What is the electric potential energy in joules of two electrons

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SUMMARY

The electric potential energy between two electrons separated by 2.43 nm is calculated to be 1.185 J, while doubling the separation results in a potential energy of 0.95 J. The formula used is derived from the electric potential energy equation, which is V = kq/r, where k is Coulomb's constant, q is the charge of the electrons, and r is the separation distance. The correct approach involves using kq²/r to account for both charges in the system.

PREREQUISITES
  • Understanding of Coulomb's Law and electric potential energy
  • Familiarity with the constants involved, specifically Coulomb's constant (k)
  • Knowledge of the charge of an electron (approximately -1.602 x 10^-19 C)
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Research the derivation of the electric potential energy formula kq²/r
  • Learn about Coulomb's constant and its significance in electrostatics
  • Explore the effects of distance on electric potential energy in multi-charge systems
  • Investigate the implications of electric potential energy in practical applications, such as capacitors
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to electric potential energy and Coulomb's Law.

BrianR84
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Homework Statement



(a) What is the electric potential energy in joules of two electrons separated by 2.43 nm?
(b) What would be it be if the separation was doubled?


Homework Equations



i think v=kq/r

The Attempt at a Solution



(a) 1.185 J
(b) 0.95 J

i can't figure out what I am doing wrong, I am following the book example.
 
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I think it would be kq^2/r. kq/r is the electric potential from a single charge. You want to multiply by the other charge to get the potential energy. But in either event, I don't get what you are getting. Maybe you should tell us what numbers you are putting in.
 

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