Calculating the work done in an electric field

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SUMMARY

The discussion focuses on calculating work done in an electric field, emphasizing the necessity of calculus for non-uniform electric fields. Key equations mentioned include \( W = Fd \), \( E = \frac{F}{q} \), and \( V = \frac{kQ}{r} \). The participant highlights that the formula for work cannot be applied directly when the electric field is not constant, necessitating the use of calculus to derive the correct expressions. The integration of force over distance is crucial for understanding potential differences in varying electric fields.

PREREQUISITES
  • Understanding of electric fields and forces, specifically \( E = \frac{F}{q} \)
  • Basic knowledge of potential energy in electric fields, particularly \( V = \frac{kQ}{r} \)
  • Familiarity with calculus concepts, especially integration and derivatives
  • Ability to apply the work-energy principle in physics
NEXT STEPS
  • Study calculus fundamentals, focusing on integration techniques
  • Learn about electric field variations and their impact on force calculations
  • Explore the relationship between electric potential and electric field strength
  • Investigate advanced applications of the work-energy theorem in non-uniform fields
USEFUL FOR

Students of physics, particularly those studying electromagnetism, as well as educators seeking to clarify concepts related to work done in electric fields and the application of calculus in physics.

Biker
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Homework Statement


It is not an actual questions, Just misconceptions
For example,
Lets find the work done over a distance in an electric field

Homework Equations


v = PE/q
E = F/q
F = k q1 q2 /r^2
W = fd

The Attempt at a Solution


I didn't study calculus yet but I am going to because I need to get to know these equations.

I know that I can't use the formula w = f d because obviously the f varies if I don't have a uniform EF
these equations only take place when the electric field is constant otherwise I need to use calculus to find the equation because the forces varies with distance.

So in V = PE/C
I can't just substitute PE with F d... so if you are able to put the proof of V = kq/r (If it is related to calculus) that would be great so I can check it when I am done with calculus.
 
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Perhaps you already know a bit about derivatives ? If you know that $${d\over dx} {1\over x} = -{1\over x^2}$$you can see that that kind of matches ## V = {kQ\over r} ## and ##E = - k {Q\over r^2 }##.

Indeed $$V(r_2) - V(r_1) = \int_{r_1}^{r_2} F(r)\, dr = -k Q \int_{r_1}^{r_2} {1\over r^2} \, dr = -kQ \left [- {1\over r} \right ]_{r_1}^{r_2} = k{Q \over r_2} - k{Q\over r_1}$$
 
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