Calculating Electrostatic Force Between Pennies with Slight Charge Differences

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SUMMARY

The discussion focuses on calculating the electrostatic force between two copper pennies with a slight charge difference of 0.0001%. The relevant equation used is Coulomb's Law, F = kQ1Q2/r², where 'k' is the electrostatic constant, 'Q1' and 'Q2' are the charges of the pennies, and 'r' is the distance between them (1.0 m). Participants emphasize estimating the number of protons and electrons in a penny to determine the charges accurately. The discussion highlights the importance of understanding charge differences in electrostatic calculations.

PREREQUISITES
  • Coulomb's Law (F = kQ1Q2/r²)
  • Basic knowledge of atomic structure (protons, electrons)
  • Periodic table for element properties
  • Understanding of charge quantification and estimation
NEXT STEPS
  • Calculate the electrostatic force using varying charge differences
  • Explore the implications of charge differences on electrostatic interactions
  • Research the properties of copper and its electron configuration
  • Learn about the applications of electrostatics in real-world scenarios
USEFUL FOR

Students studying physics, particularly those focusing on electrostatics, as well as educators looking for practical examples of charge interactions in atomic structures.

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Homework Statement


We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, however, that these magnitudes differed from each other by as little as 0.0001%. With what force would two copper pennies, placed 1.0 m apart, repel each other? Estimate any quantities you need that you do not know.

Homework Equations


F=kQ1Q2/r^2

The Attempt at a Solution


I do not know from where to start. I drew a picture of the charges. I do not know how to find the charge from 0.0001% condition. Can you help me please?
 
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You could start by estimating the numbers of protons and electrons in a penny. You may need to consult the periodic table...
 

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