Integrating an inverse square to find U

Click For Summary
SUMMARY

The discussion centers on the integration of the work equation to derive Electric Potential Energy, specifically the expression kQq(1/r_a - 1/r_b). The user seeks clarification on the origin of the term (1/r_a - 1/r_b) and its relation to calculus, particularly the power rule and anti-derivatives. The conclusion emphasizes that taking the anti-derivative of r^-1 leads to the definite integral resulting in the specified expression, with a sign switch necessary due to the relationship between Work and Potential Energy.

PREREQUISITES
  • Understanding of calculus, specifically integration and the power rule.
  • Familiarity with the concepts of Electric Potential Energy and work in physics.
  • Knowledge of the constants k, Q, and q in the context of electrostatics.
  • Basic grasp of definite integrals and their application in physics.
NEXT STEPS
  • Review the power rule in calculus for integration.
  • Study the derivation of Electric Potential Energy in electrostatics.
  • Learn about the relationship between Work and Potential Energy in physics.
  • Explore advanced integration techniques relevant to physics applications.
USEFUL FOR

Students of physics, particularly those studying electrostatics, and individuals looking to refresh their calculus skills related to integration and its applications in physical concepts.

Imabioperson
Messages
5
Reaction score
0
Hello everyone,

This is probably going to come off as a very silly question. However, I have not had calculus in several years. I was angry that my physics textbook did not have a derivation of Electric Potential Energy. So, I finally came across it, and I see that the integration of the work equation from some point, r_a to another point, r_b yields, kQq (1/r_a - 1/r_b). Can someone explain to me where, (1/r_a - 1/r_b) is coming from?
 
Physics news on Phys.org
Sounds like you realize it's a calculus issue. Review the power rule.
 
So, if you take the anti-derivative FIRST... we yield r^-1 in the numerator. And then, we will take the definite integral, leaving us with (1/upper limit - 1/lower limit)? And then do I switch the sign because of the relation between Work and Potential energy?
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 28 ·
Replies
28
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K