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To find the capacitance of 2 concentric spherical shells

  • Thread starter hermy
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  • #1
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Homework Statement



There are 2 concentric shells. The outer shell of radius a is given a charge, while inner shell of radius b is earthed. We need to find the capacitance of the system.

Homework Equations



Delta V = - integral E.dl (for some funny reason, latex is not working here)

C = [tex]Q/ V[/tex]

The Attempt at a Solution



Both the shells are at the same potential.
So, [tex]\Delta[/tex]V = 0

So, capacitance should be infinite.

Where am I going wrong?
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement



There are 2 concentric shells. The outer shell of radius a is given a charge, while inner shell of radius b is earthed. We need to find the capacitance of the system.

The Attempt at a Solution



Both the shells are at the same potential.
So, [tex]\Delta[/tex]V = 0

So, capacitance should be infinite.

Where am I going wrong?

Why should both shells be at the same potential?

ehild
 
  • #3
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The potential inside a conducting spherical shell is equal to the potential at its surface.
 
  • #4
ehild
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The potential inside a conducting spherical shell is equal to the potential at its surface.
You have an inner shell, grounded, so at zero potential. Outside of it, it is surrounded by an other conducting shell, isolated from it, and charged. Why should it be at zero potential? You mean that no work is needed to add any amount of excess charges to the outer shell?

ehild
 
  • #5
collinsmark
Homework Helper
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Hello Hermy,

Follow Ehild's advice, but allow me to give you one more pointer:

We know that the inner sphere is earthed (grounded). But don't fall into the trap thinking that just because the inner sphere is earthed it has no charge. As a matter of fact in this case it is quite the opposite. Charge can and does flow from a ground line. And in this problem, the inner sphere does have charge merely because it is earthed.

Think about your quote for a minute,

The potential inside a conducting spherical shell is equal to the potential at its surface.
That quote is true so long as there are no other charges inside the sphere. So for a moment imagine that the inner sphere has zero charge on it. If that were the case the inner sphere would have the same potential as the outer shell.

Suppose for a moment that we define the inner sphere's radius as a, and the outer shell's radius as b.

Also suppose that the inner sphere has no charge. But if the outer shell had charge (and the inner shell didn't), there would still be a potential with respect to infinity of the outer shell (you can use Gauss' law to determine the electric field, and integrate from infinity to b [with a minus sign] to find the potential at b). That would mean the inner shell would have this non-zero potential too, with respect to infinity.

But wait! we know that the inner sphere is earthed! It has zero potential with respect to infinity (imagine a conducting wire attached to the inner sphere going all the way out to infinity). So which is it? Zero or nonzero? Well, we know it has to be zero by definition (because it is earthed), so what's going on here?

The way to resolve this conundrum is to realize that the inner sphere must be charged. By definition of being earthed, it must have zero potential. Something must be bringing its potential down to zero from the potential of point b. And that something is the inner sphere's charge which got there via its earth connection. And that charge flowed into the inner sphere (via the earth connection) when the outer shell was given its charge. :cool:

[Edit: The inner sphere's charge is not necessarily the same polarity as the outer shell's charge. But you can figure all that out. :smile:]
 
Last edited:
  • #6
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i got it. thanks, ehild, and thanks collins, it was a wonderful explanation.
 

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