Gauss's Law Problem: Sphere and Conducting Shell

[@calthar13]

Why do physics classes that cost hundred of dollars and demand a commitment of 5 hours a week, less helpful than this 18min youtube video? thank god for you

[@PhysicsNinja]

Thanks for the support. You’ve got this

[@quynhpham8596]

thanks man u just saved my uni life

[@tiernanomahoney3925]

University...., I'm learning this in 9th grade.

[@akbaer60]

@@tiernanomahoney3925 least overstudied south korean

[@mrtom-a-hawk6732]

Wow, this is a lot easier than my professor made it out to be. Thank you so much!

[@hasimrutkayboyac7246]

You just saved a human being from just loosing his mind thank u for existing.

[@mohammadmaarefi3393]

Beautifully made, I love how you mention all possible cases of this question. It makes perfect sense.

[@PhysicsNinja]

Thank you so much!

[@mahdialkak5579]

Who watched this in the last week of 2020 then you are a legend, thanks dr

[@farhanbinfaisal135]

This is far far the best explanation I got till now. Thanks man

[@rickastleysmicrophone7544]

Gosh! This video is perfect. I have my finals in 2 days and you saved me! This topic makes a lot more sense now. Thank you very much sir!

[@PhysicsNinja]

You got this!

[@blessbrian1295]

Watched you video and went back to what my professor did and it made a lot of sense to me. Thank you very much and I must say my professor teaches well too.

[@ice_the_kicker]

Brilliant explanation. I have never actually been into the world of physics, mostly programming and mathematics. Your video triggered me into learning electromagnetic fields.

[@PhysicsNinja]

Thanks you. Good luck with your studies.

[@aaronlopez3852]

This is awesome. Thanks for being thorough, yet concise. Keep it up!

[@kabandajamir9844]

The world's best teacher

[@kommsk]

You did a very good job explaining Gauss' Law through a real example...Thanks man , you're a Ninja...:)

[@jennaalqdah9538]

This is a super useful video and extremely well explained. Thank you so much, you clarified everything for me!

[@PhysicsNinja]

Thank you so much. These types of comments motivate me to make more videos.

[@carsontaylor8771]

It all makes sense now. You are amazing good sir!

[@madnad4284]

Thank you so much!! I was so confused and now I solve for my upcoming mid exam.

[@geniusblack4796]

Wow... It's now that I understand this completely... Thank you.... 😊

[@kimberlygonzalez4543]

Excellent video! Made me understand Gauss's Law perfectly.

[@MultiSubjector]

Thanks for the clear and concise lesson. I haven't seen this since taking physics, and revisiting the concept for one of upper div EE courses has been rough with my professor

[@PhysicsNinja]

thanks for the comment. I'm happy the video was helpful for you. Good luck with your course.

[@johnychannellifestyle942]

Awesome video I was upset when saw gauss law that make confuse but finally I find this video that clearly explain to me.🙏🙏🙏 Bro

[@eccedentesiast4647]

Superb and amazing 👌🙌😍 explanation. Infinitely many thanks to you from my bottom of the heart.

[@d3m3nt3d_t3acup_]

Gotta say, this is the explanation that made sense to me. Keep this up!

[@Bullets_ricochet]

You're a life saver. Thank you!!!!!!!!!

[@ironears]

Hi there, you're just awesome. This was simple and clear. Thank you

[@alexfontaine6233]

dude this video helped me out immensely. Thanks!

[@dilettanter]

I agree with the comments - this is great!!!!👍 👍👍👍🙏🙏🙏🙏💜💜💜

[@AblieFatty-pb3ds]

thank you so much, this video really helps me to understand this topic

[@abdullahxaero]

قمه الجمال والله و قمه الفهم ف 18 دقيقه بس loved that 💛💛💛💛💛

[@beuz6626]

You're amazing. Thanks for the video!! It's very helpful.

[@PhysicsNinja]

Glad it was helpful!

[@princerenaud8790]

You are the best bro. Thank you so much!!!!!!!!

[@THEVEGEfr]

Thanks for the helpful video sir!

[@borjalarrain6335]

Thank you so much! This really helped

[@endlessmijot]

Online learning moment. So much simpler than my professor made it out to be.

[@markkennedy9767]

At 2:00 are you saying that the only way E=0 everywhere inside is if q_enclosed=0. I'm pretty sure that isn't true. As you know, the integral (which evaluates to q_enclosed) can be zero but the integrand E could be non-zero in places. The implication the other way is true: if E=0 inside everywhere is zero, then q_enclosed is zero. My question is: Can you explain why in an electrostatic situation on a conducting sphere, that the electric field inside is zero necessarily. I know that in an electrostatic situation, the electric field has to be zero where the charges are (otherwise the charges would move), but why does the electric field have to be zero where the charges are not. The usual explanation in texts is by contradiction: i.e. if we had an electrostatic arrangement and we were to place another charge inside the sphere it would move if the electric field was non-zero inside, supposedly contradicting the original electrostatic situation. But by placing another charge inside the sphere aren't we changing the original electrostatic arrangement that we assumed. So we're not contradicting the original electrostatic arrangement. So it's not really a contradiction. Can you tell me where I'm going wrong with this. Thanks.

[@faisalali2001]

Great work

[@RAYAMUSTAFA]

can you please explain how you so easily plugged in 4/3pi r^3 / 4/3pi a^3?

[@nashithrahmy2934]

It's considering the total charge 3 Q is spread on sphere with radius. So for our sphere made with radius R it should be a fraction of voles right. Because the charges are uniformly distributed. Think like a clay ball. The portion we want us a fraction of total

[@rayamustafa5454]

@@nashithrahmy2934Thank you for the understanding!

[@andrewjustin256]

14:40 Would you, sir, clarify how the charge outside be positive when the charge is negative on the shell?

[@AhmedKhaleelAhmedAhmed]

thank you keep updating please

[@boisteranto9275]

you just seem like you'd be a cool professor to have

[@FaridCenreng-]

Why the outer sphere dont affect in the white field (a

[@PhysicsNinja]

It’s only what’s on the inside that matters. If you would calculate the total fuel from outside shell you would find 0. It’s takes a bit of calculus to show but that’s also what you get from Gauss’ Law

[@juannunez8080]

great video

[@kpraveenkumarkpraveenkumar7090]

thanks man well explained

[@mostafae3139]

absolute legend

[@vivekyedge4125]

Can't thank you enough 🙏😭

[@sabarafiei9334]

thank you for your amazing channel♥♥

[@kina4288]

When we say a sphere, are we automatically talking about a hollow object or not? I know a spade is a spade but when it comes to shpere, is it solid or hollow becomes important. Also, if the central charged object is off centre ie not at the centre of the metal shell, how it will change the calculation? Thank you ninja

[@REASON_REALM]

you are the best

[@MuhammadAli-tw9ht]

Sir , can i please know about the potential inside the shell? As electric field is zero inside shell , the potential will be constant but would it be same as due to inner sphere or the conducting shell.?

[@Bestofchatgpt]

This is what I been looking for

[@classiccode3694]

thanks for the video. helped me a lot to understand

[@lisajose9219]

Excellent

[@iliastompoulidis]

You really saved me!

[@mininuwanbandara624]

thanks you ......... thanks you..... nice explanation.

[@lazarbeam5593]

what if the outside shell is nonconducting charged, do you just do the same thing as the interior sphere?

[@faraday7879]

you are the man

[@amlhassan7090]

hi how can i calculate the surface charge density on the inner surface of the outer spherical conductor and on the outer surface of the outer spherical conductor?

[@nashithrahmy2934]

Surface charge density is epsilon=Q/area. So to find the charge multiply by area

[@archieli4556]

Thank you Sir!!!! I understand now！

[@nashhm7693]

If it is metal sphere. The r

[@PhysicsNinja]

Yes

[@dilipmahalingam]

Thank you so much sir

[@antonyrasan5667]

Tks nice explanation

[@BlueRaz21]

Thank you!

[@AngelaAngel-v6j]

It was awesome But I have a doubt, what about at point b? I mean how to find E right at b point?

[@sudhanshub999]

See case 2 heading there clearly written that a

[@harbinger9231]

Thanks a lot but how do you find the potential across two points in such a setup?

[@PhysicsNinja]

Delta V= -integral( E.dl)

[@Clintaso]

Any reason you didn't do an example in the case of an insulating shell? Is it just not common?

[@dakshigoel9543]

sir, if a=b then how will we find the electric field anywhere inside the spherical shell??

[@PhysicsNinja]

If a=b it looks like it would simply be a conducting sphere with radius c. E=0 inside conductors. The charge would be at the surface c.

[@JB-tj2ot]

so if im asked to get the EF inside the sphere what would the answer be=(? 0?

[@omniamohamed404]

for region two, it is still inside the outer sphere which is conducting, so how is it possible that region two does not have zero electric field?

[@PhysicsNinja]

Because the is charge on the inner sphere

[@ahmadirshaid6497]

Is the electric field on the outer shell when (r=c) is zero ???? and when (r= a) is zero too??

[@liambrennan4902]

How would we find potential at the center of both spheres?

[@batuyild]

Helpful..Thanks..

[@PhysicsNinja]

Most welcome!

[@walterdiaz2003]

Do licensed electrical contractors know this or should?. I know some physics 1/2/3, Digital Systems 1 and so on. I have a BS in Computer Science but I am interested in getting the plumbing and electrical licenses (Homeowner issues....). Any advise?

[@PhysicsNinja]

Absolutely not!

[@sabrinanovayanti3395]

so helpful

[@alzahraz]

thanksss!!!

[@kleinerflugel65]

Thanks.

[@jadenjohnson4850]

How would E change in the case of r being between b and c if the shell is an insulator?

[@PhysicsNinja]

If the charge is spread uniformly in the outer shell then you can still calculate using Gauss' Law. The field will be different!

[@omercankilickaya9965]

If inner -Q and outside 3Q how is the solution change

[@ELENKTRIK]

allah razı olsun reiz

[@slendermangaming6414]

I love you so much

[@hotroadcol4166]

todos te amamos

[@bshdjdjfnfjd8379]

I wonder why the electric field at r=a ( conducting sphere ) equals to zero🤔

[@SADDAMHUSAIN-ly4mm]

S I am saddam husain from india . SirMy small talk with you is that you should solve more problems like this in the book of DavyJ Griffth by putting limit in electric field and solved more such problems sir please......and please reply sir

[@yuvanm359]

Tanks man

[@isaacverdeciacarrion145]

thanks

[@philip-zr5my]

you are awsome :3

[@Wave_equation.]

nice

[@hazimaznan2077]

Tq for the video

[@Alidaher2468]

But for a conductor the electric field inside is zero and therefore in the case 2 the charge is zero

[@Alidaher2468]

Or the inside is dielectric?

[@ahmedalmoola8191]

good video but I hated the many ads poped and distracted me.

[@eraysona]

jesus. now I can show off the girls in my uni!

[@PhysicsNinja]

Chix will dig it.

[@bautistabaiocchi-lora1339]

ty

[@vikramnagarjuna3549]

Hi sir this is Vikram from India, please do some problems on coupled oscillators in SHM with simple mathematics or advanced Mathematical methods.

[@dawwnbrk3r653]

🐐

[@lawrencelam2333]

Best video ! kzbin.info/www/bejne/pqutkqGeYrSfask Can we say (-1 Q ) charge distributed evenly in both inner and outer shell or just (-3Q)charge in inner shell and (+ 2Q) charge in outer shell.

[@efeturan3278]

so bad wroong

[@PhysicsNinja]

Lol

[@amlhassan7090]

hi how can i calculate the surface charge density on the inner surface of the outer spherical conductor and on the outer surface of the outer spherical conductor?

[@PhysicsNinja]

Total charge on that surface / (4*pi*R^2) The term in denominator is the total surface area of the sphere.

[@amlhassan7090]

@@PhysicsNinja ok thanks

[@amlhassan7090]

@@PhysicsNinja is there is any chance / way that i can send you a question to help me with ?