I don’t think this kind of approach would work for an ellipsoid, because the superposition of dipole and uniform field doesn’t have any ellipsoidal equipotential surfaces. To find the electric field in this case may require solving Laplace’s equation directly, which would be a lot more work!
I’ll add it to my to-do list! For now, the brief explanation is that it works basically the same way as solving a linear ODE. You take a trial solution of the form uₙ = rⁿ, substitute it into the recurrence relation and solve the resulting polynomial for r. The general solution is a linear combination of the rⁿ terms you get from solving the polynomial (so uₙ = Ar₁ⁿ + Br₂ⁿ + …). You can determine the constants A, B, … if you’re given a few specific terms of the sequence - the idea is the same as determining the coefficients in the solution of an ODE using initial conditions or boundary conditions.
❤ from india Sir can u pls make a video to find electric field at edge(vertex) of a triangular sheet(equilateral) with surface charge density sigma and electric field at a point p at a distance above(at z axis) centroid of a equilateral triangular sheet of surface charge density sigma My sir solved this problem by flux method but i wanted a classical integration method
Good question! Consider adding the potential derived in the video to the potential of a uniform spherical distribution of charge, i.e. Q/4πε₀r where Q is the total charge. The resulting potential is still constant over the surface r = a, so by the uniqueness theorem we can place a spherical conductor of total charge Q at r = a. We can therefore conclude that the electric field of a conducting sphere of charge Q placed in a uniform electric field is a superposition of the field derived in the video and the field of a uniform spherical charge distribution of charge Q.
Reminds me of the irodov problem,where we can solve this by placing 2 spherical charged bodies together and displacing their centre by small distance 😅
Indeed it is, so our conclusion is that the electric field outside the sphere is equivalent to the field that would be produced by a very small dipole at the centre of the sphere.
@@DrBenYelverton exactly. But the charges are at the circumference of the sphere so doesnt that make the size of the sphere and the distance between the charges comparable?
The actual charges here are indeed on the surface of the sphere, but the result of the video shows that the overall field you get is indistinguishable from the field that would be produced by a small dipole at the centre.
That's true, in that we're basically solving ∇²V = 0 in both cases. The difference is in the boundary conditions - in the electrostatic case the E field must be purely radial at r = a, whereas in the potential flow case the velocity field must be purely tangential there.
Now that just clarified 3 sections of Purcell so neatly! Thanks mate
Thanks for watching, good to hear that it made things clearer!
THANK YOU SO MUCH SIR,THIS VIDEO IS NICE AND EXPLANATION IS VERY GOOD👍👍👍👍👍💚💚💚💚💚
Thanks for watching!
Loved your work
Thank you!
I love your videos. Keep it up with the great uploads. 😀😀😀
Thanks for saying so! I have some new videos planned and hope to find time to record them soon.
this approach is nice
What if a conducting ellipsoid whose long axis is parallel to z? what is the value of r which makes total potential V=0? Thank you:)
I don’t think this kind of approach would work for an ellipsoid, because the superposition of dipole and uniform field doesn’t have any ellipsoidal equipotential surfaces. To find the electric field in this case may require solving Laplace’s equation directly, which would be a lot more work!
@@DrBenYelverton Thank you. Your videos help me a lot!
@@たこわさ-e3f Happy to hear that, and thanks for watching!
Thank you! 🙏
ben can you go over solving recursion relations using auxiliary equation?
I’ll add it to my to-do list! For now, the brief explanation is that it works basically the same way as solving a linear ODE. You take a trial solution of the form uₙ = rⁿ, substitute it into the recurrence relation and solve the resulting polynomial for r. The general solution is a linear combination of the rⁿ terms you get from solving the polynomial (so uₙ = Ar₁ⁿ + Br₂ⁿ + …). You can determine the constants A, B, … if you’re given a few specific terms of the sequence - the idea is the same as determining the coefficients in the solution of an ODE using initial conditions or boundary conditions.
Just uploaded an example showing how to use this method: ruclips.net/video/14pBNzI-yuw/видео.html
❤ from india
Sir can u pls make a video to find electric field at edge(vertex) of a triangular sheet(equilateral) with surface charge density sigma and
electric field at a point p at a distance above(at z axis) centroid of a equilateral triangular sheet of surface charge density sigma
My sir solved this problem by flux method but i wanted a classical integration method
Interesting idea, I'll see what I can do!
Great explanation
Thank you!
How about if this conducting sphere is charged? Is the situation the same? Thank you！
Good question! Consider adding the potential derived in the video to the potential of a uniform spherical distribution of charge, i.e. Q/4πε₀r where Q is the total charge. The resulting potential is still constant over the surface r = a, so by the uniqueness theorem we can place a spherical conductor of total charge Q at r = a. We can therefore conclude that the electric field of a conducting sphere of charge Q placed in a uniform electric field is a superposition of the field derived in the video and the field of a uniform spherical charge distribution of charge Q.
@@DrBenYelverton This is exactly what I came to learn. It all makes so much sense now. Thanks for the explanation.
@@starstuff11 Excellent - the uniqueness theorem is powerful but not always intuitive!
@@DrBenYelverton why superpostion makes sense,why it is a linear system?
@@超级悠米 Electric fields (and hence potentials) are ultimately governed by Maxwell's equations, which are themselves linear.
🙏
Reminds me of the irodov problem,where we can solve this by placing 2 spherical charged bodies together and displacing their centre by small distance 😅
Yes, I like that result as well!
You should have shown us the point at which you have found the potential
It's an arbitrary point.
haha mich IB content
stressing for our IB results on july 8th
I remember that wait!
The result you have used for dipole is for a far away point, I dont see how it makes sense?
Indeed it is, so our conclusion is that the electric field outside the sphere is equivalent to the field that would be produced by a very small dipole at the centre of the sphere.
@@DrBenYelverton what do you mean when you day that the dipole is verysmall? Do you mean the moment of dipole is very small?
Not necessarily the dipole moment - the separation between the two charges is much smaller than the size of the sphere.
@@DrBenYelverton exactly. But the charges are at the circumference of the sphere so doesnt that make the size of the sphere and the distance between the charges comparable?
The actual charges here are indeed on the surface of the sphere, but the result of the video shows that the overall field you get is indistinguishable from the field that would be produced by a small dipole at the centre.
Man you're cool
How can I contact you I am from India and I want to have a friend in London will you be a friend of mine 🙏 namaste from India
+ analogous to the potential flow across a sphere
That's true, in that we're basically solving ∇²V = 0 in both cases. The difference is in the boundary conditions - in the electrostatic case the E field must be purely radial at r = a, whereas in the potential flow case the velocity field must be purely tangential there.