Equation for the energy of an electron

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SUMMARY

The discussion centers on the energy of an electron in relation to its distance from the nucleus, specifically addressing the apparent contradiction between the Coulomb force and work done. As the electron moves away from the nucleus, its energy increases, which aligns with the work equation indicating more work is required to pull it away. The confusion arises from the integration of the Coulomb force, which is inversely proportional to the square of the distance. Contributors emphasize the importance of recognizing that the work done is the negative of the change in potential energy, clarifying the relationship between distance and energy.

PREREQUISITES
  • Coulomb's Law
  • Integration techniques in calculus
  • Concept of potential energy
  • Work-energy theorem
NEXT STEPS
  • Study the derivation of the Coulomb potential energy equation
  • Learn about the work-energy principle in electrostatics
  • Explore integration of force functions in physics
  • Investigate the relationship between force, work, and energy in electric fields
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Physics students, educators, and anyone interested in understanding the principles of electrostatics and energy dynamics related to electrons and atomic structure.

compuser123
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Hello,

I would like to thank all of the contributors on this site. You have helped me in more ways than I can count. I am struggling with the following concept and was wondering if anyone could clarify this.

As the electron gets further away from the nucleus, its energy increases. This makes sense when we look at the work equation, we do more work to pull it further away from where it wants to be.

What I am struggling with is the coulomb equation, where the the force is inversely proportional to distance squared. As the distance increases the force should decrease. Then, if we were to integrate that force to get work, the work would be inversely proportional to the distance, which tells us that work or energy should decrease as you get further away. I am not sure where my mistake is. Any help is much appreciated
 
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Just write down the actual integral, perform the integration from an initial distance to a final distance from the nucleus, and you will see why. Remember that the work done is the negative of the change in potential energy
 
Chandra Prayaga said:
Just write down the actual integral, perform the integration from an initial distance to a final distance from the nucleus, and you will see why. Remember that the work done is the negative of the change in potential energy

Thank you very much, will do it right away.
 

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