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Electrostatic Potential Concept

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formal
#19
Aug7-11, 09:15 AM
P: 27
Quote Quote by Studiot View Post
Potential energy in any force system is independent of time


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

http://www.physicsforums.com/showthr...ight=potential

.
thank you for the link.
but the point is: is the definition given in the thread ,(you said it is correct), just a formal definition, meaning nothing in concrete.? or we can really use a test charge to measure Δ PE? or if we moved the two conductors of a capacitor?

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, independently of its velocity. Is that correct?
Dickfore
#20
Aug7-11, 09:20 AM
P: 3,014
It is a formal definition. Potential energy is not operationally defined physical quantity, but a derived quantity.
formal
#21
Aug7-11, 09:24 AM
P: 27
Quote Quote by Dickfore View Post
It is a formal definition.
Potential energy is not operationally defined physical quantity, but a derived quantity.
Thank you!
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?)
Dickfore
#22
Aug7-11, 09:27 AM
P: 3,014
Quote Quote by formal View Post
Now could you comment the two examples I made.
Is KE = 2.25x 10 ^7/ 12? what is final v?
I can, but I won't.

Quote Quote by formal View Post
And in a synchrotron why is KE increase the same at every passage?
Because it always passes through the same potential difference between the duants.
formal
#23
Aug7-11, 09:31 AM
P: 27
Quote Quote by Dickfore View Post
Because it always passes through the same potential difference between the duants.
That is why I asked for your comment on the other case.
In that case, in vacuum, an electron at rest gets less KE than an electron with v>0

Where is the difference?
Dickfore
#24
Aug7-11, 09:34 AM
P: 3,014
The difference is that moving an electron in the field of another point charge has nothing to do with the concept of a cyclotron.
formal
#25
Aug7-11, 10:40 AM
P: 27
Quote Quote by rbnvrw View Post
This is because you are dealing with electrostatics. If the charge is moving rapidly, you are dealing with electrodynamics and magnetic fields. I don't really know what exactly is slow enough to be described by electrostatic theorems, though.
Thank you,
could you, please , explain how in the example at post #18, a rapidly moving charge's magnetic field would influence work done on it?
Naty1
#26
Aug7-11, 11:09 AM
P: 5,632
Going back to the OP:

This is the answer you are looking for...

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 none of the work done against the electric force is converted to kinetic energy - it is all stored as electric potential energy.
This condition gives validity to the calculation described above.
It's just an operational definition for convenience. The reference could have been chosen as -100 volts at infinity; very often in circuit problems earth is assigned "zero" volts as the reference potential...
formal
#27
Aug7-11, 11:16 AM
P: 27
Quote Quote by Naty1 View Post
It's just an operational definition for convenience. The reference could have been chosen as -100 volts at infinity; very often in circuit problems earth is assigned "zero" volts as the reference potential...
Thank you,
the value at infinity is really irrelevant.
The point is the remainder of the definition:

is velocity important or not? if it is not , why bother?
if it is (as it figures), in what way is it relevant? is it a matter of MF?

please see previous post #25, and also 18,19
Naty1
#28
Aug7-11, 11:35 AM
P: 5,632
is velocity important or not? if it is not , why bother?
if it is (as it figures), in what way is it relevant?
It is important NOT to change the velocity of the test particle much. That's because in the definition of potential energy you want to measure the work done against the static potential...NOT the work done to accelerate the particle related to kinetic energy...which must be kept small by comparison.

The work done in the static case is not dependent on the path taken.

If you move a test particle with a positive charge to closer proximity with a positvely charged particle it takes positive work....If of opposite charge, negative work....negative potential results...analogous to a (attractive) gravitational field where particles come closer together.
formal
#29
Aug7-11, 11:53 AM
P: 27
Quote Quote by Naty1 View Post
1) It is important NOT to change the velocity of the test particle much.
2)The work done in the static case is not dependent on the path taken.

3)If of opposite charge, negative work....negative potential results...analogous to a (attractive) gravitational field where particles come closer together.
I think there is a misunderstanding, probably I haven't made it clear enough
I am not talking of (1) changing v or (2) changing path.
I am talking of absolute value of v, I stated it repeatedly

if a charge as v= 1, 10, 100 , does it make any difference?
is post #6 correct or false?

3) if a body in a gravitational field is at rest or has v = 1 or = 10 it matters! and how!
the greater is v, the greater is KE acquired (work done)
a charge in vacuum behaves differently from the same charge between two Dee's ? why?
DeltaČ
#30
Aug7-11, 12:18 PM
P: 450
Quote Quote by formal View Post
but I surely know that the longer you expose anything to a force the greater is the KE acquired. Think about gravity!
No this is not true either. As long as the field is not time- varying the force that does the work will not be time-varying either so the work done by the force as defined by

[tex] \int_{C} F(r)dr[/tex] will not depend on time, it will depend only on the path C (and not how long it takes to travel through that path). Moreover since the electrostatic field is conservative it will depend only on the start and end points of the path C.
Dadface
#31
Aug7-11, 12:42 PM
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P: 2,031
There is a simple high school analogy.If you lift a mass M through a height h against a uniform gravitational field of field strength g then the gain of gravitational potential energy is Mgh.This change is dependant on M and g and the initial and final positions only and is independant of the method of lifting.If the mass is lifted infinitely slowly at an angle the change is Mgh.If it is fired vertically up at a billion metres per second then as the mass pases through a height h the gain of PE is again Mgh.Whatever method is used energy is conserved.
thebiggerbang
#32
Aug7-11, 12:51 PM
P: 70
OP here. I guess that the velocity of the charge that is being moved should tend to zero (vanishingly small) as if it is not so, then it implies that at some point of time the applied force was more than the force due to the electric field, there was no equilibrium.

Also, as we can't calculate the absolute energy of a system, we always consider the change in the energy of the system. For all practical purposes, we always use potential difference, and not the absolute potential. This difference will be always independent of the initial position of the charge (infinity on this case).

Mathematically, Potential difference to move a charge from A to B is given by

[itex]\Delta[/itex]P=P[itex]_{AB}[/itex]=P[itex]_{\infty B}[/itex]-P[itex]_{\infty A}[/itex]
=P[itex]_{B}[/itex]-P[itex]_{\infty}[/itex]-P[itex]_{A}[/itex]+P[itex]_{\infty}[/itex]
=P[itex]_{B}[/itex]-P[itex]_{A}[/itex]

I hope the formulation is correct
Studiot
#33
Aug7-11, 01:14 PM
P: 5,462
OP here. I guess that the velocity of the charge that is being moved should tend to zero (vanishingly small) as if it is not so, then it implies that at some point of time the applied force was more than the force due to the electric field, there was no equilibrium.
Not exactly but almost.

Yes, as I said before, it is way of stating that none of the electric potential energy is converted to kinetic energy or that the electric force does not speed up the test charge.

However some other agent may be causing motion,independent of the electric effect.

Consider.

I float up to 10000ft in a balloon and stay there in equilibrium.

The balloon and contents have a certain PE due to the altitude..

If I now start the motor and propel the balloon horizontally at 1 mile/hr does the PE change if the altitude remains constant?

If I accelerate the balloon to 10 miles/hr does this change the PE?
formal
#34
Aug8-11, 12:32 AM
P: 27
Quote Quote by Studiot View Post
1) it is way of stating that none of the electric potential energy is converted to kinetic energy or that the electric force does not speed up the test charge.

Consider.
I float up to 10000ft in a balloon and stay there in equilibrium.
The balloon and contents have a certain PE due to the altitude..

2) If I now start the motor and propel the balloon horizontally at 1 mile/hr does the PE change if the altitude remains constant?
3) If I accelerate the balloon to 10 miles/hr does this change the PE?
1) (EB) the explanation one reliable text gives for the motion being slow is:
"...this is the only case in which motion does not, of itself, cause work to be done elsewhere in the universe" ...." the vector curl E (del x E) must be zero"


2) , 3) If we move (a balloon or) a charge (horizontally or) in a normal direction to force E, PE , of course, does not change because work is not done against the force (as r0 -r = 0)(4*)

( 4*) text says work: W = q(o) * q/ 4π ε0 (1/ r0 -1/ r ) )

now if definition EB is correct, could you or someone help to interpret it?
in your previous post (#17) you interpret slowly as steadily, but if they wanted to mean steady they would just say steady.
If it is so, (and slow does not mean 'not accelerated') the absolute value of v is relevant:
OP pinpoints his previous 'without acceleration' to 'vanishingly small'. That is correct!That is what the definition is all about!

But the best way to explain it is to say what happens if v is greater, I suppose!
everyone has his own view: is post #6 correct? is post #24 correct? is post #20 correct?
Now consider this:

2,3) if you deflate your (balloon A) charge q(a) (if q(o)= 0.0000184 C)
will drop vertically and after 1 second it will gain
acc= v = 9.8 m/sec and KE = W(A)
if another (balloon B) charge q(b) is already (falling) moving alongside it at v 9.8 m/ sec it will change
its v from 9.8 to 19.6 m/ sec gaining KE W(B) > W(A) (= 4 W(A)) , while
in a cyclotron W(A)= W(B)=..W(C)...

In conclusion we have a (formal?, hypothetical?, meaningless?) definition of Electrostatic PE which states
that absolute value of v is relevant, whereas (as stated correctly in post #31)
it is not at all relevant in a gravitational field and
it is not at all relevant in an Electric field between two Dee's
Dadface
#35
Aug8-11, 04:01 AM
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I think the main reason the OP got confused is that he assumed,incorrectly, that there was an implication that the work done in moving from A to B is independant of the route taken and the method used to do that work.The key point is that it is the work done on (or by) the field that is independant of the route taken and the method used to do that work.
Studiot
#36
Aug8-11, 06:06 AM
P: 5,462
Good morning, formal.

I am not sure what you are trying to achieve here.

For the purposes of the definition of electric potential the universe comprises a single point charge and the vector field (E)surrounding it.

What do you know about curl(E) in relation to velocity or time?


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