Confusing definition of electrical potential energy

[vcsharp2003]

Homework Statement

I came across the following definition in my book for electrical potential energy, which is highly confusing. The definition is as pasted below.

How can the charge move through a displacement if external force is equal and opposite to the electrical force i.e. qE and -qE forces when applied to a charge will not make it move but produce equilibrium?

Only if the applied force is -(qE +f) where f is some arbitrary small force will the charge move and have some displacement.

Relevant Equations

$\vec F = q \vec E$ which gives us the force on a charge q placed at a point in an electric field

$W = \vec F \cdot \vec s$ which is the work done by a force

If the book had said that electrical potential energy is the negative of work done by electrical force on a charge, then the definition would be very clear and easy to understand. So, why should the book give this confusing definition instead.

 

[PeroK]

It doesn't look particularly confusing to me. Considering an external force simply adds some justification to the conclusion. If forces are balanced there is no change in the KE of a particle; the particle need not be at rest to begin with.

 

[vcsharp2003]

It doesn't look particularly confusing to me. Considering an external force simply adds some justification to the conclusion. If forces are balanced there is no change in the KE of a particle; the particle need not be at rest to begin with.

Only if the charge was not at rest at start would this definition be true. So, we are assuming that the particle is not at rest at its initial position of A?

 

[vcsharp2003]

It doesn't look particularly confusing to me. Considering an external force simply adds some justification to the conclusion. If forces are balanced there is no change in the KE of a particle; the particle need not be at rest to begin with.

If we need to lift our mobile phone from its rest position A to a higher position B with zero change in its KE, we would need to exert an upward force on the mobile phone to lift it slowly from A to B and make sure its again at rest at B. In this scenario, the palm of our hand lifting the mobile phone must exert a variable force on mobile phone so it moves up from A to B, such that the variable force is greater than the weight of the mobile phone during some parts of this journey from A to B.

 

[PeroK]

If we need to lift our mobile phone from its rest position A to a higher position B with zero change in its KE, we would need to exert an upward force on the mobile phone to lift it slowly from A to B and make sure its again at rest at B. In this scenario, the palm of our hand lifting the mobile phone must exert a variable force on mobile phone so it moves up from A to B, such that the variable force is greater than the weight of the mobile phone during some parts of this journey from A to B.

That's irrelevant. If an elevator is moving up at a certain speed, then it will continue to move up at that speed under balanced forces.

There is no reason to assume that a particle must start and stop at rest.

 

[vcsharp2003]

That's irrelevant. If an elevator is moving up at a certain speed, then it will continue to move up at that speed under balanced forces.

There is no reason to assume that a particle must start and stop at rest.

I understand the elevator example when it has an initial velocity. But my example was for the scenario when object in question was at rest initially. What if the elevator was at rest initially and needed to be at rest finally?

Sorry, but your explanation adds more to the confusion since the original definition posted is not really bothering about whether the charge was at rest or in motion initially.

 

[vcsharp2003]

Actually, the definition mentioned in the question is implicitly assuming that the charge is at rest initially, because it clearly says "if you want to move the charge...".

 

[vcsharp2003]

After thinking through this, I have come up with following reasoning, if we apply work energy theorem to moving a charge from A to B in an electric field.

$$W_{net}= \Delta {K} $$
$$W_{ex} + W_{el}= \Delta {K} $$
If there is no change in kinetic energy as charge is moved from A to B, then we have
$$W_{ex} + W_{el}= 0$$
$$\therefore W_{ex} = -W_{el}$$
From above relationship, we can surely say that work done by external force on the charge $W_{ex}$ is always going to be negative of work done by electrostatic force $W_{el}$ irrespective of whether the external force is going to be constant or varying as charge gets moved from A to B in an electric field or whether the charge has some very small velocity during its motion from A to B. But, the kinetic energy at A must equal the kinetic energy at B when we define electrical potential energy.

If work done by external force is the change in electrical potential energy $\Delta U$, according to the definition mentioned in question, then
$$\Delta U = W_{ex} = -W_{el} = - \int_A^B q \vec E \cdot \vec {ds} $$

 

[Steve4Physics]

On a separate issue I believe there’s a mistake in the textbook definition (equation 22.1 in the Post #1 attachment).

A change in a quantity is (final value) - (initial value). It is incorrect to write $\Delta U = U_A – U_B$.

It should say:
$\Delta U = U_B – U_A = \int_A^B \vec F \cdot \vec {ds} = - \int_A^B q \vec E \cdot \vec {ds}$

 

[robphy]

I would write it this way, highlighting the chain of reasoning,
$$\begin{align*} U_B - U_A \stackrel{\Delta}{\ \equiv\ } \Delta U \stackrel{\rm{cons} }{\ \equiv\ } -W_{Elec} \stackrel{ W_{E}}{\ \equiv\ } -\int_A^B \vec F_{Elec} \cdot \vec {ds} \stackrel{eqbm}{\ =\ }-\int_A^B (-\vec F_{you}) \cdot \vec {ds} \equiv W_{you} \end{align*}$$

In some intro textbooks, they use a similar balancing force to quasi-statically (i.e., "maintaining equilibrium with $\Delta K\approx 0$") deform a spring to store elastic energy.
It's annoying because it seems to be done to "explain" [or hide?] the minus sign in the real definition of
the change-in-potential-energy as minus-the-work-done-by-the-conservative-force.

To me, the minus sign is there so that we can add
the change-in-potential-energy to the change-in-kinetic-energy [in the Work-Energy Theorem]
and write
the change-in-total-mechanical-energy (which is now equal to the work-done-by-nonconservative forces).

 

[vcsharp2003]

On a separate issue I believe there’s a mistake in the textbook definition (equation 22.1 in the Post #1 attachment).

A change in a quantity is (final value) - (initial value). It is incorrect to write $\Delta U = U_A – U_B$.

It should say:
$\Delta U = U_B – U_A = \int_A^B \vec F \cdot \vec {ds} = - \int_A^B q \vec E \cdot \vec {ds}$

Yes, that's another issue I never noticed.

Also, I think saying external force is equal and opposite to electrical force as charge moves from A to B is also incorrect. It should have mentioned that external force (without stating it's magnitude) moves a charge from A to B such that kinetic energy at A and B are equal.

 

[vcsharp2003]

I would write it this way, highlighting the chain of reasoning,
$\begin{align*} U_B - U_A \stackrel{\Delta}{\ \equiv\ } \Delta U \stackrel{\rm{cons} }{\ \equiv\ } -W_{Elec} \stackrel{ W_{E}}{\ \equiv\ } -\int_A^B \vec F_{Elec} \cdot \vec {ds} \stackrel{eqbm}{\ =\ }-\int_A^B (-\vec F_{you}) \cdot \vec {ds} \equiv W_{you} \end{align*}$

Could you please explain what is meant by the terms $cons$, $W_E$ and $eqbm$ in your mathematical equation?

 

[robphy]

“Definition of conservative force (there exists a function U, such that …)”
“Work done by the electric field”
“Equilibrium (zero net force)”
I should have put $W_{you}$ on the last one for the definition of the “work done by you”

I needed short descriptions to fit over the equal signs. This tries to indicate
how the immediate left hand side LHS
of an equals sign is related to
the immediate right hand side (RHS).
Thus, the far-LHS equals the far-RHS when all of these equalities between them hold.

 

[vcsharp2003]

The change in electrical potential energy of an electrostatic system as a charge $q$ is moved in the system's electric field $\vec E$ from A to B is simply the work done by the external force that is moving the charge without any gain in kinetic energy.
Due to work-energy theorem, this external force work would also be the negative of work done by electrostatic force on charge.

$\vec F_{ex}$ need not be equal and opposite to $\vec F_{el}$, however $\vec F_{ex}$ will always be opposite in direction to $\vec F_{el}$.

$\Delta U =U_b-U_a$ $=W_{ex} = \int_A^B \vec F_{ex} \cdot \vec {ds}$ $= -\int_A^B \vec F_{el} \cdot \vec {ds}$
$= -\int_A^B q\vec E \cdot \vec {ds}$

If this external work is +ve then there is an increase in potential energy i.e. $U_b > U_a$ and if -ve then there is a decrease in potential energy i.e. $U_b < U_a$.

The above definition sounds more logical to me.

 

[robphy]

Ok, so if you write

Due to work-energy theorem, this external force work would also be the negative of work done by electrostatic force on charge.

$\vec F_{ex}$ need not be equal and opposite to $\vec F_{el}$, however $\vec F_{ex}$ will always be opposite in direction to $\vec F_{el}$.

$\Delta U =U_b-U_a$ $=W_{ex} = \int_A^B \vec F_{ex} \cdot \vec {ds}$ $= -\int_A^B \vec F_{el} \cdot \vec {ds}$
$= -\int_A^B q\vec E \cdot \vec {ds}$

then it would be good to explicitly say that
$\int_A^B \vec F_{ex} \cdot \vec {ds}= -\int_A^B \vec F_{el} \cdot \vec {ds}$
holds even though $\vec F_{ex} \neq \vec F_{el}$.
Their works are equal because there is no net-change in the kinetic energy somehow
(like for example if one of the vectors had a component perpendicular to $d\vec s$).
So, yeah, I guess you could generalize it.
With my methods, I would decorate that equal sign with something conveying the above.

To me, the use of external work in this setup distracts from
the main idea that
the primary physical notion of
potential energy is associated with a conservative force,
which,
secondarily, we can use [take advantage of, use engineering] to store work that an external agent does.
In my opinion,
the external work doesn't deserve the same attention that the work-by-a-conservative-force does.

I would introduce the external agent later, after
first establishing the relation between $\Delta U$ and $W_{Elec}$.

UPDATE: My main point said another way.
An external agent can do work, yes.
But does that imply that this external-agent-work
can be recovered because of a potential-energy-function?
NO.
Think about about friction... no such potential-energy-function for friction.
The potential-energy-function arises because the underlying vector-field is conservative.
(Saying "electrostatic field" already implies the electric vector field is conservative.)

 

[vcsharp2003]

it would be good to explicitly say that
$\int_A^B \vec F_{ex} \cdot \vec {ds}= -\int_A^B \vec F_{el} \cdot \vec {ds}$
holds even though $\vec F_{ex} \neq \vec F_{el}$

Did you mean that when the line integrals are equal, then it does not necessarily mean that $\vec F_{ex} = -\vec F_{el}$?

 

[robphy]

Did you mean that when the line integrals are equal, then it does not mean that the integrands are necessarily equal i.e. $\vec F_{ex} = -\vec F_{el}$?

Yes, exactly.
I was trying to restate, in reverse, your sentence:

$\vec F_{ex}$ need not be equal and opposite to $\vec F_{el}$

However, if one had
"when the line integrals are equal for an arbitrary path,"
then I think in that case, we can claim equality of the integrands.

So, in thinking more about your more general statement,
I think in writing
$\int_A^B \vec F_{ex} \cdot \vec {ds}= -\int_A^B \vec F_{el} \cdot \vec {ds}$ even when $\vec F_{ex} \neq \vec F_{el}$,
there is a fixed set of paths being implicitly assumed.
So, if you only consider (say) paths along the x-axis,
then there could be unequal y-components of $\vec F_{ex}$ and $\vec F_{el}$
(rendering the forces unqeual) but have their works be equal
since those $y$-components do no work along these "$x$-axis paths".

Thus, in general, for arbitrary paths, we need the external force to balance the electric force,
that is, [quasi-static] equilibrium.

 

[vcsharp2003]

However, if one had
"when the line integrals are equal for an arbitrary path,"
then I think in that case, we can claim equality of the integrands.

A line integral with limits is interpreted as area under the integrand graph (if f(x) is integrand then the graph is f(x) against x). So if two line integrals with limits are equal then all we're saying is area of graph within the limits are equal, and areas could be equal even when integrands are not equal. Does that sound right? If yes, then no matter what, the integrands may or may not be equal. Unequal integrands can give the same area between certain limits from A to B.