AP Physics C - Gauss's Law

[@ShilpitaMitraBehura]

Hi Mr. Fullerton! We used your videos in my physics class all year, and your method of teaching physics helped so much. I ended up doing well on both my AP Physics C exams this year, so I just wanted to thank you for your videos -- they definitely helped me during the test. Thank you again!!!

[@DanFullerton]

Thrilled to hear it, and you're very welcome!

[@jesseji7544]

Wow I’m finally at the topic that you posted this comment for!!

[@KapilKumar-yi3uu]

A

[@nathanbotello9023]

"We've got some pretty slick cancelations coming up here" I love it

[@calebbean2573]

best pickup line ever

[@Chelsea2009FC]

lmao

[@DescartesRenegade]

incredible video, by far and wide one of the most intuitive and concise explanations of Gauss' Law. thank you

[@DanFullerton]

You're very welcome!

[@umarpatel3517]

"We didn't really do any of the actual integration. We used our common sense" LOLOLOLOLOLOLOLOLOLOLOL You really are a legend man. These videos are super helpful

[@DanFullerton]

So glad you're finding them useful!

[@AnonymousIdiott]

You are more qualified to teach than any of my engineering professors! I appreciate your existance!

[@wingzerofor]

After scouring my coursebook for class and still not comprehending electricity, came across your videos and now it makes sense. 👍

[@DanFullerton]

Glad to hear it!

[@monolaboratories]

Thanks for the videos! I got a bad flu or something and was out for two weeks and missed all of my lectures. A friend's notes and your videos has definitely saved me from a certain failure.

[@jameslu689]

holy, this vid just made me understand gausses law that I've been struggling with all year

[@DanFullerton]

I'm thrilled to hear the video helped!

[@wesker121212]

Just wanted to say man, I'm a UCLA senior in a calculus based EM physics class and your explanation of Gauss's Law was fucking beautiful and concise (midterm in t-7 hours)!

[@DanFullerton]

Thanks, and good luck on your mid-term!!!

[@KuroHinata0321]

My Intermediate E&M professor didn't bother reviewing the 'basics,' even though it's been two years since just about anyone in the class (almost all juniors, and phys majors) took freshman E&M. This has been a really helpful refresher. Thank you, sir!

[@DanFullerton]

You're very welcome!

[@rainbowcreations4396]

One of the most efficient explanation of Gauss's Law..It is simple as well as worth understanding

[@DanFullerton]

So glad you found it useful!

[@CoyoteKitsune]

Considering I'm studying for a test right now, I greatly appreciate your video. It's making the week of confusion suddenly start making sense. I keep going "Oooh. That's how!" Thank you -so- much for this. Seriously.

[@BBVictini1]

If, by some miracle, I manage to pass this AP Physics C Electricity and Magnetism exam, it will be SOLELY because of you! These videos are the only things giving me any sort of understanding of these topics. Thank you!!

[@DanFullerton]

I prefer to think of it as electric flux on the left-hand side, but the answer to your question is yes to both. Because any charge not enclosed creates flux (field lines) that must both enter and leave the closed surface, their net flux is zero, so yes, the flux is due to the enclosed charges, but it's also due to the charges outside the closed surface, its just that charges outside the closed surface have a net flux contribution of zero.

[@DanFullerton]

In this case it doesn't matter. Charge would be distributed evenly across the shell either way. The electric field inside the shell is zero because there is no enclosed charge (the whole point of Gauss's Law). For more detailed explanations, please post questions on the APlusPhysics site and I have much more leeway and space to respond to questions there. Thanks!

[@wingjaigaming8240]

this 23 min video makes more sense than the 3 hour lecture i attended last week. Thanks for this educational video, everything is clear and easy to understand

[@DanFullerton]

Glad you found it helpful!

[@TranOfficial]

It's like Christmas all over again! Thanks Dan, for getting this E&M series up in conjunction with the pace of high school classes.

[@DanFullerton]

Absolutely -- looks like I got a touch sloppy in my notation here -- outside the sphere, the electric field is Q/(4*Pi*e0*r^2), where r is the distance from the center of the charge distribution to the point where you want to know the field.

[@DanFullerton]

Wow, that made my day (been a rough day). Thanks, and I'm glad you're finding these helpful. I'll do my best to keep up with the AP-C curriculum this spring...

[@soliderarmatang5664]

hi, i want to say thank you for the videos, they are very clear, i appreciate your effort, and i highly appreciate your professional look, i know that takes time and i acknowledge that and appreciate it, thanks once again

[@DanFullerton]

You're very welcome, and thank you for the kind words!

[@avinashvenugopal9059]

Oh man. This video is amazing, i actually watched and rewatched it. Honestly, i dont do that for many videos. Great work!

[@eday00]

Oh man, your face at 7:25 is epic. In my mind I heard you say "BAM! That's what's up." Thank you for this video, clearest I've found so far.

[@fakedude5439]

Thanks for all the time and effort you put into these videos. Your explanations are clear and informative!!

[@DanFullerton]

You are very welcome!

[@aloischipfurutse5079]

I liked this video and subscribed after watching the first one minute. Your method of teaching is incredible. I have not finished the video but I am sure it will be easy to understand.

[@DanFullerton]

Glad you're finding it helpful!

[@DanFullerton]

Hi Martin -- the net flux through the sides is going to be zero due to symmetry arguments. If it's an infinite plane, we have the same amount of charged plane to the left, right, front, and back, and they're all going to cancel out, so the only place we have a NET flux is through the caps of our Gaussian cylinder. Helpful?

[@publiclyfunded5054]

Dude, fantastic video. I have an exam on this tomorrow and your explanations are by far the best resources on Gauss's Law that I could find online.

[@DanFullerton]

Glad you found it helpful -- good luck on your exam!

[@debghoshila]

11.38 - I think it should be 4 pi epsilon r0^2 instead of R^2?

[@DanFullerton]

Thrilled to hear it helped, and make it a great day yourself!

[@briangonzalez6432]

I feel so blessed to have someone like you teaching physics. Very understandable

[@DanFullerton]

Awww, thanks!

[@alieeldeen660]

Hi Dan At 9:26 when write electric flux = EA you write A 4pi R2 why you write R the radius of the original sphere not the radius of the Gaussian surface And also Electric field due to infinite plane you don't cancel electric flux up and down as you do in Electric field due to infinite line charge you cancel electric flux right and down Explain that please and thanks in Advance :D

[@davidf2769]

Great video, saving my butt on an Electricity and Magnetism final!

[@DanFullerton]

Good luck on your exam!

[@fdist04]

great butt, saving my video on the E & M AP exam!

[@DanFullerton]

Hi Alie -- poor notation on my part, I didn't differentiate between symbols for radii. For the infinite plane, the electric field at the top or bottom surface doesn't cancel out as I'm trying to find the electric field at that point. When I'm looking at the left and right sides of the cylinder I note that by symmetry the net will be zero at any edge of the cylinder.

[@alieeldeen660]

Dan Fullerton I want to check that Look Dan when we talk about the infinite plane we put cylinder as a closed surface so when divide the body into three parts top and bottom and remain part .. so the electric field when it +ve the electric field is outward and as we know the A vector is perpendicular to surface so the angle between E and A in up and down is zero and the angle between the E and A in the remain part equal 90 so the it cancelled In infinite line Left and right make Area with E 90 degree and the remain make Angle zero so we cancelled the left and right .................... Am I right ???

[@lancelovecraft5913]

Electric flux is the product of the magnitude of the electric field through a surface. Therefore we are only concerned with the change in surface area(dA). By looking at the shape in question you know that the sum of all those changes in area will equate to the surface of a sphere.

[@michaelliangzhu]

when i read the giancoli college physics textbook of this chapter , i was totally confused . then i watched this video and everything written on the book makes more sense. thx for your effort to make the class fun and intuitive .

[@DanFullerton]

Michael Leung Tsu My pleasure Michael, thanks!

[@alex1993seattle]

Wow, really helpful, my gratitude is beyond description. I think i will be fine on the exam

[@antiview]

so... I have a test tomorrow on the entire unit 1. Your videos are very helpful. Will update with my score.

[@DanFullerton]

Hope it went well!

[@TheRealTonicus]

Fantastic video. You explain Gauss's law much better than my textbook or professor.

[@MrLatinhulk]

I don't know why but seeing his face while hes lecturing actually helps me understand the material so much better. crazy anyway thank you Dan Fullerton

[@InteractiveEducation]

This is absolutely brilliant!!! its brilliant to see fellow tutors on youtube teach so well!!!!

[@schoe1600]

you taught this infinite times better than my first year physics professor.

[@DanFullerton]

Glad to hear this helped you out!

[@jessicamannix4486]

Thank you, Mr. Fullerton. This stuff is tough, and you made it simple yet still explained everything I need to know. Thank you!!!

[@DanFullerton]

So glad you found it helpful!

[@martin19951374]

just one brief question. in 15:00 you said that flux through the surface is going to be zero. is it because the dot product has a cos (90)? if there's another explanation please mention it thank you.

[@DanFullerton]

Wish I'd thought to say that when recording the video! :-) Glad you found the video helpful. Make it a great day!

[@DanFullerton]

I'm hoping so... good luck -- would love to hear how it goes!

[@DanFullerton]

Gauss's Law is always true, that the net electric flux is equal to the sum of the enclosed charge divided by the permittivity; its usefulness in finding the electric field, however, is basically limited to situations of symmetry. The left hand side of Gauss's Law always gives you the net electric flux... does that help? Good luck to you!

[@poodook]

Dr. Fullerton: I have a question regarding the placement of our imaginary Gaussian surface. For the infinite line of charge (19:36), what dictates our decision to place the cylinder as shown in the video? What's stopping us from instead situating the surface in the vertical direction (perpendicular to horizontal orientation of infinite line)? I would appreciate some general discussion on the criterion one should consider when placing this Gaussian surface... it's not quite so intuitive for me yet.

[@aqibsoomro2702]

good

[@dr.j.a.ebenezerassociatepr7796]

Wow. Your piece of information, demonstration and pictures inspire me. Nice contribution.

[@patrickmahomes6580]

YOU DA MAN!! I have a physics test in a couple hours and I know fully understand the equations and how they work!

[@DanFullerton]

Good luck!

[@ezginingunlugu3247]

hi, thank you for the videos they are incredible. I didn't understand just one thing. At 16:18 you're saying "no dependence on the distance same thing we saw previously" but you show on the pre example the more distance means less E (11:58) can you please help me I'm so confused

[@darkryuful]

Great, clear video. Thanks for taking the time.

[@DanFullerton]

You're very welcome!

[@theodoreberbach9562]

I don't get you at 11:17 at all, what do you mean it doesn't matter if you are outside the field, and what does the graph suppose to mean

[@DanFullerton]

If you are outside the shell, you get the same electric field from an enclosed sphere of charge as if you had a point charge in the center.

[@jacobvandijk6525]

How can epsilon-nought (in the denominator) be combined with q-enclosed (in the numerator)? Epsilon-nought is the permittivity of free space. In a vacuum there are no particles, so there is no charge! So when do you combine these two?

[@zacay5717]

awesome video. Can you elaborate a little bit more on the idea of the flux on the sides of the cylinder being 0? thank you.

[@DanFullerton]

Because it's an infinite sheet of charge, there is the same amount of charge on any side of the cylinder, so the next flux will total zero. I can elaborate further if you let me know whether you're looking at the line of charge problem or the infinite plane problem...

[@Cub994]

Can I ask a question? about 16:21, if we choose a point that an infinitely remote from the plane, would flux be the same?

[@DanFullerton]

It would be the same, because it's an infinite plane. Note that there is no dependence on distance from the plane (kinda cool, huh?)

[@ben_swain]

It took me a while to figure out that, in your formulas for electric field, you meant R to represent ri and ro instead of the R as used in the diagram for the radius of the charged shell. I'm guessing that has confused a lot of people, but overall a great video!

[@DanFullerton]

Thanks Ben, but in this video the R does indeed represent the R in the diagrams. It is quite possible in some of the other videos I become a bit more lax in notation as we become more familiar with the concepts, but in looking over this one I think this video is OK...

[@jackysub1941]

@8:35, do the fluxes on the left and the right 'cancel out' or are they just 0? I assume they are zero because the direction of the Electric field and their dA is perpendicular?

[@DanFullerton]

I think you have the wrong time-stamp here... 8:35 isn't talking about fluxes.

[@jackysub1941]

@@DanFullerton I think I was thinking about @20:28

[@markdave2456]

At 9:13, why is the area 4(pi)(R^2)? To be more clear why do you use radius "R" and not "r inside". In the last example in this video you used the area of the Gaussian surface but in the example at 9:13 you didn't use the area of the Gaussian surface, you instead used the area of the thin hallow shell (I'm assuming this is a spherical hallow shell)?

[@markdave2456]

My question also applies to finding the electric field outside the thin hallow shell, the radius being used is "R" and not "r outside". When we use Gauss' law do we use the area of the Gaussian surface or the area of the object in question?

[@DanFullerton]

Typo, it should be r_inside (see previous comments below). :-)

[@rahulbball9395]

Thanks a lot. You are a really good teacher. Keep doing your thing bro, im sure it helps thousands of people around the world.

[@DanFullerton]

Mighty kind of you to say, and I certainly will. Thanks!

[@akshanshgupta3731]

In the last problem about the infinitely line charge, why does fluxL and fluxR cancel out?

[@DanFullerton]

My pleasure, and you're welcome!

[@ezrapotter4631]

The surface area of a cylinder is (2*pi*r*L)+(2*pi*r^2)

[@DanFullerton]

True... which is why when we look at just the sides of the cylinder, cut it flat and look at the area, you get the 2*Pi*r*L noted at 21:06. No flux through the caps, so we don't worry about the 2*pi*r^2.

[@DanFullerton]

Glad to hear you're feeling better!

[@dennistu]

at 11:11, shouldn't the R be the radius of the Gaussian surface instead of the radius of the shell charge distribution?

[@samanthaspano3652]

When you derive the E formula for a Gaussian surface outside a hollow sphere of charge, shouldn't the radius used be the distance from the charge itself to the Gaussian surface (a.k.a. r0 minus R)?

[@jackmurphy5614]

Really helpful video! The only part that I was confused about was with the spherical Gaussian surface example where r0>R. The radius of the charged spherical shell was labeled R, and the radius of the larger spherical Gaussian surface was r0. The final formula you arrived at was E = Q/r*pi*Eo*R^2. Is the R supposed to be referring to the radius of the charged spherical shell as implied here, or to the radius of the larger gaussian spherical surface?

[@DanFullerton]

Hi Jack -- For all radii less than the radius of the charged shell, E=0. For all radii greater than the radius of the charged shell, E=Q/(4*pi*Eo*R^2), where I'm using R as a general radius (didn't mean to set R as a constant equal to the radius of the charged shell.) Hope that helps, and sorry for any confusion!

[@jackmurphy5614]

Oh ok thanks that makes sense!

[@zhenoob]

Thank you for the video! Could it be explained that the flux inside the hollow sphere is zero because by vector addition, the electric fields from individual charges cancel out at every point in the sphere?

[@DanFullerton]

Glad you're finding the videos helpful!!!

[@tanalpha123]

when we use the guss's law to find the electric field at a point in the gaussian surface and when there are different charges out side of the gussian surface, giving different electric fields to that point, does the answer from the guss's law gives the electric field occurs due to the enclosed charges or the net electric field due to the all charges? thank you

[@JossinJax]

Hi. In your second example, with the infinite plane of charge density, why is there E perpendicular to the surface? Aren't we talking about a plane that has charge? So are you saying that E radiates out of the plane in both directions instead of E existing only on the plane? Gosh, no wonder I've been confused. I had the whole thing inverted in my mind.

[@nafichowdhury54]

Very clear information Sir. More clear than my academic books. It helps me. Thank You :)

[@DanFullerton]

+Nafi Chowdhury So glad it helped you!

[@pengrunhuang3270]

Thank you for the video! I have a question, if the electric flux represents the amount of electric field penetrating a surface, why should use we cosine but not sine (at 2:36)? If a electric field is 90 degree to the plain, the electric flux is 0, it doesn't make sense to me. Can you explain that, thank you.

[@DanFullerton]

Sure. Think of it like blowing air across a paper. The air is the flux, the amount that hits the paper is what counts. If the paper is at 90 degrees to the moving air, it all hits the paper (cos 0 = 1). Keep in mind the direction of dA vector is perpendicular to the surface. Tons of flux. If the paper is parallel to the air (cos 90=0) lots of wind, but it just goes along with the paper, not into it. Take careful note of the direction of the dA vector, that's where people tend to get flummoxed.

[@pengrunhuang3270]

Oh, that makes sense if the direction of dA vector is perpendicular to the surface. Well, but A is the area, why will dA be perpendicular to the area?

[@DanFullerton]

That is how the dA vector is defined, otherwise all the following formulas would need to be re-worked.

[@tvedi23]

You are a fantastic human being, sir!

[@DanFullerton]

Thanks!

[@tanalpha123]

Thanks for the reply.Actually what i dont undestand is when we are using the equation EA=Q/£ does the ''E'' stand for the electric field occurs due to only the enclosed charges inside the gaussian surface or is it the resultant electric field due to all the charges inside and out side of the gaussian surface. Thank you

[@01TheViewer]

I'm still not really understand in 9:49. the shell is conductor or non-conductor?? Why the electric field inside the shell is zero but it has a E field outside....?Could you explain more details? thank you.

[@benjaminbridge5975]

Thanks a lot, you make it easier to connect the calculus with the physics

[@littledivergirl1105]

Do you have a video that would explain how to treat a cylinder with a smaller cylinder cut out- to find the E inside that hollow space. I know I it has to do with a superposition of Gaussian surfaces but not sure how to go about the actual calculations. Thanks!

[@MnMDigital]

Dan, great videos and a life saver! Just a question, at 11:00 , were we supposed to change the R in the denominator to r0? Because doesn't the electric field depend on the radius of the point at which we are trying to determine the electric field? Thank you!

[@DanFullerton]

Glad you liked it -- would still recommending reading the textbook though. :-)

[@DanFullerton]

Glad you liked it!

[@desborah]

Hello, Doctor Fullerton, thank you so much for this video, everything is so clear right now, I was trying to understand this by reading my textbook, but it didnt work. Now that I saw your video, its so much more clear!! Greatings from Brazil! keep up the good job!! :D

[@DanFullerton]

+Déborah Andrade So glad to hear the video helped. Good luck, and make it a great day!

[@andrewchang7194]

Just a fun question for you, what would the electric flux penetrating a mobius strip be? It's a non-orientable surface and it's impossible to find a normal vector. Are there work-arounds?

[@DanFullerton]

+Andrew Chang That question makes my brain hurt. Sounds like a great research project!

[@vrowniediamond6202]

This video explained the concept so much better than my teacher ;-; thank you so much

[@DanFullerton]

Glad you found it helpful!

[@111.saisrikanthsattenapall9]

In finding the electric field of a thin metallic sphere inside and outside of the sphere, mustn't be the radius for calculating area and electric field be taken as the radius of the gaussian surface?

[@Hardly_Human]

Could someone explain why the theta is always zero a little bit more.

[@sasa2825]

Thank you so much. You are a spectacular teacher

[@ibobble17]

how come the flux will not equal zero for the cylinder crossing the infinite plane? Won't the flux pointing towards the top surface area and the flux pointing away from the bottom surface area always cancel out to be zero since one is positive and one is negative?

[@DanFullerton]

They will, but that's not what we're calculating with Gauss's Law. With Gauss's Law we're looking for the net flux going from the inside to the outside of our Gaussian Surface. At both the top surface and the bottom surface, the flux is positive (coming out of the surface) so it doesn't cancel out, it adds.

[@si-onkim4149]

thanks dan you just saved my exam grade!

[@DanFullerton]

+Si-On Kim I'm betting you saved your exam grade, but glad I could help!

[@DanFullerton]

Glad you like it!

[@Karlwojo1]

when your'e writing the equations for the gaussian surface inside and outside the shell, wouldn't you use ri and ro and not R, because the idea behind the gaussian surface is to determine the charge on that surface a certain distance away

[@Karlwojo1]

actually i think i see what your saying, ri and ro are infinetly close to R, but just on the inside and outside so you just sub in R for those values right?

[@DanFullerton]

Karlwojo1 If I understand what you're asking (and not quite 100% sure I do), I think what you're saying is correct. Remember, the shell is infinitely thin.

[@DanFullerton]

The flux through the left is equal in magnitude and opposite in direction to the flux through the right, so when you add them up to find the net flux, you get a total of zero.

[@kpraveenkumar5444]

I really like the video thanks but I couldn't get this How did the flux cancelled in the infinite line charge but not in infinite plane

[@tigerzhao2861]

How do you choose a Gaussian Surface? I'm confused as to what shape you choose and how you would decide which shape is the best fit.

[@DanFullerton]

Tiger Zhao Lots of practice. Your choices are typically spheres or cylinders, but look for a shape in which the electric field through the Gaussian surface will be the same on the surfaces of interest.

[@tywer27]

great video. definitely beats reading the textbook

[@joshuasen3017]

Mr. Fullerton, how do you figure out the Gaussian surface to use when using Gauss's law?

[@DanFullerton]

Practice and experience. Look for a Gaussian surface where the electric field will be constant with respect to what you're trying to solve for. This guide sheet may help: www.aplusphysics.com/courses/ap-c/tutorials/APC-EField.pdf

[@IshraqIslam]

Why cant I pull back the video timeline in youtube?

[@jackysub1941]

A quick question. When solving for the electric field outside the sphere, why do we still use 4PIr^2 as the total area?

[@DanFullerton]

Our Gaussian surface is a sphere, so to apply Gauss's Law, we need the area of our Gaussian surface.

[@jackysub1941]

Shouldnt the radius of the gaussian surface be different from the given sphere?

[@sogdavid8016]

I like the way you drew that cylinder to enclose the arrow 👍🏿👌🏿

[@DanFullerton]

Nope... the R variable is the point at which we're trying to determine the electric field. If we wanted to know the electric field at R=r0, we could replace it, but I'm just finding the "generic" solution in this problem.

[@TheCheetah2022]

For the last problem, why'd substitute 2pi RL for dA instead of pi R^2 L? And thanks for the video, it helped me a lot!

[@DanFullerton]

Pi*r*r*L would give you the volume (Pi*r^2 is the area of a circle, multiplied by its length is volume). Instead, imagine we take the surface of the cylinder, like a soda can, and cut it so as to spread it out into a rectangle. The length will by L, and the width will be the perimeter of the circle, 2*pi*r

[@johnsullivan1606]

Great video! I love your playlist. I assume though, when you chose R for the outer radius you meant that as the general variable not R for the original sphere, right?

[@DanFullerton]

Correct