I would like to thank you from the bottom of my heart, Griffiths did a terrible job of explaining how to get to this point and I really appreciate you.
@himanshujha6910
6 жыл бұрын
Sir, It would be more helpful if u perform the integral part. That is the important part.
@ciarahendricks2003
Жыл бұрын
This is a physics lesson , not a calculus lesson. You’re supposed to already know the calculus before you get here, because it is the most important part, it’s a prerequisite. If you don’t, get an integral table or use an online calculator.
@tomoalla
Жыл бұрын
@@ciarahendricks2003 Typical Physics Professor answer lol
@brendanlarson3432
4 жыл бұрын
Very helpful, but the static gets to be annoying.
@tonmoydeka7319
4 жыл бұрын
griffiths is not understandable,,,its so direct
@apotpan6715
8 жыл бұрын
Greetings Sir,I think in V(r) you have abused the symbol θ' which isn't the same for Pcosθ and dθ'
@jamesvanhowe1718
8 жыл бұрын
I don't edit these, so occasionally there are small errors. These videos are made to go with a face-to-face classes where my students and I can follow up in-person. It could be that prime is missing in one of the equations. However, there shouldn't be many errors and my hope is that the viewer is following along well enough to catch them when obvious. However, please use as fitting! This particular video should follow closely the notation in Griffiths Introduction to Electrodynamics. Thanks for your note, I will look into this for future versions.
@arushsingh5471
5 жыл бұрын
the integral part can be easily done if u take the whole denominator as some r. Then the numerator consists of its derivative.
@sadiatahir3825
7 жыл бұрын
Griffiths writes Cos theta in potential relation.Why?
@woutgevaert4864
7 жыл бұрын
Sadia Tahir because Griffiths also considers places where theta≠0, i.e. not only on the z-axis
@RAJDEEP.SARKAR.
8 ай бұрын
Thank you sir 🎉🇮🇳
@maujo2009
6 жыл бұрын
Why does Gauss' law fail when using it in this problem?
@jamesvanhowe1718
6 жыл бұрын
Conceivably you could use Gauss's law if the Qenc = Qfenc + Qbenc, that is you take into account all free and bound charge, AND you draw a Gaussian surface on which all field lines are normal to the surface AND simultaneously where the E-field is the same value on this surface (in Gauss's law we can only un-dot the dot product in (E dot da) if this is true; we can only pull the E out of the integral if it is constant on the surface we draw). There is no obvious symmetry (at least in Cartesian, Cylindrical, or Spherical coordinates) that allows this here.
@brycejohnson9291
3 жыл бұрын
Thank you
@quantaali543
5 жыл бұрын
can you please tell me how to solve problem 4.10 where we have a non-uniform polarization. In the solution manual, they have used gausses law but you have said in @MauJo reply it cannot be used.
@sangwan.ajay89
2 жыл бұрын
Tell me what is that problem 4.10
@xoxo7940
7 ай бұрын
Thank you from india❤
@Sg0224
4 жыл бұрын
for field due to uniformly polarized sphere using gauss law i failed terribly ! plz help
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