Origin of potentials

  • #1
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).
 

Answers and Replies

  • #2
BvU
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How did you derive (think ?!?) ##V(r)=\dfrac{kq}{r}## for this charge configuration ?
 
  • #3
haruspex
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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).
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.
 
  • #4
rude man
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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.
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 ##?
 
  • #5
haruspex
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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.
 
  • #7
ehild
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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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  • #8
Adesh
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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.
 
Last edited:
  • #9
BvU
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@Guillem_dlc : what do you know of the gauss theorem in connection with electric fields ?
 

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