Gauss's Law Problem: Sphere and Conducting Shell

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  • Опубліковано 4 січ 2025

КОМЕНТАРІ • 124

  • @calthar13
    @calthar13 Рік тому +49

    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
      @PhysicsNinja  Рік тому +4

      Thanks for the support. You’ve got this

  • @quynhpham8596
    @quynhpham8596 4 роки тому +68

    thanks man u just saved my uni life

    • @tiernanomahoney3925
      @tiernanomahoney3925 2 роки тому +1

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

    • @akbaer60
      @akbaer60 Рік тому +13

      @@tiernanomahoney3925 least overstudied south korean

  • @mohammadmaarefi3393
    @mohammadmaarefi3393 Рік тому +8

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

  • @mrtom-a-hawk6732
    @mrtom-a-hawk6732 3 роки тому +36

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

  • @hasimrutkayboyac7246
    @hasimrutkayboyac7246 Рік тому +2

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

  • @mahdialkak5579
    @mahdialkak5579 4 роки тому +3

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

  • @farhanbinfaisal135
    @farhanbinfaisal135 2 роки тому +4

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

  • @rickastleysmicrophone7544
    @rickastleysmicrophone7544 Рік тому +3

    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!

  • @aaronlopez3852
    @aaronlopez3852 4 роки тому +18

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

  • @blessbrian1295
    @blessbrian1295 Рік тому +1

    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
    @ice_the_kicker Рік тому +2

    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
      @PhysicsNinja  Рік тому

      Thanks you. Good luck with your studies.

  • @kommsk
    @kommsk 4 роки тому +5

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

  • @jennaalqdah9538
    @jennaalqdah9538 Рік тому +3

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

    • @PhysicsNinja
      @PhysicsNinja  Рік тому

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

  • @johnychannellifestyle942
    @johnychannellifestyle942 2 роки тому +1

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

  • @MultiSubjector
    @MultiSubjector 2 роки тому +1

    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
      @PhysicsNinja  2 роки тому

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

  • @carsontaylor8771
    @carsontaylor8771 4 роки тому +5

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

  • @madnad4284
    @madnad4284 Рік тому +2

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

  • @kimberlygonzalez4543
    @kimberlygonzalez4543 3 роки тому +3

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

  • @kabandajamir9844
    @kabandajamir9844 2 роки тому +1

    The world's best teacher

  • @eccedentesiast4647
    @eccedentesiast4647 4 роки тому +2

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

  • @geniusblack4796
    @geniusblack4796 4 роки тому +20

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

  • @d3m3nt3d_t3acup_
    @d3m3nt3d_t3acup_ 3 роки тому +1

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

  • @dilettanter
    @dilettanter 11 місяців тому

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

  • @abdullahxaero
    @abdullahxaero 10 місяців тому +1

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

  • @ironears
    @ironears 3 роки тому +1

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

  • @andrewjustin256
    @andrewjustin256 11 місяців тому

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

  • @AblieFatty-pb3ds
    @AblieFatty-pb3ds Рік тому

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

  • @alexfontaine6233
    @alexfontaine6233 11 місяців тому

    dude this video helped me out immensely. Thanks!

  • @Bullets_ricochet
    @Bullets_ricochet 4 роки тому +4

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

  • @RAYAMUSTAFA
    @RAYAMUSTAFA 3 місяці тому +1

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

    • @nashithrahmy2934
      @nashithrahmy2934 2 місяці тому +1

      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
      @rayamustafa5454 2 місяці тому

      @@nashithrahmy2934Thank you for the understanding!

  • @beuz6626
    @beuz6626 Рік тому +1

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

  • @markkennedy9767
    @markkennedy9767 Рік тому

    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.

  • @AhmedKhaleelAhmedAhmed
    @AhmedKhaleelAhmedAhmed 5 місяців тому

    thank you keep updating please

  • @kina4288
    @kina4288 Рік тому +1

    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

  • @THEVEGEfr
    @THEVEGEfr Рік тому +1

    Thanks for the helpful video sir!

  • @faisalali2001
    @faisalali2001 4 роки тому +2

    Great work

  • @endlessmijot
    @endlessmijot 2 роки тому +1

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

  • @FaridCenreng-
    @FaridCenreng- 3 роки тому +1

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

    • @PhysicsNinja
      @PhysicsNinja  3 роки тому

      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

  • @princerenaud8790
    @princerenaud8790 4 роки тому +1

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

  • @kpraveenkumarkpraveenkumar7090

    thanks man well explained

  • @borjalarrain6335
    @borjalarrain6335 2 роки тому +1

    Thank you so much! This really helped

  • @juannunez8080
    @juannunez8080 3 роки тому +2

    great video

  • @sabarafiei9334
    @sabarafiei9334 2 роки тому

    thank you for your amazing channel♥♥

  • @boisteranto9275
    @boisteranto9275 4 роки тому +2

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

  • @mostafae3139
    @mostafae3139 Рік тому +1

    absolute legend

  • @lazarbeam5593
    @lazarbeam5593 Місяць тому

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

  • @vivekyedge4125
    @vivekyedge4125 3 роки тому +2

    Can't thank you enough 🙏😭

  • @MuhammadAli-tw9ht
    @MuhammadAli-tw9ht 4 роки тому +2

    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.?

  • @lisajose9219
    @lisajose9219 Рік тому +1

    Excellent

  • @REASON_REALM
    @REASON_REALM Рік тому +1

    you are the best

  • @JB-tj2ot
    @JB-tj2ot 4 роки тому +1

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

  • @mininuwanbandara624
    @mininuwanbandara624 3 роки тому +1

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

  • @Bestofchatgpt
    @Bestofchatgpt 3 роки тому

    This is what I been looking for

  • @ahmadirshaid6497
    @ahmadirshaid6497 2 роки тому

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

  • @nashhm7693
    @nashhm7693 2 роки тому +1

    If it is metal sphere. The r

  • @amlhassan7090
    @amlhassan7090 4 роки тому +4

    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
      @nashithrahmy2934 2 місяці тому

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

  • @classiccode3694
    @classiccode3694 4 роки тому

    thanks for the video. helped me a lot to understand

  • @iliastompoulidis
    @iliastompoulidis 3 роки тому +1

    You really saved me!

  • @omercankilickaya9965
    @omercankilickaya9965 4 роки тому +2

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

  • @omniamohamed404
    @omniamohamed404 Рік тому

    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
      @PhysicsNinja  Рік тому +1

      Because the is charge on the inner sphere

  • @archieli4556
    @archieli4556 3 роки тому

    Thank you Sir!!!! I understand now!

  • @AngelaAngel-v6j
    @AngelaAngel-v6j Рік тому

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

    • @sudhanshub999
      @sudhanshub999 Рік тому

      See case 2 heading there clearly written that a

  • @antonyrasan5667
    @antonyrasan5667 3 роки тому +1

    Tks nice explanation

  • @Clintaso
    @Clintaso 4 роки тому

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

  • @BlueRaz21
    @BlueRaz21 11 місяців тому

    Thank you!

  • @batuyild
    @batuyild Рік тому +1

    Helpful..Thanks..

  • @faraday7879
    @faraday7879 4 роки тому +1

    you are the man

  • @harbinger9231
    @harbinger9231 2 роки тому

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

  • @dilipmahalingam
    @dilipmahalingam 4 роки тому +1

    Thank you so much sir

  • @liambrennan4902
    @liambrennan4902 3 роки тому

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

  • @dakshigoel9543
    @dakshigoel9543 4 роки тому +1

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

    • @PhysicsNinja
      @PhysicsNinja  4 роки тому +1

      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.

  • @jadenjohnson4850
    @jadenjohnson4850 3 роки тому

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

    • @PhysicsNinja
      @PhysicsNinja  3 роки тому

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

  • @walterdiaz2003
    @walterdiaz2003 3 роки тому

    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?

  • @sabrinanovayanti3395
    @sabrinanovayanti3395 3 роки тому +1

    so helpful

  • @alzahraz
    @alzahraz 2 роки тому +1

    thanksss!!!

  • @kleinerflugel65
    @kleinerflugel65 2 роки тому +1

    Thanks.

  • @bshdjdjfnfjd8379
    @bshdjdjfnfjd8379 2 роки тому

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

  • @hotroadcol4166
    @hotroadcol4166 Рік тому +1

    todos te amamos

  • @ELENKTRIK
    @ELENKTRIK Місяць тому

    allah razı olsun reiz

  • @slendermangaming6414
    @slendermangaming6414 2 роки тому +1

    I love you so much

  • @SADDAMHUSAIN-ly4mm
    @SADDAMHUSAIN-ly4mm 3 роки тому

    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

  • @philip-zr5my
    @philip-zr5my Рік тому +1

    you are awsome
    :3

  • @isaacverdeciacarrion145
    @isaacverdeciacarrion145 8 місяців тому

    thanks

  • @Wave_equation.
    @Wave_equation. Рік тому +1

    nice

  • @yuvanm359
    @yuvanm359 3 роки тому

    Tanks man

  • @hazimaznan2077
    @hazimaznan2077 4 роки тому

    Tq for the video

  • @Alidaher2468
    @Alidaher2468 6 місяців тому

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

    • @Alidaher2468
      @Alidaher2468 6 місяців тому

      Or the inside is dielectric?

  • @ahmedalmoola8191
    @ahmedalmoola8191 4 роки тому

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

  • @eraysona
    @eraysona 9 місяців тому +1

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

  • @bautistabaiocchi-lora1339
    @bautistabaiocchi-lora1339 3 роки тому

    ty

  • @dawwnbrk3r653
    @dawwnbrk3r653 3 роки тому

    🐐

  • @vikramnagarjuna3549
    @vikramnagarjuna3549 4 роки тому +1

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

  • @lawrencelam2333
    @lawrencelam2333 Рік тому +1

    Best video ! ua-cam.com/video/pzJanh-Ol4g/v-deo.html 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
    @efeturan3278 Рік тому

    so bad wroong

  • @amlhassan7090
    @amlhassan7090 4 роки тому +1

    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
      @PhysicsNinja  4 роки тому +1

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

    • @amlhassan7090
      @amlhassan7090 4 роки тому

      @@PhysicsNinja ok thanks

    • @amlhassan7090
      @amlhassan7090 4 роки тому

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