Origin of Potential Energy - 65 Characters

In summary, the potential at infinity need not be zero, and you can derive the potential by integrating the electric field along a path and between two points.
  • #1
Guillem_dlc
184
15
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).
 
Physics news on Phys.org
  • #2
How did you derive (think ?!?) ##V(r)=\dfrac{kq}{r}## for this charge configuration ?
 
  • #3
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.
 
  • #4
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 ##?
 
  • #5
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.
 
  • Like
Likes Adesh
  • #7
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"
 
  • Like
Likes Delta2 and Adesh
  • #8
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:
  • Like
Likes Delta2
  • #9
@Guillem_dlc : what do you know of the gauss theorem in connection with electric fields ?
 

Related to Origin of Potential Energy - 65 Characters

What is potential energy?

Potential energy is the energy an object possesses due to its position or configuration. It is stored energy that can be converted into other forms of energy, such as kinetic energy.

What are the different types of potential energy?

The three main types of potential energy are gravitational potential energy, elastic potential energy, and chemical potential energy. Gravitational potential energy is related to an object's position in a gravitational field, elastic potential energy is related to the deformation of an object, and chemical potential energy is related to the chemical bonds within a substance.

How is potential energy calculated?

The formula for calculating potential energy depends on the type of potential energy being considered. For gravitational potential energy, the formula is PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object. For elastic potential energy, the formula is PE = 1/2kx^2, where k is the spring constant and x is the displacement from equilibrium. For chemical potential energy, the formula is PE = nRT, where n is the number of moles of the substance, R is the gas constant, and T is the temperature in Kelvin.

What is the relationship between potential energy and kinetic energy?

Potential energy and kinetic energy are two forms of energy that can be converted back and forth between each other. When an object has potential energy, it has the potential to do work and therefore has the potential to gain kinetic energy. When an object is in motion, it has kinetic energy, but it also has the potential to gain potential energy if it changes its position or configuration.

How does potential energy relate to the conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. This means that potential energy can be converted into other forms of energy, such as kinetic energy, but the total amount of energy in a closed system remains constant. Therefore, potential energy plays a crucial role in the conservation of energy.

Similar threads

Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
428
  • Introductory Physics Homework Help
Replies
1
Views
957
Replies
1
Views
216
  • Introductory Physics Homework Help
Replies
1
Views
909
  • Introductory Physics Homework Help
Replies
23
Views
406
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
Back
Top