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- Thread starter thebiggerbang
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- #31

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- #32

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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

- #33

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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?

- #34

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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

"...this is the only case in which motion does not, of itself, cause work to be done elsewhere in the universe" ...." the vector curl

2) , 3) If we move (a balloon or) a charge (horizontally or) in a normal direction to force E,

(

now if definition

in your previous post (#17) you interpret

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)

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?,

that

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

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- #35

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- #36

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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 (

What do you know about curl(

- #37

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I am afraid that I have a really difficult time communicating with you.

*(these words were the welcome to the forum)*

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- #38

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I know my basics very well, that it is a conservative force :)

- #39

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the work done on the particle doesnt depend on its velocity )

That's exactly what

......the velocity of the charge

Why can't we just

Do you think EB is incorrect? can you improve or explain it?

THAT is what the thread is all about!

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- #40

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hi tBB,...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.

(if this is not it) ..could you please quote the explanation given in your textbook?

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1) It is important NOT to change the velocity .

2) If of opposite charge, negative work....negative potential results...

3)analogous to a (attractive) gravitational field where particles come closer together.

Hi Naty,

suppose we never change the velocity, but keep it steady at , say 20 cm/ sec, how would this v affect the measurement of PE?

2)if we assume a positive charge at the origin, we must move an electron from r= to infinity,right?

3) I assume a positive charge to make comparison betwwen gravitational vs Coulomb force easier.

Thanks

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