How Is Electric Potential Energy Calculated in Electron Orbit Problems?

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SUMMARY

The calculation of electric potential energy for an electron orbiting a nucleus with a charge of +9.6E-19 C and an orbit radius of 2.0E-10 m is performed using the formula Ep = kQ1Q2/r. The correct calculation yields an electric potential energy of -6.9E-18 J, while the answer key incorrectly states -6.9E-19 J. Additionally, the voltage (V) is calculated as 43.2 V, confirming the relationship V = Ep/Q. The discrepancy in the answer key is attributed to a potential error in exponent notation.

PREREQUISITES
  • Understanding of electric potential energy calculations
  • Familiarity with Coulomb's law and constants (k = 9E9 N m²/C²)
  • Knowledge of charge values for an electron (-1.6E-19 C) and a proton (+9.6E-19 C)
  • Basic grasp of voltage and its relationship to energy and charge
NEXT STEPS
  • Review the derivation and application of Coulomb's law in electric potential energy calculations
  • Study the concept of electric potential (V) and its significance in electrostatics
  • Explore common errors in exponent notation and how to avoid them in calculations
  • Practice additional problems involving electric potential energy and charge interactions
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to electric potential energy and charge interactions.

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Homework Statement


An electron orbits a nucleus which carries a charge of +9.6E-19 C. If the electron's orbit radius is 2.0E-10 m, what is its electric potential energy?


Homework Equations


Ep = kQ1Q2/r
V = kQ/r
V = Ep/Q


The Attempt at a Solution


I tried using Ep = kQ1Q2/r
Ep = (9E9)(9.6E-19C)(-1.6E-19C) / (2.0E-10m)
Ep = -6.9E-18 J
the answer key is saying -6.9E-19 J

and V = kQ/r , V = Ep/Q
V = (9E9)(9.6E-19C) / (2.0E-10m)
V = 43.2 V
V = Ep/Q
Ep = 43.2 V x (1.6E-19)
Ep = -6.9E-18 J
the answer key is saying -6.9E-19 J

Am I doing something wrong, or is the answer key off by an exponent?

Thanks,
 
Physics news on Phys.org
you seem to be very correct to me

marlon
 
ok, thank you
 

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