Definition of potential energy

[gracy]

In my book definition of potential energy difference between two points as the work required to be done by an external force in moving without accelerating charge q from one point to another in electric field of any arbitrary charge configuration.I want to know why without accelerating ?

 

[Nugatory]

In my book definition of potential energy difference between two points as the work required to be done by an external force in moving without accelerating charge q from one point to another in electric field of any arbitrary charge configuration.I want to know why without accelerating ?

Because if you accelerate it, you also change the kinetic energy so that the work you do ends up being equal to the potential energy difference plus the change in kinetic energy.

 

[gracy]

you also change the kinetic energy so that the work you do ends up being equal to the potential energy difference plus the change in kinetic energy.

I did not understand.

 

[Nugatory]

I did not understand.

Consider a charged particle sitting somewhere in an electric field, at rest. Because it is at rest its kinetic energy is zero.
I do some work on it to move it somewhere else.

Case 1: It's at rest when I'm done moving it. This is the no-acceleration case, and the work that I did to move it is the change in the potential energy. There's no change in kinetic energy because it's at rest when I'm done moving it so the kinetic energy is still zero.

Case 2: It's not at rest when I'm done moving it, instead it's moving at speed $v$. This is the case in which the particle did accelerate, and the particle now has kinetic energy $\frac{mv^2}{2}$. The work that I did is the change in the potential energy from moving the particle, PLUS the work that I did to accelerate the particle and increase its kinetic energy.

 

[gracy]

The work that I did is the change in the potential energy from moving the particle, PLUS the work that I did to accelerate the particle and increase its kinetic energy.

I understood now.
It becomes $W= ΔU+ ΔKE$
But As per definition

definition of potential energy difference between two points as the work required to be done by an external force in moving without accelerating charge q from one point to another in electric field of any arbitrary charge configuration

$W=ΔU$
That's why no acceleration.

 

[ehild]

In my book definition of potential energy difference between two points as the work required to be done by an external force in moving without accelerating charge q from one point to another in electric field of any arbitrary charge configuration.I want to know why without accelerating ?

Potential energy is attributed to conservative forces. The work of a conservative force on an object does not depend on the path taken. It depends only on the initial and final positions of the object. Force such forces, a potential energy function is defined, which depends on the position only.
Introducing an external force and zero acceleration into the definition causes lot of confusion. A simpler and less confusing definition of potential energy difference U_B-U_A between two points A and B is the negative work done by the force while the object moves from A to B: U_B-U_A=-W_AB, or the work done by the force is the difference between the initial and final potential energy.
The potential energy at a point A is equal to the work the force does when the object moves from A to the place where the potential is zero. In case of the field of a point charge Q, the potential energy is zero at infinity.

 

[gracy]

The work that I did is the change in the potential energy from moving the particle, PLUS the work that I did to accelerate the particle and increase its kinetic energy.

But work energy theorem says $W$=$ΔKE$

 

[HallsofIvy]

No, it does not. The work done is the change in kinetic energy, \Delta KE, minus the change in potential energy, plus the work done against friction.

(Strictly speaking the "potential energy" only exists for a "conservative" force, in which there is no friction but if you are given a "potential energy" formula, you might have divided the force into a conservative part (as gravitational force), which has a potential energy, and a non-conservative part.)

 

[gracy]

Ok.
As per my understanding
$Wnet$=$ΔKE$
(According to work energy theorem)

$Wconservative$= $-ΔP.E$
$Wnon conservative$=$ΔP.E$+$ΔK.E$
Am I correct?

 

[jbriggs444]

Ok.
As per my understanding
$Wnet$=$ΔKE$
(According to work energy theorem)

$Wconservative$= $-ΔP.E$
$Wnon conservative$=$ΔP.E$+$ΔK.E$
Am I correct?

Equations by themselves do little without a definition of the terms in use.

Yes, the work done by a conservative force on an object moving in its field is equal and opposite to the change in potential energy between the starting point and the ending point of the path through which it moves.

However, the work done by a non-conservative force on an object moving in its field can have nothing whatsoever to do with the change in potential energy between the starting point and the ending point for the simple reason that there is no such thing as potential energy in a non-conservative field.

 

[gracy]

Is it every time applicable
$Wconservative$=$-ΔP.E$=$ΔKE$

 

[jbriggs444]

Is it every time applicable
$Wconservative$=$-P.E$=$KE$

Again, I urge you to define your terms before simply writing down equations. What does "KE" in the above equation denote? Be specific. Do not simply say "Kinetic Energy". The kinetic energy of what?

 

[gracy]

The kinetic energy of what?

Depends on context.That's what I am asking is this applicable always?That is irrespective of any scenario or situations?

 

[gracy]

For example During a free fall there is change in KE of the body in the expense of PE.

 

[nasu]

Is it every time applicable
$Wconservative$=$-P.E$=$KE$

No. This will work only if there are only conservative forces doing the work. So that W is the net work.

 

[jbriggs444]

For example During a free fall there is change in KE of the body in the expense of PE.

Again, I urge you not to reply twice to the same posting. And again, I ask you to define the terms in your equation. "It depends" is not a definition.

What I am looking for is something that relates "W_conservative" to "-P.E." and to "KE"

Is KE, for instance supposed to mean the change in kinetic energy of an object subject [only] to a particular conservative force as it moves between a pair of endpoints? And is P.E. supposed to mean the change in potential energy as that object moves between the same endpoints? And is W_conservative supposed to denote the work done by the conservative force on that object over some path between those same endpoints?

 

[gracy]

Is it every time applicable
Wconservative=−ΔP.E-ΔP.E=ΔKE

No. This will work only if there are only conservative forces doing the work. So that W is the net work.

In my OP there is only conservative force but it says without acceleration that means $ΔKE$=0
But $ΔP.E$ has nonzero magnitude which is equal to work done by conservative (electric)force
So even though it is only conservative force the below equation is not applicable
$W$=$-ΔP.E$ =$ΔKE$
Right?

 

[jbriggs444]

In my OP there is only conservative force but it says without acceleration that means $ΔKE$=0
But $ΔP.E$ has nonzero magnitude which is equal to work done by conservative (electric)force
So even though it is only conservative force the below equation is not applicable
$W$=$-ΔP.E$ =$ΔKE$
Right?

In your original post there are two forces. There is the force from the field whose potential is being evaluated. And there is an external force.

 

[gracy]

Is KE, for instance supposed to mean the change in kinetic energy of an object subject [only] to a particular conservative force as it moves between a pair of endpoints?

Yes

is P.E. supposed to mean the change in potential energy as that object moves between the same endpoints?

Yes.

And is Wconservative supposed to denote the work done by the conservative force on that object over some path between those same endpoints?

Yes.
According to me.You decide this is the
question

A charge s moved in an electric field of a fixed charge distribution from point A to another point B SLOWLY.The work done by external agent in doing so is 100J.

 

[jtbell]

Ok.
As per my understanding
$Wnet$=$ΔKE$
(According to work energy theorem)

OK. Now, what is W_net also equal to? You're lacking one step necessary to get to your last equation below.

$Wconservative$= $-ΔP.E$
$Wnon conservative$=$ΔP.E$+$ΔK.E$
Am I correct?

 

[Chestermiller]

Ok.
As per my understanding
$Wnet$=$ΔKE$
(According to work energy theorem)

$Wconservative$= $-ΔP.E$
$Wnon conservative$=$ΔP.E$+$ΔK.E$
Am I correct?

I don't know. Personally, I wouldn't look at it this way. I would let the math resolve the issue for me. Here's a one dimensional example. Suppose I have a charged particle of mass m and charge q in an electric field E (directed along the positive x axis), and I exert a force F along the x-axis to move the charge from x₁ to x₂. If a write a force balance on the system, I get:
$$m\frac{dv}{dt}=F+qE$$
Since an electric field is conservative, it can be represented as minus the derivative of the potential U. So, I have:
$$m\frac{dv}{dt}=F-q\frac{dU}{dx}$$
If I substitute dt=dx/v into this equation, I get
$$mv\frac{dv}{dx}=F-q\frac{dU}{dx}$$
Integrating this equation from x₁ to x₂ yields
$$\left(m\frac{v^2}{2}\right)_{x_2}-\left(m\frac{v^2}{2}\right)_{x_1}=\int_{x_1}^{x_2}Fdx-q(U(x_2)-U(x_1))$$
The terms on the left hand side of this equation is the change in kinetic energy. The first term on the right hand side is the external work done on the charge. The second term on the right hand side is minus the change in potential energy. So
$$Δ(KE)=W-Δ(PE)$$or
$$W=Δ(PE)+Δ(KE)$$
If the work is done in such a way that the change in kinetic energy is zero, then the work is equal to the change in potential energy.

Chet

 

[gracy]

Wnet also equal to?

$Wnet$=$Wconservative$+$Wnonconservative$

 

[jtbell]

OK, so with that step, and your first two equations, you should be able to verify that your final equation is correct, with a tiny bit of algebra.

 

[gracy]

In your original post there are two forces. There is the force from the field whose potential is being evaluated. And there is an external force.

Oh,yes.Please look at my post 19

 

[gracy]

, you should be able to verify that your final equation is correct,

Are all my equations correct?

 

[gracy]

If the work is done in such a way that the change in kinetic energy is zero, then the work is equal to the change in potential energy.

No matter what the force is conservative or non conservative?

 

[jtbell]

Non-conservative forces don't have potential energy associated with them. Chestermiller's statement applies if there are no non-conservative forces acting on the object; that is, if the only forces are conservative.

 

[gracy]

Chestermiller's statement applies if there are no non-conservative forces acting on the object.

so it's only for conservative forces,right?

 

[jtbell]

Right.

 

[gracy]

So conservative forces can contribute to kinetic energy also?
$Wconservative$ΔPE$+$ΔKE##
Is the formula above correct?

 

[jbriggs444]

A charge s moved in an electric field of a fixed charge distribution from point A to another point B SLOWLY.The work done by external agent in doing so is 100J.

The electric field results from a fixed charge distribution. That means that the electric field will be conservative. Only two forces seem relevant -- the force on the charge from the electric field and the unspecified external force.

Since the charge is moved slowly, we can neglect kinetic energy. We can also neglect concerns about magnetism or electromagnetic radiation. In order to maintain the condition of slow movement, the external force must be at all times equal and opposite to the electric field. [Though we may need a nudge to get things started]

If the external agent does 100J of work and no kinetic energy is gained then there must be an increase of 100J in potential energy in the electric field.

100 Joules of potential difference for a charge of s coulombs. You should be able to figure out how many volts that is.

 

[Chestermiller]

No matter what the force is conservative or non conservative?

Please tell me your definition of a "conservative force."

Chet

 

[gracy]

definition of a "conservative force."

conservative force is a type f fprce which has potential energy associated with it and it is path independent.
please tell me is my post 30 correct

 

[jtbell]

So conservative forces can contribute to kinetic energy also?
$Wconservative$ΔPE$+$ΔKE##
Is the formula above correct?

Something seems to be missing from that equation.

 

[Chestermiller]

conservative force is a type f fprce which has potential energy associated with it and it is path independent.
please tell me is my post 30 correct

According to your definition, a conservative force is one that is controlled in such a way that it differs only slightly from the force of the electric field over the entire path, such that over the entire path, the particle is only slightly removed from always being in equilibrium. Under these circumstances, the changes in kinetic energy along the path are negligible, and the work done by the force will be equal to the change in potential energy at every point along the path.

 

[gracy]

Something seems to be missing from that equation.

what?please guide

 

[jtbell]

Where's the = sign?

 

[gracy]

Wconservative=ΔPE+ΔKE##

 

[jtbell]

That doesn't agree with the second equation in your post #9.

I advise you to never use W_conservative and ΔPE in the same equation, except this one: W_conservative = -ΔPE, which gives the relationship between them.

In effect, we use ΔPE and the concept of PE in general, as a replacement for W_conservative.

 

[gracy]

Please tell me if there is only conservative force present i.e $Wconservative$=$Wnet$
in such conditions the below formula is always applicable
$Wconservative$=$-ΔPE$=$ΔKE$

 

[jbriggs444]

Please tell me if there is only conservative force present i.e $Wconservative$=$Wnet$
in such conditions the below formula is always applicable
$Wconservative$=$-ΔPE$=$ΔKE$

If you slide down a hill on a sled with rusty runners do you have the same kinetic energy as when you slide down on one with freshly waxed runners?

 

[gracy]

If you slide down a hill on a sled with rusty runners do you have the same kinetic energy as when you slide down on one with freshly waxed runners?

I THINK, NO.

 

[jbriggs444]

I THINK, NO.

I apologize. My snarky comment had ignored your caveats that no non-conservative force was involved. So no rusty runners for your sled. And the equation does hold.

 

[gracy]

in such conditions the below formula is always applicable
WconservativeWconservative=−ΔPE-ΔPE=ΔKE

Mark (notice) my word "always"
is your answer still yes?

 

[gracy]

Please answer

 

[gracy]

However, the work done by a non-conservative force on an object moving in its field can have nothing whatsoever to do with the change in potential energy between the starting point and the ending point for the simple reason that there is no such thing as potential energy in a non-conservative field.

But
$Wnet$=$Wconservative$+$Wnonconservative$
$Wnet$=$ΔKE$
(WORK ENERGY THEOREM)
$Wconservative$=$-ΔPE$
(definition of potential energy)
hence we can derive from above equations
$Wnnconservative$=$Wnet$ - $Wconservative$
=$ΔKE$ - ($-ΔPE$)
=$ΔKE$+$ΔPE$
so there seems to be relation between Work due to non- conservative forces and potential energy atleast mathematically (as we know mathematics is language of physics)

 

[jtbell]

so there seems to be relation between Work due to non- conservative forces and potential energy atleast mathematically

An object's PE depends only its location. For example, gravitational PE depends only on elevation (height). If we fix the starting and ending locations, then ΔPE is also fixed, and changing W_{non-conservative} can change only ΔKE. I think that's the kind of situation that jbriggs444 is thinking about.

However, if we keep the starting location fixed but allow the ending point to vary, then by changing W_{non-conservative} we can also change the ending location. If we do more work on a ball while lifting it up in the air, we can raise it up further (increasing its ΔPE).

 

[gracy]

$Wnc$=$ΔPE$+$ΔKE$
Is only applicable when starting and /or ending locations are not fixed,right?

 

[jtbell]

No, it applies regardless. Can you describe a specific situation in which you think it doesn't apply?

 

[gracy]

Can you describe a specific situation in which you think it doesn't apply?

No.