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Topic: Earthed conductor shell  (Read 11398 times)

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Offline Hunt

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Earthed conductor shell
« on: April 10, 2006, 08:58:45 AM »
A conductor spherical shell has a radius 20cm concentric with another conducting spherical shell with a radius 40cm. The outer shell has a charge of 4 uC , and the inner shell is earthed.

Our Physics professor gave us a hint that the total potential of the system ( small shell + large shell ) is zero , does anyone have any idea why? I know that the potential of the inner shell ( earthed ) is zero.

Offline Borek

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Re: Earthed conductor shell
« Reply #1 on: April 10, 2006, 09:07:08 AM »
Is there any question asked, or is it just a statement to analyze?
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Offline Hunt

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Re: Earthed conductor shell
« Reply #2 on: April 10, 2006, 09:22:29 AM »
Actually there are 3 simple questions. My problem is not getting what the term 'earthed' is supposed to mean. I think it simply means 'grounded' i.e the small shell is the ref for V , but perhaps there's more to it.

i-What is the potential on the surface of the inner shell?
ii-Find the charge on the surface of the inner shell.
iii-Determine the potential on the surface of the outer shell

i-0 V

ii-If VT = 0 , then the charge on the surface of the inner shell can be calculated q1 = - r1 q2 / r2

iii-I dont think we need to take an infinitesimal charge here, V = K q2 / r2

I only need to know a precise physical meaning for 'earthed' conductor, and why VT is 0.
« Last Edit: April 10, 2006, 09:26:08 AM by Vant_Hoff »

Offline lemonoman

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Re: Earthed conductor shell
« Reply #3 on: April 10, 2006, 06:17:30 PM »
I'm not an electrician, but dictionary.com tells me that "Earth" can refer to the ground of an electrical circuit.

Hope that helps!

Offline Hunt

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Re: Earthed conductor shell
« Reply #4 on: April 11, 2006, 04:16:37 AM »
Quote
I'm not an electrician, but dictionary.com tells me that "Earth" can refer to the ground of an electrical circuit.

Hope that helps!

Thanks, lemonoman. Any help will just do...

But I still can't seem to understand why the total potential of the system would be zero if the small spherical shell is the reference here. I'll get back later to it, I guess ...

Offline Hunt

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Re: Earthed conductor shell
« Reply #5 on: April 12, 2006, 07:05:25 PM »
I think I get it now, VT, the potential of the system is not really zero. The small sphere is earthed , its potential is zero. However, it is charged and we need to find its charge q1.

Since the two spheres are charged, there will be an induced electric field opposite to the direction of flow of electrons. The potential V1 at the surface of the small sphere is Kq2 / r2 + Kq1 / r1. V1 = 0 ( earthed )

Kq2 / r2 + Kq1 / r1 = 0

q1 = - q2 r1 / r2

Quote
iii-I dont think we need to take an infinitesimal charge here, V = K q2 / r2

I also made a mistake here. The potential  of the larger spherical shell is due to q2 and q1.

Vq2 = Kq2 / r2 + Kq1 / r2 = Kq2 / r2 - K q2 r1 / r22 = Kq2 / r2 ( 1 - r1 / r2 )

I think this might be the correct answer.
« Last Edit: April 13, 2006, 05:55:17 PM by Vant_Hoff »

teteamigo

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Re: Earthed conductor shell
« Reply #6 on: April 20, 2006, 12:28:15 AM »
V=0 in the inner sphere, then the charge is null in this sphere, or by other mean the work to reach ground is null. Then the induced charge in inner side of great sphere is null. The electric field between the spheres is null, then the energy stored in the field is null (system of charges).

The charge in outside of great sphere don't influence the electric field between spheres.

You have to read about Gauss Law to understand this. Or ask your teacher: What matter with conductor to shield the effect of surface charge, inside of conductor, and  in  particular case of two sides conductor, the conductor blinds?

Offline Hunt

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Re: Earthed conductor shell
« Reply #7 on: April 23, 2006, 08:28:31 PM »
Thanks for your post,  teteamigo.

The problem has already been solved by our professor, and there was a slight mistake in it.

Quote
Then the induced charge in inner side of great sphere is null

According to Gauss' law , the charge on the inner face should be equal to -Q of the large sphere.

Quote
The electric field between the spheres is null, then the energy stored in the field is null (system of charges).

The charge in outside of great sphere don't influence the electric field between spheres.

True

Quote
You have to read about Gauss Law to understand this

Trust me, I've done a lot more than reading about it.

teteamigo

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Re: Earthed conductor shell
« Reply #8 on: April 24, 2006, 12:35:38 AM »

Quote
Then the induced charge in inner side of great sphere is null

According to Gauss' law , the charge on the inner face should be equal to -Q of the large sphere.

Trust me, I've done a lot more than reading about it.

This is true, But if we have E=0 V/m then closed area integral around the part that includes +Q and -Q is null.

But if we know that +Q=0 C then -Q=0 C.

Right?

Offline Hunt

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Re: Earthed conductor shell
« Reply #9 on: April 27, 2006, 06:10:58 PM »
Quote
This is true, But if we have E=0 V/m then closed area integral around the part that includes +Q and -Q is null.

But if we know that +Q=0 C then -Q=0 C.

Right?

100% correct

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