Voltage of a charged conducting sphere

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SUMMARY

The voltage of a charged conducting sphere remains constant throughout its interior and at its surface due to the electric field being zero inside the sphere. While the electric field exists on the surface, the potential does not change significantly as one approaches the surface, as any change in voltage is infinitesimal. To understand the relationship between electric field and voltage, one must integrate the electric field (E) to derive the voltage (V). This concept can be illustrated using a step function to demonstrate the area under the curve, emphasizing the continuity of potential despite changes in electric field.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with integration in calculus
  • Knowledge of charged conductors and their properties
  • Basic concepts of step functions in mathematics
NEXT STEPS
  • Study the relationship between electric field and electric potential in electrostatics
  • Learn about the mathematical integration of electric fields to find voltage
  • Explore the properties of charged conductors and their behavior in electrostatic equilibrium
  • Investigate step functions and their applications in calculus and physics
USEFUL FOR

Students of physics, electrical engineers, and anyone interested in understanding electrostatics and the behavior of electric fields and potentials in charged conductors.

maccha
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I've recently learned that the voltage of a charged conduction sphere remains constant inside the sphere all the way to the surface, as the electric field inside is zero. What I don't understand is how the voltage can be the same on the surface as it is inside- since there is an electric field on the surface, wouldn't the potential change?
 
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maccha said:
I've recently learned that the voltage of a charged conduction sphere remains constant inside the sphere all the way to the surface, as the electric field inside is zero. What I don't understand is how the voltage can be the same on the surface as it is inside- since there is an electric field on the surface, wouldn't the potential change?

As you increase r to the surface, the potential does change, but infinitesimally with any infinitesimal increase in r. The electric field can change dramatically, as you have pointed out. But you need to integrate E to get V.

If it helps, let me pose you this question. Consider a step function where y jumps straight up from 0 to 1 at x = 10. What's the area under the curve from x = 0 to x = 9.999999? What's the area under the curve from x = 0 to x = 10.0001?
 

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