How to get electric field from electric potential

[yuanxiang]

Thank you so much! My professor have never taught us about gradient and expected us to just know what it means out of thin air. This is a life saver!

[stephanm.tjaden]

This was very, very helpful. I have been struggling to no end on this because the book poorly explains it. This explanation is perfect and very clear cut, and easy to understand.

[cjterhark]

Clear and concise. I like that you clarified what the gradient is and what it does. Helped me a lot. Thanks!

[zachgreenwood]

For those of you looking for bare bones answer to this question. Just take the negative gradient of your electrical potential equation, that is take the derivative of the equation with respect to x, then y of the original, and then z of the original and assign x hat, y hat, and z hat respectively.

[danijelumicevic]

You just made my day, now I can go to sleep because I found a guy with decent explenation

[JustinDBKidrauhl]

You sir are an angel fallen from above. Crazy how I love physics when I'm being taught by YouTube professors :)

[MikeHawk-sg]

Thank you so much! I was really scared of those partial derivatives until now!

[Daniela-izec]

Sir thank you so much !!!!
You just made my life so much easier and happier !!!
Thank you !!!!

[ashewood]

kinda random but the way he wipes the markers off when they're not dry yet is frqn mesmerizing

[jamesbedortha]

good video, help me understand what's going on in my physics class

[kosmos]

Thank you so much sir, tommorow is my physics test, your videos helped me a lot.

[jessicaquemado]

I got stuck in studying this in my book, I did not know how this equation was derived: E = -dV/dr until I decided to search the internet and found your video hahahahaha thank you so much Dr.

[ethioupdatedinfo.]

Thank you very much professor Anderson 🙏you are really help me a lot.

[jahansaid]

Thank you very much, It was so much helpful, you opened my mind :)

[davemcintosh]

Thanks. Slightly confused by why the absolute value of the potential due to the charged sheet is proportional to z. Wouldn't it reduce with greater displacement from the sheet? i.e. wouldn't it be proportional to 1/z, in the same way as potential due to a point charge is proportional to 1/r?

[yasir.]

Hello prof, hope hale and hearty during this quarantine


Cheers,
Yasir

[spaghettigal]

Where could I find some harder examples? Thanks in advance!

[shalomfadullon]

great video, but couldnt pay attention because i was distracted by the fact that HE HAS TO write everything BACKWARDS???? or flip it??? mind blown. he is so good

[seannibecker]

are you drawing everying in reverse? cant figure it out

[shubhamghadai]

Sir, why have you considered two equipotential surfaces in the second case?