Charge distribution on concentric spheres

AI Thread Summary
The discussion revolves around determining the charge distribution on two concentric conducting spherical shells, where the inner shell is earthed and the outer shell is charged. Participants clarify that the inner shell cannot be assumed to have zero charge simply because it is grounded, as it will still influence the potential. The potential at the center of the inner shell is discussed in relation to the charge on both shells, leading to the conclusion that the potential of the inner shell must be set to zero due to grounding. An equation is derived to express the relationship between the charges on the inner and outer shells. The conversation emphasizes the importance of understanding electric potential and charge interactions in this configuration.
Abhimessi10
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Homework Statement


Two thin conducting spherical shells have radii R1 and R2.Outer shell is charged to q and inner shell earthed.Find charge appearing on both the shells.

Homework Equations



The Attempt at a Solution


Isnt the charge on inner shell 0 and charge on outer shell remains Q as it doesn't involve any connection?
 
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The inner shell is connected to ground so you cannot make the assumption that the charge on it is zero.
 
Abhimessi10 said:
So how do i approach this problem?
Sorry isn't the inner shell at potential zero as it is connected to the Earth and thus holds 0 charge

If i am wrong please correct me and tell me how to do this?
 
Abhimessi10 said:
Sorry isn't the inner shell at potential zero as it is connected to the Earth and thus holds 0 charge

If i am wrong please correct me and tell me how to do this?
Picture a capacitor and a battery connected in parallel, and the bottom connection for both is Earth grounded. What is the charge on each plate of the capacitor? :smile:
 
berkeman said:
Picture a capacitor and a battery connected in parallel, and the bottom connection for both is Earth grounded. What is the charge on each plate of the capacitor? :smile:

berkeman said:
Picture a capacitor and a battery connected in parallel, and the bottom connection for both is Earth grounded. What is the charge on each plate of the capacitor? :smile:
Still i think its 0 .Sorry i was born dumb.
 
Last edited:
Abhimessi10 said:
Still i think its 0 .Sorry i was born dumb.
Forget the inner shell for the moment. If you have a shell radius r with charge q, what is the potential on its surface? What is the potential at the centre of the shell?
 
haruspex said:
Forget the inner shell for the moment. If you have a shell radius r with charge q, what is the potential on its surface? What is the potential at the centre of the shell?
KQ/R right?
 
Abhimessi10 said:
KQ/R right?
Right.
So now put another shell inside that has no charge. How will that change potential at the centre?
 
haruspex said:
Right.
So now put another shell inside that has no charge. How will that change potential at the centre?
It will not change?
 
  • #10
Abhimessi10 said:
It will not change?
And so cannot become zero.
Do you still think that a grounded sphere placed inside the shell would have no charge?
 
  • #11
Ok
haruspex said:
And so cannot become zero.
Do you still think that a grounded sphere placed inside the shell would have no charge?
But how do i find charges when the inner shell is earthed?
 
  • #12
Abhimessi10 said:
Ok

But how do i find charges when the inner shell is earthed?
Suppose it has charge q'. What is the potential at the centre now?
 
  • #13
haruspex said:
Suppose it has charge q'. What is the potential at the centre now?
KQ'/R1+kQ/R2 ?
 
  • #14
Abhimessi10 said:
KQ'/R1+kQ/R2 ?
Right.
And if the inner shell is grounded what can you say about that potential?
 
  • #15
haruspex said:
Right.
And if the inner shell is grounded what can you say about that potential?
Equating potential of inner shell to 0?
 
  • #16
Abhimessi10 said:
Equating potential of inner shell to 0?
Yes. So what equation do you get?
 
  • #17
haruspex said:
Yes. So what equation do you get?
Kq'/r1+kq/r2=0?
 
  • #18
Abhimessi10 said:
Kq'/r1+kq/r2=0?
Right.
 

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