Why the change in potential energy is the same in all cases?

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  • #1
Frigus
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Definition of conservative field I use:-it is a field in which potential energy of system is independent of path taken.

I understand that it is independent because whenever we take some path than all the perpendicular displacements with respect to force are not counted and if we go further than the point for which we are finding potential energy and come back to that point than additional potential energy due to further displacement is canceled by the reverse path and thus energy is conserved.
But I can't understand the case I have posted,
Work done in moving particle from A to C is equal to B to C but I don't know why.
I thought it should be same because force on point at b is more than point at c thus it should but i want to know why work done in both cases are exactly similar.
Edit:- both points are at infinity
 

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  • #2
kuruman
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Are you moving a negatively charged particle in a uniform electric field? Is the direction of the force on the particle represented by the arrow labeled F in your drawing?

Work from A to C is generally not equal to work from B to C because the starting points are different. Work fro A to C along the straight path is equal to work from A to B plus work from B to C. The end and start points must be the same.
 
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  • #3
Frigus
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Are you moving a negatively charged particle in a uniform electric field?
No,I am moving these negative charges in a field created by positively charged particle which I have shown in figure.
the direction of the force on the particle represented by the arrow labeled F in your drawing?
Yes,it is force and I also have represented its component in its displacement direction.

Work from A to C is generally not equal to work from B to C because the starting points are different. Work fro A to C along the straight path is equal to work from A to B plus work from B to C. The end and start points must be the same.
I was talking about the case in which both b and c particles are very away so they have negligible interactions with positive particle(thus at infinity).

Sorry for bad diagram representation.
 
  • #4
kuruman
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No,I am moving these negative charges in a field created by positively charged particle which I have shown in figure.
Yes,it is force and I also have represented its component in its displacement direction.


I was talking about the case in which both b and c particles are very away so they have negligible interactions with positive particle(thus at infinity).

Sorry for bad diagram representation.
Can you post a concise "Statement of the problem" describing as best as you can the physical situation and the question you want answered? If a field is created by a positively charged particle, state its position and magnitude or describe the electric field it creates. You have these clear in your mind, but others do not. In your drawing I see two charges labeled ##-q## near the top left and right and charge ##q_{m?}## at bottom right. What am I to make of this?
 
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  • #5
Frigus
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Can you post a concise "Statement of the problem" describing as best as you can the physical situation and the question you want answered? If a field is created by a positively charged particle, state its position and magnitude or describe the electric field it creates. You have these clear in your mind, but others do not. In your drawing I see two charges labeled ##-q## near the top left and right and charge ##q_{m?}## at bottom right. What am I to make of this?
Sorry for erroneous explanation,
I tried to reframe my question and I hope I have now done it better. their are 3 particles +q₁,+q₂,+q₃ in the figure I have posted and we brought particle q₂ and q₃ from their initial position to their final position.
Now some work has been done on particle +q₃ and +q₂(that work is done against the force generated by particle q₁)
and these work add upon with their respective potential energies(I am talking about the potential energy of pair q₁,q₂ and q₁,q₃) and now their new potential energies are same.
Now I can't understand how there potential energy became same after reaching final position.
 

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  • #6
jbriggs444
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Sorry for erroneous explanation,
I tried to reframe my question and I hope I have now done it better. their are 3 particles +q₁,+q₂,+q₃ in the figure I have posted and we brought particle q₂ and q₃ from their initial position to their final position.
Now some work has been done on particle +q₃ and +q₂(that work is done against the force generated by particle q₁)
and these work add upon with their respective potential energies(I am talking about the potential energy of pair q₁,q₂ and q₁,q₃) and now their new potential energies are same.
Now I can't understand how there potential energy became same after reaching final position.
So we have two particles (##q_2## and ##q_3##) which start with different potential energies in the field of stationary charge ##q_1##.

The two particles are moved on different paths that take different amounts of work but wind up with the same potential energy as a result.

You express dismay at this result. What do you find startling about it?
 
  • #8
Frigus
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So we have two particles (q2q2q_2 and q3q3q_3) which start with different potential energies in the field of stationary charge q1q1q_1.

The two particles are moved on different paths that take different amounts of work but wind up with the same potential energy as a result.
Yes this is what i am trying to say
You express dismay at this result. What do you find startling about it?
I tried to answer why their potential energy got equal when they brought to same point but it just seems magical to me that it happens and I am unable to even find a bit why it work.
Please give me some hint for where to start thinking about it.
I think it is a stupid question and I am just thinking very complex.
 
  • #9
jbriggs444
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Yes this is what i am trying to say
I tried to answer why their potential energy got equal when they brought to same point but it just seems magical to me that it happens and I am unable to even find a bit why it work.
Please give me some hint for where to start thinking about it.
I think it is a stupid question and I am just thinking very complex.
Do the ##q_2## and ##q_3## have the same charge?

If so, then their potential energy "at infinity" is conventionally defined to be zero. The path-independence of the field means that their potential at the final point is fixed regardless of path.

Which means that their potential at the starting points is guaranteed to be just right so that the work done bringing them to the final point gets the right increment to potential energy.
 
  • #10
Frigus
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Do the ##q_2## and ##q_3## have the same charge?
Yes,I have also written it in picture posted in post 5.
 
  • #11
jbriggs444
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Yes,I have also written it in picture posted in post 5.
Draw two paths. One from infinity to the final point with a stop at the starting position for ##q_2## and the other from infinithy to the final point with a stop at the starting position for ##q_3##.

If the field is conservative, the work done on both paths must be equal.
 
  • #12
Frigus
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Which means that their potential at the starting points is guaranteed to be just right so that the work done bringing them to the final point gets the right increment to potential energy.
How can we prove it if we didn't knew that it is a conservative field?
Please give some hint only so I can use my brain to tackle it.
 
  • #13
jbriggs444
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How can we prove it if we didn't knew that it is a conservative field?
Please give some hint only so I can use my brain to tackle it.
If the field is not conservative then it does not even have a potential. So you cannot prove it.
 
  • #14
vanhees71
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As far as I understand the question, you talk about the Coulomb field. Since this can be described by a potential (for a particle sitting in the origin of the reference frame)
$$\Phi(\vec{x})=\frac{q}{4 \pi \epsilon_0 |\vec{x}|}$$
The work of a test charge is independent of the path. Then there's Poincare's lemma telling you that in any simply connected region of space the path-independence of the line integral implies that existence of a scalar potential and that's equivalent to have ##\vec{\nabla} \times \vec{E}=0## everywhere in the simply-connected region.
 
  • #15
Frigus
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As far as I understand the question, you talk about the Coulomb field. Since this can be described by a potential (for a particle sitting in the origin of the reference frame)
$$\Phi(\vec{x})=\frac{q}{4 \pi \epsilon_0 |\vec{x}|}$$
The work of a test charge is independent of the path. Then there's Poincare's lemma telling you that in any simply connected region of space the path-independence of the line integral implies that existence of a scalar potential and that's equivalent to have ##\vec{\nabla} \times \vec{E}=0## everywhere in the simply-connected region.
Sorry but the only word I am able to understand in your post is coulomb field😅.
This is beyond my knowledge and I haven't listened a single word you have written.
Will I have to study all this stuff like poincare to understand this?
And also I am a medical student so I don't know too much about maths,I know only basics and some easy calculus.
 
  • #16
vanhees71
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To really understand physics you need to study mathematics (linear algebra and vector calculus for field theory).
 
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  • #17
sophiecentaur
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And also I am a medical student so I don't know too much about maths,I know only basics and some easy calculus.
There's no easy way into this. At least, in this case, the bare 'facts' and consequences about the PE difference are relatively easy to understand. Without some more Maths you may just have to accept that but it's a very useful principle to use when figuring out such problems qualitatively.
A 'black box' approach can open a lot of doors. In fact everybody does it to some extent so there's no need to feel guilty that you aren't doing it all 'properly'.
 
  • #18
jbriggs444
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How can we prove it if we didn't knew that it is a conservative field?
Please give some hint only so I can use my brain to tackle it.
Perhaps I should understand the question as @vanhees71 did:

"How do we know that the Coulomb force is conservative?"

There is a theorem: If you have a force that always points toward or away from a fixed center and if the magnitude of that force only depends on the distance from the center then that force field is conservative.

This makes good intuitive sense. The work you do against this force only depends on how much progress you make toward (or away from) the center. It is independent of how you move tangentially around the center. The potential of such a field will be purely a function of the distance from the center.
 
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  • #19
wrobel
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Definition of conservative field I use:-it is a field in which potential energy of system is independent of path taken.
that is senseless, read a textbook first
 
  • #20
jbriggs444
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Definition of conservative field I use:-it is a field in which potential energy of system is independent of path taken.
that is senseless, read a textbook first
This criticism is reasonable and good advice, albeit a bit harsh.

One of the benefits of learning mathematics in a mathematics classroom rather than in internet discussion forums is that you pick up on little details. Patterns of speech. Ways of thinking. Terminology. Background assumptions and information that can be called upon without conscious thought. Proper mathematical discourse has rules. One of which is that you should not reference a thing without first having grounds on which to assert its existence. (Objection, your honor: foundation).

The "potential energy of a system" is not quite the right term. We are talking here about the potential of a "vector field". But as long as the force pattern is unchanging (that is, as long as the object that is sourcing the field is fixed) we can let that slide.

The bigger problem is that the "potential" of a "vector field" is only defined if the field is conservative. One cannot use potential to define conservative because even having a potential depends on the field being conservative in the first place.

So instead of talking about "potential energy of a system" you need to be talking about "work done along a path". One could then support that by defining "work done" in terms of the path integral between two end points along a particular path.

So the definition of "conservative" you need to be using is, roughly speaking:

A field is conservative if and only if the work done on a path between two endpoints depends only on the end points and not on the path taken between them.
 
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  • #21
Frigus
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The work you do against this force only depends on how much progress you make toward (or away from) the center. It is independent of how you move tangentially around the center. The potential of such a field will be purely a function of the distance from the center.
It makes a lot sense in case of uniform field but it is very hard for me to understand the cases like i have drawn in figure,
From path a to b it is very hard to apply this logic.
After thinking for long time about this I thought now I should not waste time and I opened YouTube then I searched path independence of Line integral and I started watching video about it and In video they start talking about partial derivatives e.t.c and then at this point I understood why it is better to remember it as a fact 😅.
I will try to get some time for learning vector calculus and linear algebra so as to understand more clearly about this topic.
Thanks to everyone for again spending your precious time on me.
This criticism is reasonable and good advice, albeit a bit harsh.
I have no problem when someone criticize me although in past I have done some mistakes but I have problem when someone praises me so as to get rid of me when I asks questions.
A 'black box' approach can open a lot of doors. In fact everybody does it to some extent so there's no need to feel guilty that you aren't doing it all 'properly'.
I think you are right,if we just start digging about the reason about everything and start spending time which has to be given to course then we will end up with nothing but with useless and chaotic information but problem is that whenever I accepts this ideology and start studying then my heart starts wiggling and say to me that I should not do this.🥺
 

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  • #22
Frigus
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A field is conservative if and only if the work done on a path between two endpoints depends only on the end points and not on the path
Thanks for correcting me,now I will use this definition.
 
  • #23
vanhees71
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You cannot understand it with the appropriate math. Some people in physics didactics seem to think it's helping students to avoid math at any cost, but that's misleading and cheating the students. The most ideosyncratic thing that ever came to my attention was a textbook at a library title "Non-calculus physics". This is a contractio in adjecto. There is no physics to be possible to discussed without calculus to begin with.

What you need to learn here is some vector calculus. Your case of the Coulomb field is very easy, because it is very simple to see that the Coulomb field
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0} \frac{\vec{r}}{|\vec{r}|^3}$$
is derivable from a potential,
$$\vec{E}=-\vec{\nabla} \phi \quad \text{with} \quad \phi(\vec{r})=\frac{Q}{4 \pi \epsilon_0 |\vec{r}|}.$$
To see this, just take the derivatives
$$\partial \phi/\partial x_1=-\frac{Q}{4 \pi \epsilon_0 |\vec{r}|^2} \frac{\partial |\vec{r}|}{\partial x_1}.$$
Now
$$|\vec{r}|=\sqrt{x_1^2+x_2^2+x_3^2} \; \Rightarrow \; \frac{\partial |\vec{r}|}{\partial x_1}=\frac{x_1}{\sqrt{x_1^2+x_2^2+x_3}^3}=\frac{x_1}{|\vec{r}|}.$$
So finally, doing the same calculation also for the derivative wrt. ##x_2## and ##x_3##, you indeed get ##\vec{E}=-\vec{\nabla} \phi##.

Now if you have a vector field that has a potential, the line integral connecting two points ##\vec{r}_1## and ##\vec{r}_2## does not depend on the shape of the integration path but only on the points connected by this path.

To see this, parametrize the curve ##C## by ##\vec{r}=\vec{r}(\lambda)## with ##\lambda \in [\lambda_1,\lambda_2]## such that ##\vec{r}(\lambda_1)=\vec{r}_1## and ##\vec{r}(\lambda_2)=\vec{r}_2##. Then the line integral is
$$\int_C \mathrm{d} \vec{r} \cdot \vec{E}=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \frac{\mathrm{d} \vec{r}(\lambda)}{\mathrm{d} \lambda} \cdot \vec{E}[\vec{r}(\lambda)],$$
but
$$\vec{E}[\vec{r}(\lambda)]=-\vec{\nabla} \phi[\vec{r}(\lambda)]$$
and thus
$$\frac{\mathrm{d} \vec{r}(\lambda)}{\mathrm{d} \lambda} \cdot \vec{E}[\vec{r}(\lambda)] = -\frac{\mathrm{d} \vec{r}(\lambda)}{\mathrm{d} \lambda} \cdot \vec{\nabla} \phi[\vec{r}(\lambda)].$$
According to the chain rule you thus finally have
$$\frac{\mathrm{d} \vec{r}(\lambda)}{\mathrm{d} \lambda} \cdot \vec{E}[\vec{r}(\lambda)] =-\frac{\mathrm{d}}{\mathrm{d} \lambda} \phi[\vec{r}(\lambda)].$$
Plugging this into the integral then leads to
$$\int_{C} \mathrm{d} \vec{r} \cdot \vec{E}=-[\phi(\vec{r}_2)-\phi(\vec{r}_1)],$$
which is indeed independent of the path connecting the points ##\vec{r}_1## and ##\vec{r}_2##.
 

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