-Q induces a total charge +Q onto the inner surface of the conducting sphere. The net charge of the sphere is 0. So -Q is induced on the outer surface of the conducting sphere. The inducex charge distribution is not spherically symmetric though since -Q is off center. The field lines are normal to the surface of the conductor. For r> R2, the electric field is E=-k*Q/r^2. k is coulombs constant. Gauss law states: I can compute the E field, as if -Q were in the center of the sphere. a) E=-k*Q/(R1/2+a)^2. E points radially inward. b) If B is just inside the charging sphere: E=0. If just outside the sphere: E=-kQ/R2^2, pointing radially inward. c) same as B. d) V_A=-kQ/(R1/2+a). V_A=0 if r goes to infinity in order not to get an "insane" result 😊. e) V_C=-k*Q/R2 f) V_D=V_C=-k*Q/R2. Potential on conducting sphere is constant. V_B = V_C as well.
Sir, from where do you get the motivation that you study through videos at the age of 88 while other professors prefer to rest at this age. Sir, what is your motivation, please share it with us.❤
Hey sir, here's my go at it: The -Q inside indues a +Q charge on the inner shell (though not of uniform distribution) and that in turn induces a -Q on the outer shell (of uniform distribution according to your 8.02 lecture (I think it was lecture 3 or 4 ). Applying Gauss's theorem, we find: A. E is radially inward with magnitude E = -kq/(a+r1/2)² B. Radially inward with magnitude E = -kq/(R2)² C. Radially inward with magnitude E = -kq/(R2)² The effect of the -q and the +q induced on the inner shell cancel out in the potential, so the potential here is caused by the -q on the outer shell, which means it is -kq/(a+r1/2) for D and -kq/R1 for E
Hi sir. The charge induced on the inner sphere makes it as if we have a charge -Q at the center and the outer sphere makes no defference at all A. The electric field has magnitude KQ/(R1/2+a)^2 and is radially inwards towards the center B. The electric field has magnitude KQ/((R1/2)^4+b^2) and it points radially towards the center C. The electric field has magnitude KQ/R2^2 pointin radially towards the center D. The electric potential at A is -KQ/(R1/2+a) E. The electric potential at C is -KQ/R2 F. The electric potential at D is -KQ/R1
I looked at assignment #2 of 8.02, in which it is shown that the charge on the outer surface of the conducting sphere is uniformly distributed (and totals -Q in this case) regardless of the position of Q inside the sphere. Thus, external to the sphere, the E field is the same as for a charge -Q at the center point with no sphere. I hope I have the answers correct now based on the above. Let k = 1/(4pi epsilon_0) a) E field at A is kQ/((a+R1/2)^2) directed toward the center of the sphere (toward the left in the diagram); b) E field at B is kQ/R2^2, directed toward the center of the sphere; c) E field at C is also kQ/R2^2, directed toward the center of the sphere; d) potential at A is -kQ/(a+R1/2) e) potential at C is -kQ/R2 f) potential at D is also -kQ/R2 since the field is zero inside the conductor
The off center charge -Q induces a charge +Q on the inside surface of the conductor such that all field line are normal to the inner surface. The field lines and potential varies inside the sphere accordingly. The conductor itself is at one potential as caused by a induced charge -Q uniformly distributed on the outer surface of the conductor. So for all r > R_2, the hollow conductor with a charge -Q inside it acts as a sphere with a uniform charge -Q on its surface. a. E at A is radially inward to the sphere = -kQ/r^2 (r = a + R_1/2) b. E at B is radially inward = -kQ/R_2^2 c. E at C is also radially inward = -kQ/R_2^2 d. V at A = -kQ/r (r = a + R_1/2) e. and f. C and D on the conductor are at the same potential = -kQ/R_2
Electric field: ---------------- a) At point A: E = Q/[4πε₀·(R₁/2 + a)²] Direction: radially inwards b) At point B: E = Q/[4πε₀·R₂²] Direction: radially inwards c) At point C: E = Q/[4πε₀·R₂²] Direction: radially inwards Electric potential: ---------------------- d) At point A: V = - Q/[4πε₀·(R₁/2 + a)] e) At point C: V = - Q/[4πε₀·R₂] f) At point D: V = - Q/[4πε₀·R₂]
Assuming Q>0, a. (Q/4πϵ₀) / (a + R₁/2)² towards the center of the sphere. b. (Q/4πϵ₀) / R₂² towards the center of the sphere. c. (Q/4πϵ₀) / R₂² towards the center of the sphere. d. -(Q/4πϵ₀) / (a + R₁/2) e. -(Q/4πϵ₀) / R₂ f. -(Q/4πϵ₀) / R₂
Sir where to find your simple pendulum lectures. By the way sir your old lectures are very good.sir i love physics very very very very very very very much and sir i eat yogurt everyday except on Fridays it worked or Einstein and you, I hope it also works for me.
Sir how to be humble because you have lot of knowledge in physics but you be so humble but sir at the same point some people know less but they will show that they know more than you how to deal with that
E(A) = kQ/(a+R_1/2)^2 E(B/C) = kQ/(R_2)^2 V(A) = -kQ/(a+R_1/2) V(C/D) = -kQ/R_2 Electic field's direction is twowards the center of the sphere. Best wishes
Sir i have a question Sir when we extract the honey from honey comb by (spinning machine ) we say that the honey came out from comb due to centrifugal force. centrifugal force is not the real force it is basically pseudo force . Is it true to say that this is due to centrifugal force
we call it a pseudo force as it is not a force in our inertial reference frame - but it is a real force in the rotating ref frame (which is not an inertial ref frame)
"Sir, I find your advice about eating yogurt every day except on Friday quite interesting. Could you please explain why Friday is excluded? I'd love to understand the reasoning behind it."
Hello sir i am engeneering students from india and a big fan of you after Albert Einstein sir i need a help from you that please give me a free book the love of physics
Yes I have read the quran and know quite a bit about islam - after reading the quran I decided to remain an atheist. Ofcoz I respect what you believe in.
k=1/(4 pi epsilon) a) kq/( (a+(R1/2))^2 ) b) kq/((R2)^2) c) kq/((R2)^2) d) -kq/(a+(R1/2)) e) -kq/(R2) f) -kq/(R2) Too many repeating values, I'm getting the usual feeling I've done something wrong. I'll try this question again and I'll mention at the top of the comment if I've made any corrections.
@@lecturesbywalterlewin.they9259 My bad, all will be radially inwards towards the center. I just noticed my comment didn't post, I had commented this the same day you replied.
Dr Lewin, I have spent the last several years taking care of my father and he passed in August. Now I would like to complete the work on a model of the dipole moment that bears out the wave function. What university or person would you recommend for collaboration? Are you available?
@@lecturesbywalterlewin.they9259 With all due respect, I think you're selling yourself short as a teacher for thousands of us here! There was no greater teacher in my life than my father and now that he's passed, and Minkowski, and Maxwell, and Lord Kelvin, and Andrew Gray, and Einstein, no greater than Dr. Lewin!
@@lecturesbywalterlewin.they9259 My physics teachers, all them, were baffled. When I read the scientific literature about the dipole moment (every subatomic particle has one) it says, as Wikipedia does and every other credible source as well, that quarks are annihilated giving rise to the dipole moment. This idea of these quarks being slaughtered, disappeared, destroyed, reminds me of the language of the sacrifice. But when I dared to put pen to paper as a youngster, and tried to map out space and time as honestly as I could, I was left with something, forty years later, that closely resembles Isaiah's explanation of Hope in a future Messiah "from the Root of Jesse". In other words people were expecting David's return, But Isaiah, being the honest broker he was, his integrity cautioned against false hope. As "Jesse" was David's father, he appears be implying that one may anticipate the root of the root returning from which our future hope in our new teacher will emerge. He gave the first account of what time is and how It works at a scale that back then nobody gave consideration to beyond it being in the Spirit realm. But the more one reads any thoughtful literature that aims for a high standard, the more clues appear that suggest that our subconsciousness extends to these unimaginably small pathways and they have geoscalar effects in the world we live in. Meaning, we can see the laws of the small world play out in our macroworld but with obviously different characteristics. In other words, in time, it's parity that is reached that leads to a new wave of expansion equal in size to its parents. It's the closest point of return between two pathways, one expanding the other orbiting that gives rise to this geometry and this dipole moment. We can stand upright because we have legs of roughly equal size, there's no sacrifice or annihilation of our legs necessary for us to stand and walk, quite the opposite. And so it is that we can bridge our understanding in a beautiful way through a love of the laws of physics.
IIT offers 38 free classes from technology department none of them physics classes. Only JEE offers a physics class. Professor Lewin is a teacher of physics who teaches here publicly at UA-cam for the benefit of anyone with internet. Do you wish him to teach physics at IIT or take physics classes at IIT which apparently offers no physics? Confused by your comment. Seems disrespectful to a man who is teaching here for all of us, out of his love for physics and nothing else.
This is a very good example. Even teaching at 88, man!
Love from India. Please respond.
Nah nah
-Q induces a total charge +Q onto the inner surface of the conducting sphere. The net charge of the sphere is 0. So -Q is induced on the outer surface of the conducting sphere.
The inducex charge distribution is not spherically symmetric though since -Q is off center. The field lines are normal to the surface of the conductor.
For r> R2, the electric field is E=-k*Q/r^2. k is coulombs constant. Gauss law states: I can compute the E field, as if -Q were in the center of the sphere.
a) E=-k*Q/(R1/2+a)^2. E points radially inward.
b) If B is just inside the charging sphere: E=0. If just outside the sphere: E=-kQ/R2^2, pointing radially inward.
c) same as B.
d) V_A=-kQ/(R1/2+a). V_A=0 if r goes to infinity in order not to get an "insane" result 😊.
e) V_C=-k*Q/R2
f) V_D=V_C=-k*Q/R2. Potential on conducting sphere is constant. V_B = V_C as well.
You're absolutely great sir
You know i'm appeared in jee exam in 2026 by your mit lectures only
great!
Coincidentally i was studying this chapter 😂😂
It is great, l have no enough word to describe your potential on physics. Wow! you are true physician.
Thank you for your kind words.
Sir, from where do you get the motivation that you study through videos at the age of 88 while other professors prefer to rest at this age. Sir, what is your motivation, please share it with us.❤
I was born to teach starting at age -1 till after my death
Prof you are very motivating❤
Hey sir, here's my go at it:
The -Q inside indues a +Q charge on the inner shell (though not of uniform distribution) and that in turn induces a -Q on the outer shell (of uniform distribution according to your 8.02 lecture (I think it was lecture 3 or 4 ). Applying Gauss's theorem, we find:
A. E is radially inward with magnitude E = -kq/(a+r1/2)²
B. Radially inward with magnitude E = -kq/(R2)²
C. Radially inward with magnitude E = -kq/(R2)²
The effect of the -q and the +q induced on the inner shell cancel out in the potential, so the potential here is caused by the -q on the outer shell, which means it is -kq/(a+r1/2) for D and -kq/R1 for E
Hi sir. The charge induced on the inner sphere makes it as if we have a charge -Q at the center and the outer sphere makes no defference at all
A. The electric field has magnitude KQ/(R1/2+a)^2 and is radially inwards towards the center
B. The electric field has magnitude KQ/((R1/2)^4+b^2) and it points radially towards the center
C. The electric field has magnitude KQ/R2^2 pointin radially towards the center
D. The electric potential at A is -KQ/(R1/2+a)
E. The electric potential at C is -KQ/R2
F. The electric potential at D is -KQ/R1
sir how did you became so great in physics?
i would love to hear your advice
I ate yogurt every day but *never on Fridays* Einstein also did that.
Hhhhhhhhhhhhh I love you professor @@lecturesbywalterlewin.they9259
I looked at assignment #2 of 8.02, in which it is shown that the charge on the outer surface of the conducting sphere is uniformly distributed (and totals -Q in this case) regardless of the position of Q inside the sphere.
Thus, external to the sphere, the E field is the same as for a charge -Q at the center point with no sphere.
I hope I have the answers correct now based on the above.
Let k = 1/(4pi epsilon_0)
a) E field at A is kQ/((a+R1/2)^2) directed toward the center of the sphere (toward the left in the diagram);
b) E field at B is kQ/R2^2, directed toward the center of the sphere;
c) E field at C is also kQ/R2^2, directed toward the center of the sphere;
d) potential at A is -kQ/(a+R1/2)
e) potential at C is -kQ/R2
f) potential at D is also -kQ/R2 since the field is zero inside the conductor
The off center charge -Q induces a charge +Q on the inside surface of the conductor such that all field line are normal to the inner surface. The field lines and potential varies inside the sphere accordingly.
The conductor itself is at one potential as caused by a induced charge -Q uniformly distributed on the outer surface of the conductor. So for all r > R_2, the hollow conductor with a charge -Q inside it acts as a sphere with a uniform charge -Q on its surface.
a. E at A is radially inward to the sphere = -kQ/r^2 (r = a + R_1/2)
b. E at B is radially inward = -kQ/R_2^2
c. E at C is also radially inward = -kQ/R_2^2
d. V at A = -kQ/r (r = a + R_1/2)
e. and f. C and D on the conductor are at the same potential = -kQ/R_2
I think for (a) (b) and (c) the E field magnitude should not have a negative sign.
Electric field:
----------------
a) At point A:
E = Q/[4πε₀·(R₁/2 + a)²]
Direction: radially inwards
b) At point B:
E = Q/[4πε₀·R₂²]
Direction: radially inwards
c) At point C:
E = Q/[4πε₀·R₂²]
Direction: radially inwards
Electric potential:
----------------------
d) At point A:
V = - Q/[4πε₀·(R₁/2 + a)]
e) At point C:
V = - Q/[4πε₀·R₂]
f) At point D:
V = - Q/[4πε₀·R₂]
Thanks
Assuming Q>0,
a. (Q/4πϵ₀) / (a + R₁/2)² towards the center of the sphere.
b. (Q/4πϵ₀) / R₂² towards the center of the sphere.
c. (Q/4πϵ₀) / R₂² towards the center of the sphere.
d. -(Q/4πϵ₀) / (a + R₁/2)
e. -(Q/4πϵ₀) / R₂
f. -(Q/4πϵ₀) / R₂
I am preparing for IIT kanpur bless me sir huge fan your autograph on your book for the love of physics
keep it up
Sir I want your autograph on your book for the love of physics
So that I can courier it to you please
How's the atmosphere there teach
Sir i am glade to see your video
Hope you see my comment
Iam in 9 grade but love your vidoe I hope i will meet you😊
Sir where to find your simple pendulum lectures. By the way sir your old lectures are very good.sir i love physics very very very very very very very much and sir i eat yogurt everyday except on Fridays it worked or Einstein and you, I hope it also works for me.
look at 8.01 playlist
Hallo Mr : l would like to ques you. Are you belive that there is other world after die?
after life and heaven were both invented by people out of fear for death.
Sir the electric field is in same Direction as point A towards right
Hi doctor is physics really mixed with mathematics or do we wantto explain it with mathematics
math is the language of physics
@lecturesbywalterlewin.they9259 Thank you professor another question what method do you suggest for a better anderstanding of physics problems
@@Always-x7t eat yogurt every day but *never on Fridays* thhat also worked well for Einstain and for me
Sir how to be humble because you have lot of knowledge in physics but you be so humble but sir at the same point some people know less but they will show that they know more than you how to deal with that
it's not important for me
Stop asking stupid questions
E(A) = kQ/(a+R_1/2)^2
E(B/C) = kQ/(R_2)^2
V(A) = -kQ/(a+R_1/2)
V(C/D) = -kQ/R_2
Electic field's direction is twowards the center of the sphere.
Best wishes
Hello Hello Hellooh!!❤
I hope you're ready for a new problem!
Sir i have a question
Sir when we extract the honey from honey comb by (spinning machine ) we say that the honey came out from comb due to centrifugal force.
centrifugal force is not the real force it is basically pseudo force . Is it true to say that this is due to centrifugal force
we call it a pseudo force as it is not a force in our inertial reference frame - but it is a real force in the rotating ref frame (which is not an inertial ref frame)
@lecturesbywalterlewin.they9259 thanks for this explanation sir
LOVE YOU SIR
Sir, I am student of grade 12 Preparing for MDCAT, I am waiting for solution.
I will post the solution in about 4 days - so far only 7 people have all answers correct
@lecturesbywalterlewin.they9259 Thanks for Reply ❤️,"Until I receive your solution, I will continue working on solving this question myself."
I am a computer science student, and i am struggling with electric physics, the teacher in exam only give use numerical question what should i do
eat yogurt every day but *never on Fridays* That also worked for Einstein and for me
"Sir, I find your advice about eating yogurt every day except on Friday quite interesting. Could you please explain why Friday is excluded? I'd love to understand the reasoning behind it."
Hello sir i am engeneering students from india and a big fan of you after Albert Einstein sir i need a help from you that please give me a free book the love of physics
fiisikis.weebly.com/uploads/5/4/9/3/54939617/for_the_love_of_physics.pdf
❤
First ❤❤ from india ❤❤
You are amazing doctor but do you know about lslam
Yes I have read the quran and know quite a bit about islam - after reading the quran I decided to remain an atheist. Ofcoz I respect what you believe in.
k=1/(4 pi epsilon)
a) kq/( (a+(R1/2))^2 )
b) kq/((R2)^2)
c) kq/((R2)^2)
d) -kq/(a+(R1/2))
e) -kq/(R2)
f) -kq/(R2)
Too many repeating values, I'm getting the usual feeling I've done something wrong. I'll try this question again and I'll mention at the top of the comment if I've made any corrections.
missing are the directions of E-fields in the first 3 questions
@@lecturesbywalterlewin.they9259 ah my bad, forgot that. All are radially inwards
@@lecturesbywalterlewin.they9259 My bad, all will be radially inwards towards the center. I just noticed my comment didn't post, I had commented this the same day you replied.
Sir i am from india and i love your physics world 🌎 🪨
Hello 👋🏻
Sir do you read biology
Yeah sir please answer
Direction at point A is rightwards
that is incorrect
What is the correct answer sir
@@lecturesbywalterlewin.they9259
And the magnitude is Qover [4πEpsilon knot (a+R1/2)²]
Direction at all the points is same as magnitude is same as well?
Sir I want your autograph on your book FOR the love of Physics how do I reach out to you ❤❤❤
From Brasil sir
Such a great explanation 😊
read more closely - *this is NOT an explanatiion* it's the latest bi-weekly Physics problem
Dr Lewin, I have spent the last several years taking care of my father and he passed in August. Now I would like to complete the work on a model of the dipole moment that bears out the wave function. What university or person would you recommend for collaboration? Are you available?
I never get involved in research or ideas of my viewers - it's a matter of scientific integrity - ask you own teachers
@@lecturesbywalterlewin.they9259 With all due respect, I think you're selling yourself short as a teacher for thousands of us here! There was no greater teacher in my life than my father and now that he's passed, and Minkowski, and Maxwell, and Lord Kelvin, and Andrew Gray, and Einstein, no greater than Dr. Lewin!
@@lecturesbywalterlewin.they9259 My physics teachers, all them, were baffled. When I read the scientific literature about the dipole moment (every subatomic particle has one) it says, as Wikipedia does and every other credible source as well, that quarks are annihilated giving rise to the dipole moment. This idea of these quarks being slaughtered, disappeared, destroyed, reminds me of the language of the sacrifice. But when I dared to put pen to paper as a youngster, and tried to map out space and time as honestly as I could, I was left with something, forty years later, that closely resembles Isaiah's explanation of Hope in a future Messiah "from the Root of Jesse". In other words people were expecting David's return, But Isaiah, being the honest broker he was, his integrity cautioned against false hope. As "Jesse" was David's father, he appears be implying that one may anticipate the root of the root returning from which our future hope in our new teacher will emerge. He gave the first account of what time is and how It works at a scale that back then nobody gave consideration to beyond it being in the Spirit realm. But the more one reads any thoughtful literature that aims for a high standard, the more clues appear that suggest that our subconsciousness extends to these unimaginably small pathways and they have geoscalar effects in the world we live in. Meaning, we can see the laws of the small world play out in our macroworld but with obviously different characteristics. In other words, in time, it's parity that is reached that leads to a new wave of expansion equal in size to its parents. It's the closest point of return between two pathways, one expanding the other orbiting that gives rise to this geometry and this dipole moment. We can stand upright because we have legs of roughly equal size, there's no sacrifice or annihilation of our legs necessary for us to stand and walk, quite the opposite. And so it is that we can bridge our understanding in a beautiful way through a love of the laws of physics.
I saw this the other day. Seems like he got a lot from your lectures :
ua-cam.com/video/24GfgNtnjXc/v-deo.htmlsi=P0ArJmTIngfera1e
this is a very bad video - my rainbow videos + demo are *MUCHHHHHHHHHHHHHHHHHHHHHHHHHHH* better
Thank you sir
Welcome
Bro went from making videos at MIT to recording his screen with a camera 😂😂😂
Sir we Indians will be grateful if you take physics classes on IIT
IIT offers 38 free classes from technology department none of them physics classes. Only JEE offers a physics class. Professor Lewin is a teacher of physics who teaches here publicly at UA-cam for the benefit of anyone with internet. Do you wish him to teach physics at IIT or take physics classes at IIT which apparently offers no physics? Confused by your comment. Seems disrespectful to a man who is teaching here for all of us, out of his love for physics and nothing else.
thanxxxxxxxxxx