Origin of Potential Energy - 65 Characters

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Homework Help Overview

The discussion revolves around the concept of electric potential related to an infinite straight wire with a given linear charge density. Participants are exploring the origin of potential energy and how it is defined in this context.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the derivation of the potential formula V(r)=kq/r for the charge configuration and discussing the implications of choosing a reference point for potential. There are attempts to clarify the relationship between potential and charge density, as well as the assumptions made regarding the potential at infinity.

Discussion Status

The discussion is active, with participants providing insights into the nature of electric potential and the assumptions involved. Some guidance has been offered regarding the integration of the electric field to derive potential, and there is an ongoing exploration of different interpretations of the problem.

Contextual Notes

Participants are addressing the complexity of the problem, including the implications of infinite charge and the need to define a reference point for potential. There is a recognition that the potential at infinity may not necessarily be zero, which is a key aspect under consideration.

Guillem_dlc
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Homework Statement
If an infinite straight wire loaded with a linear density [itex]\lambda =10^{-8}\, \textrm{C/m}[/itex] creates a potential of [itex]239\, \textrm{V}[/itex] at a distance of [itex]8\, \textrm{m}[/itex] from the wire, calculate at what distance [itex]r_0[/itex] from the wire is the origin of potentials [itex] V(r=r_0)=0[/itex]. Express the result in [itex]\textrm{m}[/itex].
a) [itex]114[/itex]
b) [itex]10.6[/itex]
c) [itex]+\infty[/itex]
d) [itex]30.2[/itex]
Relevant Equations
[tex]V(r)=\dfrac{kq}{r}[/tex]
I think the right choice is c. I'll pass on my reasoning to you:

We can think that if the formula of the potential is

[tex]V(r)=\dfrac{kq}{r}[/tex]

If [itex]r[/itex] tends to infinity, then [itex]V(r)=0[/itex].

But the correct answer is d).
 
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How did you derive (think ?!?) ##V(r)=\dfrac{kq}{r}## for this charge configuration ?
 
Guillem_dlc said:
Homework Statement:: If an infinite straight wire loaded with a linear density [itex]\lambda =10^{-8}\, \textrm{C/m}[/itex] creates a potential of [itex]239\, \textrm{V}[/itex] at a distance of [itex]8\, \textrm{m}[/itex] from the wire, calculate at what distance [itex]r_0[/itex] from the wire is the origin of potentials [itex]V(r=r_0)=0[/itex]. Express the result in [itex]\textrm{m}[/itex].
a) [itex]114[/itex]
b) [itex]10.6[/itex]
c) [itex]+\infty[/itex]
d) [itex]30.2[/itex]
Relevant Equations:: [tex]V(r)=\dfrac{kq}{r}[/tex]

I think the right choice is c. I'll pass on my reasoning to you:

We can think that if the formula of the potential is

[tex]V(r)=\dfrac{kq}{r}[/tex]

If [itex]r[/itex] tends to infinity, then [itex]V(r)=0[/itex].

But the correct answer is d).
Potentials are relative to some chosen zero. The formula [tex]V(r)=\dfrac{kq}{r}[/tex] assumes the chosen zero is at infinity, but it need not be.
You are asked to find where the chosen zero is in this case.
 
haruspex said:
Potentials are relative to some chosen zero. The formula [tex]V(r)=\dfrac{kq}{r}[/tex] assumes the chosen zero is at infinity, but it need not be.
You are asked to find where the chosen zero is in this case.
Guillem_dlc said:
We can think that if the formula of the potential is
[tex]V(r)=\dfrac{kq}{r}[/tex]
Doesn' it worry you that ## q = \infty ##?
 
rude man said:
Doesn' it worry you that ## q = \infty ##?
@BvU already addressed that aspect, but it seemed to me the OP also needed to be made aware that the potential at infinity need not be zero.
Perhaps I should have made that clearer.
 
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Guillem_dlc said:
Relevant Equations:: [tex]V(r)=\dfrac{kq}{r}[/tex]

It is a charged wire, not a point charge q.
The linear charge density is given, you have to use it instead of "q"
 
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It’s misleading that the formula for potential is $$V = \frac{1}{4\pi \epsilon_0} \int \frac{\lambda (\vec{r’})}{\mathcal{r}} dl’$$ true. It is true but only when the charge density goes to zero at infinity. The above integration formula for potential is derived from the Poisson’s Eqaution $$ \nabla^2 V =-\frac{\lambda}{\epsilon_0}$$ (NOTE: I’m writing ##\lambda## instead of ##\rho## just to make this discussion more OP oriented). That Poisson’s Equation is a second order partial differential equation, which is very tough to solve, and for cases where the line charge density goes to zero we can solve it to get the formula in that integral form.

So, my suggestion is derive the formula for potential by integrating the electrical field along some path and between two points.
 
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@Guillem_dlc : what do you know of the gauss theorem in connection with electric fields ?
 

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