Maximum potential given a charge

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SUMMARY

The discussion centers on calculating the maximum charge and potential of a hollow metal sphere with a radius of 2 cm, given the breakdown field of air at 1 MV/m. The maximum charge (Q) that can be placed on the sphere is determined to be 4.4 x 10-8 C using the equation E = Q/(4πε0r2). The maximum potential (V) on the surface of the sphere is calculated using V = Q/(4πε0r), resulting in a value of 20 kV. The relationship between electric field and potential is clarified, highlighting the difference in the dependence on radius.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with Coulomb's law
  • Knowledge of the constants ε0 (vacuum permittivity)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of electric field equations for spherical conductors
  • Learn about the relationship between electric field and potential
  • Explore the concept of breakdown voltage in different materials
  • Investigate the applications of electrostatics in real-world scenarios
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone interested in electrostatics and the behavior of charged conductors.

c1223
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Homework Statement



So the question states that we have a hollow metal sphere of radius 2cm that is charged up. We take the breakdown field of air to be 1MV/m. We are asked to:

a) Calculate the maximum amount of charge which can be placed on the sphere.
b) Calculate the maximum potential on the surface of the sphere.

Homework Equations



[itex]E = \frac{Q}{4\Piε_{0}r^{2}}[/itex]

The Attempt at a Solution



a) [itex]1\cdot10^{6} = \frac{Q}{4\Piε_{0}0.02^{2}}[/itex]
So we get Q as being [itex]4.4\cdot10^{-8}C[/itex]​

However, I'm stuck with what to do next to get the potential. I'm not entirely sure I understand the question? Is it the potential between that space on the surface and the center of the sphere?

I know the answer (or at least it is given as 20kV) but I'm not sure how to arrive at that answer.

Cheers
 
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I think I might have cracked it. I realized that I also need the equation for potential which is:

[itex]V = \frac{Q}{4\Piε_{0}r}[/itex]

Could someone please explain why it is [itex]r^{2}[/itex] for electric field but [itex]r[/itex] for potential?
 
You just need to recall how the potential and the field are related to Coulomb's force.
 

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