# Electrostatic Potential Concept

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...

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

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.

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?

Delta2
Homework Helper
Gold Member
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

$$\int_{C} F(r)dr$$ 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.

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.

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

$\Delta$P=P$_{AB}$=P$_{\infty B}$-P$_{\infty A}$
=P$_{B}$-P$_{\infty}$-P$_{A}$+P$_{\infty}$
=P$_{B}$-P$_{A}$

I hope the formulation is correct

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?

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?
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

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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.

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?

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

the work done on the particle doesnt depend on its velocity )
That's exactly what I thought, too. Now, if you still think so, could you explain why
......the velocity of the charge must be vanishingly small?

Why can't we just fire a charge up a billion m/s (as dadface says) and measure E-PE?
Do you think EB is incorrect? can you improve or explain it?

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

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