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So, as it says that there will be no acceleration, does it imply that there will be no change in the velocity of the charge that we are moving? And thus, does this imply that there is no change in the kinetic energy of the charge?

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- #1

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So, as it says that there will be no acceleration, does it imply that there will be no change in the velocity of the charge that we are moving? And thus, does this imply that there is no change in the kinetic energy of the charge?

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Yes that's correct.So, as it says that there will be no acceleration, does it imply that there will be no change in the velocity of the charge that we are moving? And thus, does this imply that there is no change in the kinetic energy of the charge?

Another way to put this is to say that for its entire journey from infinity to its location, the charge is in equlibrium or passes through a series of equilibrium states.

go well

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And thus, does this imply that there isno change in the kinetic energyof the charge?

Is the absolute value of kynetic energy not relevant?, any speed will do, as long as it remains constant?Yes that's correct.

In many textbooks I read

What is slowly?

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My point is that (KE), velocity cannot be leftThis is because you are dealing with electrostatics. ... I don't really know what exactly is slow enough to be described by electrostatic theorems, though.

because if v is half, then time t is double.

Then also total

So what is the correct Electrostatic Potential Energy?

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so, what is E-PE?

it seems we cannot calculate it!

b)And what if you don't start from infinity?

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But, look at the definition you quoted. It says taken from infinity.And what if you don't start from infinity?

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1)Electrostatic potential is the work doneon a unit charge to bring itfrom infinity to a pointfrom a given charge without acceleration, against the electric field presend due to the given charge.

2)Yes that's correct.

That's my quote 1 and 2:But, look at the definition you quoted.

It says taken from infinity.

how do we calculate E-PE from infinity to a point if time is infinite?

b) if we do not start from infinity, the absolute value of velocity is not relevant?

again, what is the threshold?

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You are doing a mistake here, the work done on the particle doesnt depend on its velocity neither on the time it takes, because the force that does this work doesnt depend on velocity(the electric field is only due to the given charge which is considered stationary, while we consider the field from the moving unit charge negligible) neither on time (it is constant with time)My point is that (KE), velocity cannot be leftvague.

because if v is half, then time t is double.

Then also totalwork done on charge(PE) is double, or more

So what is the correct Electrostatic Potential Energy?

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what does this have to do with anything?That's my quote 1 and 2:

how do we calculate E-PE

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I do not know about v, that's why I was asking,the work done on the particle doesnt depend on its velocity neither on the time it takes, because the force that does this work doesnt depend on velocity)

but I surely know that

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If one says that PE is calculated bringing a test charge from infinity to anywhere andwhat does this have to do with anything?

the time required to do this is infinite, then he is saying simply we cannot calculate it, at any speed.

Am I wrong?

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Yes. The work done by a conservative force is equal to the difference between the initial potential energy and the final potential energy:If one says that PE is calculated bringing a test charge from infinity to anywhere and

the time required to do this is infinite, then he is saying simply we cannot calculate it.

Am I wrong?

[tex]

W_{\mathrm{cons}} = (E_{p})_{i} - (E_{p})_{f}

[/tex]

The work-energy theroem tells us that the total work done on an object is equal to the change in kinetic energy. In our case, since there is no acceleration, the velocity of the object remains the same, therefore the change in kinetic energy is zero, regardless of the speed of the object.

Furthermore, there are two forces acting on the object at any time. The electrostatic force (no Lorentz force since the speed of the object is infinitely small) and the external force that counters it. Therefore, the work-energy theorem gives:

[tex]

W_{\mathrm{ext}} + W_{\mathrm{cons}} = 0

[/tex]

Substituting for the work done by the conservative (electrostatic) force:

[tex]

W_{\mathrm{ext}} + \left((E_{p})_{i} - (E_{p})_{f}\right) = 0

[/tex]

Solving for the final potential energy:

[tex]

(E_{p})_{f} = (E_{p})_{i} + W_{\mathrm{ext}}

[/tex]

Now, we choose the normalization that [itex](E_{p})_{i} = (E_{p})_{\infty} = 0[/itex]. Then:

[tex]

(E_{p})_{f} = W_{\mathrm{ext}}

[/tex]

This is the mathematical formulation of the sentence stated in the OP.

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Potential energy in any force system is independent of time and dependant solely on position.Is the absolute value of kynetic energy not relevant?, any speed will do, as long as it remains constant?

In many textbooks I read slowly

What is slowly?

That includes gravity.

We are discussing electric potential enrgy here.

The definition is specifically worded to run from infinity to the position because we cannot calculate the work of separating charges.

The calculation is easy and shown on post 40 of this thread

https://www.physicsforums.com/showthread.php?t=489731&highlight=potential

Your book says 'slowly' instead of 'equilibrium' or 'without acceleration' I expect. They are all equivalent statements.

The represent the desired condition that as the charge is brought in from infinity

This condition gives validity to the calculation described above.

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This is the maths formulation alright, and so I suppose it applies also to gravitation.Yes.

This is the mathematical formulation of the sentence stated in the OP.

and I suppose it applies also in the case we do not start from infinity. (point b)

Imagine we have in vacuum 55200 esu,0.0000148 C ,positive charge ( 4x 10^14 at 1m).

An electron is at rest at r0= 10^7m/ (^9 cm.) (acc= 4 m/sec^2, 400cm/s^2).

If we move it to r = 6.4x 10^6m/ (^8cm) (acc= 9.8 m/s^2 ,980cm/s^2)

is work done on the charge the same if v changes? or is it useless and we need only Maths and

KE= Δ PE = 2.25 x 10^ 7 ?

That is to say the same KE an electron would get

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thank you for the link.Potential energy in any force system is independent of time

The calculation is easy and shown on post 40 of this thread

https://www.physicsforums.com/showthread.php?t=489731&highlight=potential

.

but the point is: is the definition given in the thread ,(you said it is correct), just a

In a cyclotron you have a set difference in E-PE, and every time a charge makes a jump from a Dee it gets the same amount of KE,

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Thank you!

Potential energy is not operationally defined physical quantity, but a derived quantity.

That is just what worried me.It really did not make much sense.

Now could you comment the two examples I made.

Is KE = 2.25x 10 ^7 J, final v= 1.8x 10^ 4 m/s, 18 Km/ sec if electron is at rest ? and much more if it is not?

And in a cyclotron why the increase in KE is the same at every passage even if speed is always different?

(P.S. should we say derived or derivative quantity?)

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I can, but I won't.Now could you comment the two examples I made.

Is KE = 2.25x 10 ^7/ 12? what is final v?

Because it always passes through the same potential difference between the duants.And in a synchrotron why is KE increase the same at every passage?

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That is why I asked for your comment on the other case.Because it always passes through the same potential difference between the duants.

In that case, in vacuum, an electron at rest gets less KE than an electron with v>0

Where is the difference?

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Thank you,

could you, please , explain how in the example at post #18, a rapidly moving charge's magnetic field would influence

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