Charge Distribution Between Two Conducting Spheres

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SUMMARY

The discussion focuses on the charge distribution between two conducting spheres with diameters of 0.400m and 1.00m, connected by a wire and charged to a total of 7.00x10-6C. The capacitance for each sphere is calculated using the formula C = r/ke, where r is the radius and ke is the permittivity constant. The key conclusion is that when connected, the spheres reach equipotential, meaning the electric potential becomes equal across both spheres, allowing for the determination of charge distribution based on their respective capacitances.

PREREQUISITES
  • Understanding of electric potential and capacitance
  • Familiarity with the concept of equipotential surfaces
  • Knowledge of basic electrostatics and charge distribution
  • Ability to apply formulas for capacitance (C = r/ke) and charge (Q = C * DV)
NEXT STEPS
  • Study the principles of capacitance in spherical conductors
  • Learn about equipotential surfaces and their significance in electrostatics
  • Explore the relationship between charge, capacitance, and electric potential
  • Investigate practical applications of charge distribution in electrical engineering
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Students and professionals in physics and electrical engineering, particularly those interested in electrostatics, capacitance, and charge distribution principles.

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[SOLVED] Capacitance and Charge

2 conducting spheres, one of diameter .400m and the other of diameter 1.00m, are separated by a distance large compared to the diameter. The spheres are connected by a wire and the system charged to a total of 7.00x10E-6C (total charge of system). How is the charge distributed between the spheres? (ignore any charge on the wire)

Since the capacitance for each sphere is constant and independent of their charge, I was thinking of using that...just that: C = r/ke for each sphere. Also, C = Q/DV, DV as deltaV. The one thing that I keep getting messed up with is that the electric potential of a sphere changes in respect to charge...so I'm not exactly sure what I can do at that point.
 
Last edited:
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When two conductors are joined by wire, which quantity becomes equal on both of them? (Charge will flow until that thing becomes equal on both.)
 
heh...they become equipotential...solved it, thanks.
 

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