Griffiths Electrodynamics 2.4 Electric Field Above Center of Square Loop (DETAILED SOLUTION)

[@bmoreno-ri8]

Hey! Thank you so much! everyone was solving it by doing obvious subs based on the results of the previous griffiths problem and I was struggling a lot to find the r vector and establish the initial integral!, you did that and much more so thanks a lot!!

[@kirokamui]

No one else around opens the calculus like you Sir. Thanks a bazillion!!

[@miraj_sippy]

I don't know why India's engenniring entrance exam is so tough that I have read almost every good concept in physics on youtube😅

[@ashukushwaha3469]

Waw , what's a smooth idea. I respect you sir 🙏

[@anirbantalukder5281]

At 5:20 I think -a/2 will be in place of a/2 . Because a/2 lies in the negative of y axis.

[@NickHeumannUniversity]

Hi! The r vector is the "path" we must take from the charged line to the point of reference P. Thus, because we are integrating at y=-a/2, we must move +a/2 to get to y=0,which is where P is at. Hope that helps!

[@Seven4Three]

thank you so much. I have some question, When the find r vector, can I use x vector to 'x' instead '-x'?

[@yousiftop7605]

very helpful brother, ty so much ❤

[@Iqbal00123]

great work bro

[@anxiety9z]

woooow thank youu so so so so much mannn

[@ImranAli-bz6ks]

Bundles of thanks 😊☺

[@marcoolivar]

Se agradece enormemente..ahora si es dificil, más dificil me la haces en inglés...jejejej....Uno en español...please

[@NickHeumannUniversity]

se puede activar subtitulos autotraducidos en español :D si te sirve de algo

[@AnaVictoriaCampos]

Maybe try being a bit more detailed on the integrals, I really don't understand how you got dx on the tan one :,C