Sphere's Electric Potential HELP

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SUMMARY

The electric potential of a 2.04-mm-diameter ball bearing with 8.710×10^9 excess electrons is calculated using the formula V = (1/4πE₀) * (Q/r). The constants used include E₀ = 8.85 x 10^-12 F/m and the charge Q is determined by multiplying the elementary charge (1.6 x 10^-19 C) by the number of excess electrons. The final computed potential is 12308.78 V. The discussion highlights a common misunderstanding regarding the calculation of charge Q from excess electrons.

PREREQUISITES
  • Understanding of electric potential and its formula
  • Familiarity with the concept of excess charge in electrostatics
  • Knowledge of fundamental constants such as E₀ (permittivity of free space)
  • Basic proficiency in unit conversions and calculations involving scientific notation
NEXT STEPS
  • Study the derivation and applications of the electric potential formula V = (1/4πE₀) * (Q/r)
  • Learn about the significance of permittivity of free space (E₀) in electrostatics
  • Explore the concept of charge quantization and its implications in physics
  • Investigate the effects of excess charge on conductive and non-conductive materials
USEFUL FOR

Students in physics, particularly those studying electrostatics, as well as educators and anyone seeking to understand the principles of electric potential and charge calculations.

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Sphere's Electric Potential...HELP!

Homework Statement


A 2.04-mm-diameter ball bearing has 8.710×10^9 excess electrons. What is the ball bearing�s potential?


Homework Equations





The Attempt at a Solution


V = (1/4*pi*Eo) * (Q/r)
Eo = 8.85 x 10^-12
Q = 1.6x10^-19 * 8.71x10^9
r = 0.00204/2
V = 12308.78 V

Can someone tell me what I am doing wrong?
 
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Why are you multiplying your excess charge by Epsilon-not to get Q?
 
nevermind I just got it...thanks though!
 
Great!
 

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