Gauss's Law Problem: Sphere and Conducting Shell
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- Опубликовано: 25 авг 2024
- Physics Ninja looks at a classic Gauss's Law problem involving a sphere and a conducting shell. The inner sphere can be a conductor or an insulator and the outer shell is assumed to be a conductor. Charge is placed on both the sphere and shell. We use Gauss's Law to find the Electric Field everywhere in space.
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Wow, this is a lot easier than my professor made it out to be. Thank you so much!
Beautifully made, I love how you mention all possible cases of this question. It makes perfect sense.
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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.
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You did a very good job explaining Gauss' Law through a real example...Thanks man , you're a Ninja...:)
This is a super useful video and extremely well explained. Thank you so much, you clarified everything for me!
Thank you so much. These types of comments motivate me to make more videos.
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Excellent video! Made me understand Gauss's Law perfectly.
Thank you so much!! I was so confused and now I solve for my upcoming mid exam.
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
thanks for the comment. I'm happy the video was helpful for you. Good luck with your course.
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Great work
The world's best teacher
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Glad it was helpful!
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Excellent
thanks for the video. helped me a lot to understand
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?
Thanks.
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
Thank you!
Tks nice explanation
You really saved me!
Helpful..Thanks..
Most welcome!
Thank you so much sir
beautiful
Why the outer sphere dont affect in the white field (a
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
14:40 Would you, sir, clarify how the charge outside be positive when the charge is negative on the shell?
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.?
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so helpful
Thank you Sir!!!! I understand now!
If inner -Q and outside 3Q how is the solution change
todos te amamos
so if im asked to get the EF inside the sphere what would the answer be=(? 0?
If it is metal sphere. The r
Yes
thanksss!!!
nice
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.
I love you so much
Any reason you didn't do an example in the case of an insulating shell? Is it just not common?
thanks
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?
Because the is charge on the inner sphere
Is the electric field on the outer shell when (r=c) is zero ????
and when (r= a) is zero too??
It was awesome
But I have a doubt, what about at point b? I mean how to find E right at b point?
See case 2 heading there clearly written that a
sir, if a=b then how will we find the electric field anywhere inside the spherical shell??
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.
How would we find potential at the center of both spheres?
Tanks man
you are awsome
:3
Thanks a lot but how do you find the potential across two points in such a setup?
Delta V= -integral( E.dl)
Tq for the video
How would E change in the case of r being between b and c if the shell is an insulator?
If the charge is spread uniformly in the outer shell then you can still calculate using Gauss' Law. The field will be different!
I wonder why the electric field at r=a ( conducting sphere ) equals to zero🤔
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?
Absolutely not!
But for a conductor the electric field inside is zero and therefore in the case 2 the charge is zero
Or the inside is dielectric?
ty
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
Best video ! ruclips.net/video/pzJanh-Ol4g/видео.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.
good video but I hated the many ads poped and distracted me.
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Hi sir this is Vikram from India, please do some problems on coupled oscillators in SHM with simple mathematics or advanced Mathematical methods.
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Lol
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?
Total charge on that surface / (4*pi*R^2)
The term in denominator is the total surface area of the sphere.
@@PhysicsNinja ok thanks
@@PhysicsNinja is there is any chance / way that i can send you a question to help me with ?