAGE OF THIS RECENT VIDEO BY YOU TO SAY SOMETHING OFF-SUBJECT, BUT OF MUCH SIGNIFICANCE IN YOU 802 LECTURE 20 ABOUT THE RL DEMO!!!
You gave an outstanding lecture and demo of the L/R time constant with a huge 30H 4 Ohm inductor...you spent time talking about the slow charge AND DISCHARGE of the RL circuit, however, the big demo could only show the slow illumination of the light bulb and NOT the slow fade-out.
There is an easy way to do this:
Just put a fuse in series with the battery and close a switch which will blow the cheap fuse (or a circuit-breaker), taking the battery OUT of the circuit, but keep the current flowing...the bulb then will SLOWLY go dim!!! I would love to see a circuit powered by an inductor only!! To this day, I have yet to see it.
I wish I was there!!
BTW: You mentioned that later in the course you will show how such a huge inductor is made, bit I searched all of the lectures and never found it.
I can't get a response from your UA-cam lecture series.
You have really effected my life with your OCW videos!!
THANKS MUCH!!
--Dale
To find the electric field at a distance r from one ring
we take a small piece, dq, and integrate:
E = ∮ (K/r²) dq = ∮ [K/(x² + R²)] dq = KQ/(x² + R²)
Since all vertical components cancel, we only get an
E-field in the x-direction.
If we draw a line from the location of dq to the x-axis
then this line has the length r = √(x² + R²).
Between r and the x-axis we get an angle θ.
The E-field from one ring in the x-direction becomes:
E(x) = KQ/(x² + R²)·cos θ
=> E(x) = KQ/(x² + R²)·x/√(x² + R²)
=> E(x) = KQ·x/(x² + R²)³ᐟ²
To find the total electric field at a point x, we use
superposition of the fields from both rings.
Let LR denote the left ring and RR the right ring.
Angle θ₁ = θ for RR, and θ₂ = θ for LR.
Locations are x₁ for RR, and x₂ for LR.
Since none of the rings is centered in origo
but LR is located at x₂ = -L/2 and RR at x₁ = +L/2
we must rewrite the formula for E(x):
E(x) = KQ{ (x-x₁)/[(x-x₁)² + R²]³ᐟ²+ (x-x₂)/[(x-x₂)² + R²]³ᐟ²}
Answer:
(a) x = L
x-x₁= L - L/2 = L/2, x-x₂ = L + L/2 = 3L/2
=> E(L) = KQ{ (L/2)/[(L/2)² + R²]³ᐟ²+ (3L/2)/[(3L/2)² + R²]³ᐟ²}
The E-field points in the negative x-direction.
(b) x = 2L
x-x₁= 2L - L/2 = 3L/2, x-x₂ = 2L + L/2 = 5L/2
=> E(2L) = KQ{ (3L/2)/[(3L/2)² + R²]³ᐟ²+ (5L/2)/[(5L/2)² + R²]³ᐟ²}
The E-field points in the negative x-direction.
(c) x = -1.5L = -3L/2
x-x₁= -3L/2 - L/2 = -2L, x-x₂ = -3L/2 + L/2 = -L
=> E(-1.5L) = KQ{ (-2L)/[(-2L)² + R²]³ᐟ²+ (-L)/[(-L)² + R²]³ᐟ²}
The E-field points in the positive x-direction.
Note: Q is a negative charge for both rings.
I think field st hebpoint should be subtracted ie field of ring a - field of ring b as by the formula kq/r2 cos theta and we can calculate the net electric field as e1 vect- e2 vect
Im more interested in the general behaviour of electricity and electrostatics, and to quote you sir "Its really the electric forces that hold our world together" , we are electrical beings living inside a huge electrical field , and i suspect we could learn alot from the research of Royal Raymond Rife. But thats just me !!
Great problem! I think I will propose it to my students!
It's good but not that great, it's a standard problem that students can do it they know the formulas, rather give them something interesting like
There are two charges +q and -q both seperated by a distance x, an electric field line goes from +q making an angle A with the horizontal and reaches -q making an angle B.
Find the relation between angle A and B.
YEP@@Batshitcrazy6942
Note that due to symmetry, the field on the x-axis will have zero y and z components, so the direction will be either in the plus x (when x = -1.5L) or minus x direction (when x = L or 2L).
The x-direction force on the test charge will be proportional to two factors: the inverse square of the distance to the ring, and the sine of the angle made between the line from the ring to the test charge and the y-z plane.
Let k = 1/ (4 * pi *epsilonzero)
Consider only cases where x >= L/2
For the near ring, the field is
E1 = -(Q/k)(1/[(x-(L/2))^2 + R^2])((x-(L/2))/([(x-(L/2))^2 + R^2]^(1/2))
or:
E1 = -(Q/k)((x-(L/2))/([(x-(L/2))^2 + R^2]^(3/2))
For the far ring:
E2 = -(Q/k)((x+(L/2))/([(x+(L/2))^2 + R^2]^(3/2))
And the total is E = E1 + E2
CASE A: x = L; E will be in the minus x direction
E = -(Q/k) [((L/2)/([(L/2)^2 + R^2]^(3/2)) + ((3L/2)/([(3L/2)^2 + R^2]^(3/2))]
CASE B: x = 2L; E will be in the minus x direction
E = -(Q/k) [((3L/2)/([(3L/2)^2 + R^2]^(3/2)) + ((5L/2)/([(5L/2)^2 + R^2]^(3/2))]
CASE C: x = -3L/2; E will be in the plus x direction
E = +(Q/k) [(L/([L^2 + R^2]^(3/2)) + (2L/([(2L)^2 + R^2]^(3/2))]
Sanity check: let R = 0
E1 = -(Q/k)((x-(L/2))/([(x-(L/2))^2]^(3/2)) = -(Q/k)(X-(L/2))/[((X-L/2)^2)^(3/2)] = -(Q/k)/([(X-(L/2)]^2)
These are not Helmholtz coils, these are statically charged rings. No current flowing so no magnetic field.
Please tell me if I'm wrong,
I have made a general formula for these charged rings in which I can calculate the magnitude of the electric field at any general point of their axis, then using the superposition principle I can calculate the magnitude of the electric field at x= L, 2L, -1.5L.
@@hiddenbrosYou can take a look at my solution below (-21hrs, around same time as Keith's comment). Let me know if I have something wrong!
@@hiddenbros Try sanity checks with yours, like setting R to zero. Here's mine with R=0:
For x=L have:
E= (kQ/L²)(4+⁴/₉){←}
For x=2L have:
E= kQ(4/L²)(⅑+⅕){←}
= (kQ/L²)(1+¹¹/₄₅){←}
For x= -1.5L have:
E= (kQ/L²)(1¼){→}
@@hiddenbros Superposition is the right approach. Look at WL's 8.02 lectures 1 and 2.
@@The_Green_Man_OAPHey,I am getting the same value for R=0 in 1st question 🙂.
i.e( 40/9)KQ/l²
Omgggg I love himmm
Piece of cake for iitians
Re Moon landing televised walk: from APOD website:
The signals received by Parkes telescope were sent to Sydney. From there the TV signal was split. One signal went to the Australian Broadcasting Commission, the other to Houston for the international telecast.
The international signal had to travel halfway around the world from Sydney to Houston, adding a delay. So Australian audiences saw Neil Armstrong's historic first step 0.3 seconds before the rest of the world.
Sir this is old pyq of jee adv and recently asked in my mains exam as well
As you said this problem is a piece of cake for JEE students.
Lets derive the general case at a distance x....
B(x) = B due to 1st ring+ B due to 2nd ring
B(x)=[{mu°IR²/2(R²+x²)^3/2}+{mu°IR²/(R²+(L-x)²)^3/2}]
On solving
B(x)=(mu°IR²/2)[1/(R²+x²)³/²+1/{R²+(L-x)²}³/²]
read the question again m8
Don't mind me.. but I'm also a jee aspirant
....😅😅jee students " formula based hai"
bruh
bruhhhhhhh
electrostatics bruhhhhh
not moving charges bruhhhhhh
😮
Hi professor!
❤
My name is Ankit Agnihotri from India[Bharat] Uttar Pradesh. I am learning Physics and English from you. I love your way of teaching and want to become professor of physics like you. I'm highly obliged to you Sir
My doubt is specifically regarding two capillary tubes placed above one another. In case of insufficient length of one tube, will the water rising in the tube will cross the interface or would the meniscus adjust it radius on the interface for balance?
Same is kinda for b field and termed as helmotlz coil in class 12 th ij our national school book
Sir, I am student from India , everyone in the class including me felt physics is boaring only because of my physics teacher i really understood the beauty of physics by seeing pw and your lecture videos . My teacher had made us to roat learn everything. She would just read every topics by standing in a place like a rock even though we have a smart board she wont use it she wont even use the board . You had broke the myth that physics is about roat learning.
My teacher had made a beautiful subject boaring , i would consider her as a criminal as u said which is true , I really wanted to give a police complaint . Thank you for making the world to feel the beauty of physics .
Waiting for your reply.
The worse of all is that I have never seen or heard her before.
By any chance have you given lectures about the time dependent and independent schrodinger equations?
A video on my 20th birthday nice
E(r) = 1/(4*pi*e0)*Q/r²*er and for the amount component in x-direction E = 1/(4*pi*e0)*[-Q/r1²*cos(a1)-Q/r2²*cos(a2)], with sin(a1) = R / r1 and sin(a2) = R / r2.
1) x = L , so r1² = L²/4+R² and r2² = 9/4*L²+R² Direction - X - Axis
2) x = 2*L, so r1² = 9/4*L² + R² and r2² = 25/4*L² + R² Direction - X - Axis
3) x = -1.5*L, so r1² = L² + R² and r2² = 4*L² + R² Direction + X - Axis
Ive seen similar problem in my jee
E(x) = KQX/(X²+R²)³/² and superposition theoram...
I love physics 🥰
Namaste sir I heard that u are coming in iitk
-kqy/(r^2+y^2)^3/2(not to conflict with x)
individually
though
how do we do it add them?
with respect to their positions
calc at x=l
for ex left ring for x=l put y=3/2l and right y=l/2
E-fields are vectors.
x component is a cos or sin multiple
depending on the angle you use.
For points on the x axis, y & z components all cancel for each coil.
Trig functions can be +ve or -ve.
So can angles be ±, as can lengths.
Thank you sir. I want to be very good in physics. What can i do sir.
If you want to be good in physics then you have just won 1st prize in the lottery of learning; you have found this website. It's all here for those who look.
Sir can i know you which institute you belong from
Sir please reply to my problem I asked in the comment section on last video
Sir I don’t send you the answers personally but if you like I can post in the comments directly as I am a student of class 12
If not then can you please tell where can I send you the answers
Please post in the comments
Right here‼️✅😁
Please sir make a video on how to become intelligent in physics and how and from which source
eat yogurt every day but *never on Fridays* that also worked for Einstein and for me
@@lecturesbywalterlewin.they9259 I knew I went wrong somewhere, I ate yogurt on Fridays.
@@KeithandBridget 😂
@@KeithandBridget
I eat fish 🐟 on Fridays. Really‼️
Profesor are these Helmholtz coils
No
E(x=L) = [kQ(l/2)/((l/2)^2 + (R)^2)^3/2 + kQ(3l/2)/((3l/2)^2+(R)^2)^3/2] (x direction)
E(x=2L) = [kQ(3l/2)/((3l/2)^2 + (R)^2)^3/2 + kQ(5l/2)/((5l/2)^2+(R)^2)^3/2] (x direction)
E(x=-1.5L) = [kQ(l)/((l)^2 + (R)^2)^3/2 + kQ(l/2)/((l/2)^2+(R)^2)^3/2] (-x direction)
Professor Lewin, you don't have to put more difficult problems just because I'm watching your channel.
Hlo Sir
This is too tricky
I am in class 12th
How can I start to solve?
Take pictures of the problem with your phone, go to open AI chat GPT, paste the pictures and ask it to solve it, it will explain everything for you. Not intended to be a cheat, but a way to learn how to think about it.
Mind you ChatGPT is a language prompt, not a problem solver, it will only fool you that it has solved something but it will make some fundamental error in such an expert manner that it would be difficult to locate the error, not a good way to learn at all
Look at Walter Lewin's 8.02 Lectures 1 and 2
@@KeithandBridget loool now u answer like prof lewin
@@KeithandBridget say that friday dialogue too XD
I have a question not quite related to this. I never understood why people always say energy tends towards a minimum. Like why?
potential enegry goes to a minimum for obvious reasons
I’m just beginning my physics journey and these reasons are still a big mystery to me. I’m really struggling to understand this. Could you please help me
BRO IT IS MORE STABLE WHEN IT IS MINIMUM JUST LIKE A STONE WANTS TO COME CLOSER TO EARTH TO BE IN A MORE STABLE STATE.........OR MOVING FROM HIGHER TO LOWER POTENTIAL(RIGHT SIR?CORRECT ME IF NOT) AND BRO DONT FORGET ENERGY DOESNT ACTUALLY TEND TO MINIMUM ITS THE POTENTIAL ENERGY THEY ARE REFERRING TO@@flatfootedlectures9335
@@flatfootedlectures9335less energy = more stability.
General solutions:
Total E-fields are:
E=E1+E2 if x½l
For x=L , 2L & -1.5L, all have E=E1+E2.
For all below, let t=R/L
such that t⁻ⁿ:=(t⁻¹)ⁿ:=(L/R)ⁿ=0 if L=0.
For x=L have:
E= kQ(4/L²)(1/√(4t²+1)³+3/√(4t²+9)³){←}
OR:
E= kQ(4L/R³)(1/√(4+t⁻²)³+3/√(4+9t⁻²)³){←}
For x=2L have:
E= kQ(4/L²)(3/√(4t²+9)³+5/√(t²+25)³){←}
or
E= kQ(4L/R³)(3/√(4+9t⁻²)³+5/√(1+25t⁻²)³){←}
For x= -1.5L have:
E= (kQ/L²)(1/√(t²+1)³+2/√(t²+4)³){→}
OR:
E= (kQL/R³)(1/√(1+1t⁻²)³+2/√(1+4t⁻²)³){→}
~~~~~~~~~~~~~~
½,1½...1½,2½....1,2
Example solutions:
Assuming that L=2(½l)~2R
and k=1/4πεo~c²μo/4π~9*10⁹=9G
and y=0 for all x values given:
=> E-field values are approximately:
(-1.79kQ/L²)x^~{←}16.14G·Q/L², x=L(→)
(-0.537kQ/L²)x^~{←}4.834G·Q/L², x= 2L (→)
(-0.944kQ/L²)x^~{→}8.494G·Q/L²,x= -1.5L(←)
[G=10⁹ here]
Another example (zero sized rings):
t=0 (R=0)
=>
For x=L have:
E= (kQ/L²)(4+⁴/₉){←}
For x=2L have:
E= kQ(4/L²)(⅑+⅕){←}
= (kQ/L²)(1+¹¹/₄₅){←}
For x= -1.5L have:
E= (kQ/L²)(1¼){→}
Example (L=0) leads to a quandary... 🤔
(Only if you don't multiply everything by 0 )
For x=0 (=L) have:
E= kQ(4L/R³)(1/√(4+0)³+3/√(4+0)³){←}
= (kQL/R³)(4²/√4³){←}
= (2kQL/R³){←}
=0 , as L=0.
OR:
E= (21½kQL/R³){←}...⁉️😕
=0 ✅☺️ (L=0)
OR:
E= (3kQL/R³){→}...⁉️😕
=0 ✅☺️(L=0)
...The quandary may get resolved if you use the fact that you are just overlaying the rings on top of each other, so there's now only 1 ring with double the charge, but all the fields cancel out in the middle via superposition, so the real answer is: E=0 , if L=0=x.
(L is in all the above equations and it's 0)
Explanation:
Directions "-x^" are from the cosines (adj/r) of the angles that adjacents make with each ring and strengths depend on the inverse squares of the distances to each of the parts of the rings. All fields here have -Q charge and are proportional to adjacent side for each ring, so the E-fields are in directions (-1)*adj, so if adj is ←, -(←)=→ and if adj is →, it is ←.
For each ring, you just add the cosine components of each part of the E-fields (-kδQ/r²) and the sin components add to zero. Then just add the resultants for each ring at the points (x, 0) for the given x values.
(More details in reply comments below)
No need for any approximation, and I think you have misunderstood the situation.
@@KeithandBridgetThat was the shortened version!
Here's the solutions w/o assuming L=2R:
Lhs:
E1= -x^ kQ(x/r³), x^={←}.
Rhs:
E2= -x'^ kQ(x'/r'³), x'^={→}.
Total E-fields are:
E=E1+E2 if x½l
E=E1-E2 if -½l
No mention of the ring thickness, so I will assume it is near zero.
In this case the E-field components around a ring would cancel, leaving only components that are perpendicular to the ring.
Coulombs law for an element of a ring of charge -Q would be:
δE= k(-δQ/r²)cosθ for points along the x axis.
r=√(R²+(x+½l)²), lhs (x0, x L~2R
=> E~ {→}(kQ/L²)(1/√(1¼)³+2/√(4¼)³)
@@The_Green_Man_OAP Apologies. You had not misunderstood.
wait dont neglect ....yh like keith said dont approximate
coz they r comparable distances
Helmholtz coil
Not correct.
bruh
dont over reference physics galaxy XD
hemholtz was in magnetix field bro XD(or sis)
how do we send answers?
by airmail
Hey Professor, isn't there any other way to send you the answer?
Do a video? 🤔
@@The_Green_Man_OAP Good idea, but it will take a lot of time and my exams are still going on.
@@The_Green_Man_OAP😊 🙂 There is something, coming up ! 😅 May be on 6 th of March (IST ).
Sir, I wrote you a letter in Gmail but iam not sure if it is you,so can you confirm please
He doesn't give out his gmail to everyone
Sir can you start dubbing your lectures in Hindi which is useful for millions of Indian students I bet it will be worth it!
GO WATCH PHYSICS GALAXY
playing with bi weekly these days
i hate physics
My friend, just stop eating yogurts on friday
If you hate physics, you hate life.
Life _is_ physics.
Love from India
11_march my physics exam ❤
I completed most from you. થૅન્ક યુ
WAS IT EZ?XD