your videos are literally saving my academic life. thanks Michel van Biezen! You're the best!
swsj
I can't tell you how much stress you alleviated by lecturing this subject and sharing it online. Maybe my self stress is a bit self induced but I am still greatly appreciative of your effort to share knowledge like this with the general public.
baneoflife
I was completely lost in Physics 2. Then, 3 videos in I'm already understanding everything we've been going over in class for weeks. How do you explain in 30 minutes what my teacher struggles to explain in 9 hours?
pissy
I've learned more from your videos than 3 weeks of lecture about Gauss. Thank you!
moniquecastillo
Nur Bedeir,
In this example, the charge resides on the surface so the surface density was given.
Total charge = surface density * surface area
MichelvanBiezen
My main man Michel van Biezen!!!
Thank you for uploading all these tutorials. I spend hours sitting in lectures and tutorials at university but I tell you now that watching your videos has actually helped me the most.
Big Thank You !!
Psygression
2020 there are still people learning from your videos. congratulations
tbvinicius
to be honest i learn more from your videos than man teacher, before christmas i had mechanics and i passed my exam just because of your videos, now i have electormagnetism and optic, and some other math subjects. i am glad these videos are availible. thank you sir!
javadhikmati
Since the sphere is a conductor, all the charge will reside on the surface, thus the total charge will be rho * surface area. ( rho = 2.5 micro C / m^2)
MichelvanBiezen
I just want to let you know that the hard work you put into these videos is greatly appreciated. Your videos have helped me and many others quite a lot, thank you so much!
MrRdoc
You're welcome...I'm glad you found it helpful!
MichelvanBiezen
Currently I am working on waves and sound videos. I'll add some more examples of Gauss's law in the near future.
MichelvanBiezen
Prof. Van B., I believe that you have reversed the gaussian radius and the sphere's radius in your equation. The area of E.dA should equal E(4*pi *1.0^2) not E (4*pi*5.0^2). I may be wrong but to me the magnitude of E should be based on where you measure it.
obliquecolumn
Good job, very simple to understand with your explanation. Thank you!
_otunac
You are the best, These lectures are amazing, I enjoyed it reminds me with all I took in my college, it is a good preparation before teaching these lessons, please Keep up this work
MohamedGamal-rvcm
Have to agree with Javad, Your videos are amazing and without them I would be completely lost in my physics class! Thank you so much!
caseycrews
So if you have a negative charged sphere instead, would you get the same answer but negative?
cyn_knight_q
At 5:15 where 4pi is coming from I couldn't catch it
Eser_Ozbilgic
Stated simply and in a straight-forward way. Thank you for your help.
