Capacitance of spherical capacitor

  • #1
Krushnaraj Pandya
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Homework Statement


I calculated the capacitance of two concentric conducting spherical shells of radii a and b (a<b) when the inner shell has charge Q and outer has charge -Q correctly by using C=Q/(V(i)-V(o)), the problem I have is described below in "attempt at a solution"

Homework Equations


All electrostatics formulas

The Attempt at a Solution


while calculating it, I ignored that the outer shell was grounded as I have no idea what grounding does. Now I require to calculate the capacitance when the inner shell is grounded, this time- my answer was incorrect. I wish to know if it is alright to ignore earthing as the answer was correct anyway in the first case or whether I'm missing something? what would happen if the inner shell was earthed? does it become neutral as excess of charge is removed?
 

Answers and Replies

  • #2
ehild
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Homework Statement


I calculated the capacitance of two concentric conducting spherical shells of radii a and b (a<b) when the inner shell has charge Q and outer has charge -Q correctly by using C=Q/(V(i)-V(o)), the problem I have is described below in "attempt at a solution"

Homework Equations


All electrostatics formulas

The Attempt at a Solution


while calculating it, I ignored that the outer shell was grounded as I have no idea what grounding does. Now I require to calculate the capacitance when the inner shell is grounded, this time- my answer was incorrect. I wish to know if it is alright to ignore earthing as the answer was correct anyway in the first case or whether I'm missing something? what would happen if the inner shell was earthed? does it become neutral as excess of charge is removed?
Earthing does not mean that the earthed object becomes neutral. The earth means an infinite amount of available charge.
In case the outer shell is earthed and you give Q charge to the inner shell, the field of the inner charge will push some charge from the outer shell to the earth. You also know that the electric field lines must start and end in charges or at infinity. And the electric potential is continuous at interfaces. When you ground the outer shell, the potential is zero there, the same as in infinity, The electric field outside is zero, the enclosed charge must be zero, so the same amount of charge with opposite sign must accumulate on the outer shell.
It is more difficult to answer the second question. The grounded inner shell is at the same potential as infinity. Is it really neutral?
 
  • #3
Krushnaraj Pandya
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And the electric potential is continuous at interfaces.
what do you mean by this?
Also, if we ground the inner shell, it gets a potential zero and therefore becomes irrelevant, Then we're left with a single spherical conductor with radius b??
 
  • #4
ehild
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what do you mean by this?
Also, if we ground the inner shell, it gets a potential zero and therefore becomes irrelevant, Then we're left with a single spherical conductor with radius b??
Think: if you have a parallel-plate capacitor and you ground one plate, does it become irrelevant, and you will have only one plate?
But you can try, calculate the capacitance of a single shell. Is that what your book provide as solution?
The electric potential is integral of the negative electric field. The potential of the outer shell must be the same, either you integral the negative of the electric field from the inner shell to the outer one, or from infinity to the outer shell.
 
  • #5
Chandra Prayaga
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Why do you say it is irrelevant? Zero is not an irrelevant number. It means that you have an ineer shell at zero potential and some charge. The outer shell is at some potential with an equal charge of opposite sign.
 
  • #6
vela
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Why do you say it is irrelevant? Zero is not an irrelevant number.
@ehild didn't say it was irrelevant, but you said
Also, if we ground the inner shell, it gets a potential zero and therefore becomes irrelevant.
 
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  • #7
Krushnaraj Pandya
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I'm trying to understand it better. Please give me a little time, so I can respond properly, thank you
 
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  • #8
Krushnaraj Pandya
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Think: if you have a parallel-plate capacitor and you ground one plate, does it become irrelevant, and you will have only one plate?
But you can try, calculate the capacitance of a single shell. Is that what your book provide as solution?
The electric potential is integral of the negative electric field. The potential of the outer shell must be the same, either you integral the negative of the electric field from the inner shell to the outer one, or from infinity to the outer shell.
Alright, so here's what I understand. The inner sphere is earthed, so its potential is zero, but some charge q resides on its outer surface, a charge -q resides on inner surface of outer sphere thus making a capacitor, some charge Q' resides on outer surface of outer sphere and the second terminal of this capacitor is at infinity. Since potential difference is same between outer and inner sphere and outer sphere and infinity-these capacitors are in parallel connection and the total capacitance will be sum of the two. Am I correct?
 
  • #9
Krushnaraj Pandya
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Why do you say it is irrelevant? Zero is not an irrelevant number. It means that you have an ineer shell at zero potential and some charge. The outer shell is at some potential with an equal charge of opposite sign.
@ehild didn't say it was irrelevant, but you said
I understand, is my previous message correct?
 
  • #10
ehild
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Alright, so here's what I understand. The inner sphere is earthed, so its potential is zero, but some charge q resides on its outer surface, a charge -q resides on inner surface of outer sphere thus making a capacitor, some charge Q' resides on outer surface of outer sphere and the second terminal of this capacitor is at infinity. Since potential difference is same between outer and inner sphere and outer sphere and infinity-these capacitors are in parallel connection and the total capacitance will be sum of the two. Am I correct?
It looks correct. What about the details?
 
  • #11
Krushnaraj Pandya
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It looks correct. What about the details?
I got the correct expression 4(pi)(epsilon)(b^2)/(b-a). C2=capacitor formed by outer shell with other terminal at infinity=4(pi)(epsilon)(b). C1=capacitor formed by outer surface of inner shell and inner surface of outer shell=4(pi)(ab)/(b-a). Adding both gives the answer.
 
  • #12
Krushnaraj Pandya
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Thank you very much for your help :D
 
  • #13
ehild
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