Calculating the work done in an electric field

AI Thread Summary
The discussion focuses on calculating work done in an electric field, emphasizing the need for calculus when dealing with non-uniform electric fields. It highlights that the formula W = Fd is only applicable in constant electric fields, as force varies with distance in non-uniform fields. The conversation also touches on the relationship between potential energy (PE), electric potential (V), and force (F), noting that direct substitutions are not valid without proper calculus. A request for proof of the equation V = kq/r is made, linking it to derivatives and integrals. Understanding these concepts is crucial for accurately calculating work in varying electric fields.
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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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