Charge Distribution on Spherical Conductors

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SUMMARY

The discussion centers on the charge distribution between two spherical conductors, a larger sphere and a smaller sphere, connected by a thin wire. The larger sphere will accumulate more charge due to its greater surface area, which allows for a higher charge density. The relationship between charge and potential is defined by the equations V1 = k*Q1/R1 and V2 = k*Q2/R2, leading to the conclusion that both spheres will reach the same electric potential when connected, despite differing charges. The total charge remains conserved throughout the process.

PREREQUISITES
  • Understanding of electrostatics principles
  • Familiarity with the concept of electric potential
  • Knowledge of charge conservation laws
  • Basic proficiency in using equations related to electric fields and potentials
NEXT STEPS
  • Study the principles of charge distribution on conductors
  • Learn about the implications of electric potential in electrostatics
  • Explore the concept of capacitance in spherical conductors
  • Investigate the effects of connecting multiple conductors in electrostatic systems
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Students studying electrostatics, physics educators, and anyone interested in understanding charge distribution in conductive materials.

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Homework Statement



thin wire connecting two conductors, a big sphere and a small sphere. What sphere has the most charge?

Homework Equations





The Attempt at a Solution



The charge will want to distribute itself over the entire surface area of the two conductors. Therefore because of the larger sphere having more surface area, there will be more charge on the larger sphere. IS this a logical way of approaching this problem?
 
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If both spheres had a a previous charge and then were connected with the wire, both will now have the same potential put different charges, although the total charge won't change:

V1= k*Q1/R1 V2=k*Q2/R2

Now V'1=V'2 so: Q1'=V'*R1/K Q2'=V'*R2/K

and Q1 + Q2 = Q1' + Q2'
 
That makes sense! I think I was on the right track. Thanks!
 

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