Understanding Electric Potential and Energy: Tips for Physics 12 Students

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SUMMARY

This discussion focuses on the understanding of electric potential and energy for Physics 12 students, specifically addressing the conversion from voltage to charge/work and the distinction between electric fields. The formula F = kqq/r² is correctly understood, but the participant seeks clarity on why moving a charge requires 1 Joule for 1 Coulomb at 1 Volt. The conversation clarifies that while the force varies as 1/r², the potential energy is represented as 1/r, emphasizing the relationship between force, work, and potential energy through integration.

PREREQUISITES
  • Understanding of Coulomb's Law (F = kqq/r²)
  • Basic knowledge of electric potential (Voltage = J/C)
  • Familiarity with the concept of work in physics
  • Introduction to calculus (integration concepts)
NEXT STEPS
  • Study the relationship between electric potential and electric fields
  • Learn about the integral of force and its application in calculating work
  • Explore the differences between conservative and non-conservative forces
  • Review the principles of energy conservation in electric systems
USEFUL FOR

Physics students, particularly those in high school or introductory college courses, educators teaching electricity concepts, and anyone looking to deepen their understanding of electric potential and energy relationships.

Alkatran
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I'll start by apologizing for any incorrect terms. I've learned all of this in french so a few words may be misspelled or even misused.

We just started doing electricity in physics 12. I understand F = kqq/r^2 perfectly, I get that potential (electric?) energy is the same formula, multiplied by r. I can do the problems through guesswork (Well, if Voltage is J/C, and I have work and charge...) but I don't UNDERSTAND why I'm doing these things. It's like flying on autopilot.

basically what I have trouble understanding is:
Conversion from voltage to charge/work. Why does it work? It takes 1 Joule to move 1 C to the position that has 1 V electric potential right? What are the two different type of fields? (There's one that is /r^2 and one that is /r, what is each and how are they used?)

Just generally confused here. We have a quiz tomorrow and I don't want to have to do some fancy guesswork (even if it does always work).
 
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Alkatran said:
basically what I have trouble understanding is:
Conversion from voltage to charge/work. Why does it work? It takes 1 Joule to move 1 C to the position that has 1 V electric potential right? What are the two different type of fields? (There's one that is /r^2 and one that is /r, what is each and how are they used?).
No, there's only one field it is like 1/r. The other is a force. It is like pushing a car up hill. At every point, a certain force is required. This force is proportional to the slope of the hill. But the field is related to the height you've moved the car to. When you are done, the work is the change in height, but you can find the same answer by multiplying the force needed at every point by the small distance traveled and adding all these contributions together. Guess what? You'll get the same answer.

So with charges. To move two charges closer together requires force which varies with distance as 1/r^2. The amount of work done, can however be calculated much more simply by comparing the potential energy of the particles (1/r) before and after they were moved.

In short, the work is the integral of F.dx, and when you integrate 1/r^2, you get -1/r. But I don't suppose you've learned any calculus yet. There's lots of fun in store for you!
 
I'm actually doing calculus now (and having an incredibly easy time, mind you.. hit into how to find a polynomials slope at any point by myself). I already knew of the similarities between the attraction between charges and gravity but my question is very vague.

I'll just think about it for a while and it'll sort itself out... I'll associate the names with the equations tomorrow..
 

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