[Hunt]

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.

[Borek]

Is there any question asked, or is it just a statement to analyze?

[Hunt]

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 V_T = 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 V_T is 0.

[lemonoman]

I'm not an electrician, but dictionary.com tells me that "Earth" can refer to the ground of an electrical circuit.

Hope that helps!

[Hunt]

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 ...

[Hunt]

I think I get it now, V_T, 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

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.

V_q2 = Kq₂ / r₂ + Kq₁ / r₂ = Kq₂ / r₂ - K q₂ r₁ / r₂² = Kq₂ / r₂ ( 1 - r₁ / r₂ )

I think this might be the correct answer.

[teteamigo]

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?

[Hunt]

Thanks for your post,  teteamigo.

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

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.

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

You have to read about Gauss Law to understand this

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

[teteamigo]

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?

[Hunt]

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