Potential of the inner earthed sphere

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SUMMARY

The discussion focuses on calculating the electric potential of an inner earthed sphere surrounded by an outer shell with charge +Q. The potential at the surface of the inner sphere (r=a) is confirmed to be 0 due to its earthed condition, while the potential at the outer shell (r=b) requires integration of the electric field. The challenge lies in determining the induced charge on the inner sphere, which is not simply -Q, as the inner sphere is an open system. The integration method for calculating potential is emphasized, but the user struggles to achieve the expected result of 0.

PREREQUISITES
  • Understanding of electrostatics and electric potential
  • Familiarity with Gauss's Law and its applications
  • Knowledge of integration techniques in physics
  • Concept of earthed systems in electrostatics
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  • Study the principles of Gauss's Law for spherical charge distributions
  • Learn about induced charge calculations in conductive materials
  • Practice integration of electric fields to find potentials in electrostatics
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This discussion is beneficial for physics students, particularly those studying electrostatics, as well as educators seeking to clarify concepts related to electric potential and induced charges in conductive systems.

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Homework Statement


An outer shell of charge +Q is insulated by a light thread. There is an inner sphere inside which is earthed. The radii of the outer shell and inner sphere are b and a respectively. What is the potential of a point which is at a distance r from the centre of the inner sphere, where
(i) r=b?
(ii) r=a?

Homework Equations


V at r = - [tex]\int charge/ (4\pi\epsilon r^{2})[/tex] from infinity to r

The Attempt at a Solution


Before answering the questions above, I find it difficult to calculate the amount of induced charge on the inner sphere... Because the outer shell having a charge of +Q doesn't mean that the inner shell must have an induced charge of -Q.

How can I find the amount of induced charge?
My teacher gave us some hints to solve this question:
the inner sphere is an open system as it is earthed
the amount of charge of the inner sphere can be calculated

Besides, I initially thought the potential must be 0 without calculation because it is earthed, but when I tried to do the integration (V = dE/ dr), I couldn't get it as 0 (maybe it's due to the problem that I couldn't find the amount of induced charge of the inner sphere)


Can anyone help me?

thanks!
 
Last edited:
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The inner earthed sphere must have a potential of 0, because it's earthed. That means the potential of the two spheres' center must also be 0, since there's no electric field inside the inner sphere.
 
but how to prove it using mathematical calculations? i couldn't get it as 0 using the integration of electric field with respect to r.
 

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