Charge distribution over two thin concentric spherical shells

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SUMMARY

The charge distribution between two thin concentric spherical shells is defined by their respective charges and the principles of electrostatics. The inner shell, with radius 'a' and charge 'q', will have a charge of +q on its outer surface. The outer shell, with radius 'b' (where b > a) and charge 'Q', will exhibit a charge of -q on its inner surface due to the induced charge effect, resulting in a charge of +Q - q on its outer surface. The capacitance of the system can be calculated using the formula C = Q/V, where V is derived from the electric field obtained via Gauss' Law.

PREREQUISITES
  • Understanding of electrostatics and charge distribution
  • Familiarity with Gauss' Law
  • Knowledge of capacitance and its calculation
  • Basic principles of spherical symmetry in electric fields
NEXT STEPS
  • Study Gauss' Law applications in electrostatics
  • Learn about capacitance calculations for spherical capacitors
  • Explore the concept of induced charge in conductors
  • Investigate the effects of varying charge distributions on electric fields
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Students studying electrostatics, physics educators, and anyone interested in understanding charge distribution in spherical conductors.

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Homework Statement


"A thin conducting spherical shell of radius a carries a charge q. Concentric with this is another spherical shell of radius b>a carrying a net charge Q. Describe how the charge is distributed between the inner and outer surfaces of the shells. What is te capacitance of the two shells?"


Homework Equations





The Attempt at a Solution


Is it correct to say that there would be a charge +q on the outer surface of the inner shell. Due to induced charge there would then be a charge -q on the inner surface of the outer shell. Therefore there would be a charge +Q-q on the outer surface of the outer shell.

For the capacitance does it want the capacitance of the overall system. C=Q/V ??
 
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Q is the net charge, when i did this question i took it to mean that Q=-q as it is induced by the +q on the inner sphere. note: to be a capacitor the two plates have to carry equal and opposite charges.

use gauss' law to get the field and then go from there in order to get the capacitance.
 
+Q-q ?
 

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