Capacitance of Concentric Shells (different charges)

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SUMMARY

The discussion centers on calculating the capacitance of two concentric conducting shells with differing charges, specifically a thin shell of radius a with charge q and another shell of radius b with charge Q. The charge distribution is analyzed, revealing that the inner shell has no charge on its inside surface and a charge of +q on its outside surface. The outer shell is theorized to have a charge of -q on its inside surface and a total charge of q + Q on its outside surface. The capacitance is determined using the formula C = Q/V, where V is derived from the electric field E = (1/4πε₀)(q/r²) integrated from a to b.

PREREQUISITES
  • Understanding of electrostatics and charge distribution
  • Familiarity with capacitance formulas, specifically C = Q/V
  • Knowledge of electric fields and potential difference calculations
  • Basic concepts of spherical conductors and their properties
NEXT STEPS
  • Study the derivation of capacitance for spherical shells
  • Learn about the implications of charge polarity on capacitance
  • Explore the concept of electric fields in spherical coordinates
  • Investigate the effects of connecting charged conductors on potential
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Students and professionals in physics, electrical engineering, and anyone studying electrostatics and capacitance in conductive materials.

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Homework Statement



A thin conducting shell of radius a has charge q. Concentric to this is another shell of radius b (b > a) with a different charge Q. How is the charge distributed and what is the capacitance of the two shells. No hint to the relative polarity of the charges is given.


Homework Equations



[tex]C = \frac{Q}{V}[/tex]

The Attempt at a Solution



The difficulty lies in the difference in the charges (I'm sure I'm missing something). Any reference to capacitance I see requires equal and opposite charges, leading me to think the the distribution must be something like:

Inside surface of small shell: no charge
Outside surface of small shell: +q

Inside surface of larger shell: -q (?)
Outside surface of larger shell: q + Q (?)


If I assume that it the charges are -q on inner and +q on outer then I think you could find it by:

[tex]E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}[/tex]

[tex]V = -\frac{q}{4\pi\epsilon_0} \int_a^b \frac{1}{r^2} = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{a} - \frac{1}{b}\right)[/tex]

The just find the capacitance simply by [tex]C = \frac{q}{V}[/tex]

But there is no mention of Q which seems wrong. Any help or suggestions very much appreciated.
 
Physics news on Phys.org
When you connect the charged concentric shells, both must have the same potential.
The common potential is given by ( total charge/total capacity)
The capacitance of the spherical shell is proportional to its radius.
 

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