2
$\begingroup$

Electric Potential definition is as follows:

Electric potential (also called the electric field potential, potential drop, the electrostatic potential) is defined as the amount of work energy needed per unit of electric charge to move the charge from a reference point to a specific point in an electric field

But

its just a scalar field such that its gradient gives us the original vector field ,that is, the original electric field

Now , The electrostatic potential energy, U, of one point charge q at position r in the presence of an electric potential $\Phi$ is defined as the product of the charge and the electric potential

What does Electric Potential energy represent ,then ?

  1. Is Electric Potential energy just a Electric potential scalar field ,but w.r.t a charge Q, not a unit charge ,whose gradient then represent a Electric vector field that tells us the amount of force a Charge Q ,not a unit charge ,would experience and in what direction ? Is this interpretation correct ?

  2. If yes ,then can we have Electric field with respect to a non-unit charge ,and if we can ,then would be take into account the electric field of that other charge Q as its not a unit charge thus its electric force may not have negligible effect ?

$\endgroup$
1
  • $\begingroup$ The electric potential phi and electric field E are defined for a unit charge, so that the potential energy and force on the test charge q are just q*phi and q*E. Since q does not depend upon position it is true that q*grad(phi) = grad(q*phi), but there is no need to think of q*phi as some "different" field" $\endgroup$
    – Malcolm
    Jan 9 at 14:51

4 Answers 4

1
$\begingroup$

The ideas are closely related.

One thing about potential energy and potential difference is that they tell you about the difference in energy between two positions.

Potential energy of a charged object is the work done by electric forces to move from one position to the other.

Potential (or Voltage) is the work done by electric forces to move a unit charge from one position to the other. Depending on your units, a unit charge may be an electron, in which case energy is measured in eV or electron-Volts. Or it may be a Coulomb, in which case energy is in Joules.

Potential looks different when applied to charges moving in fields in a vacuum and charges moving in a circuit, but it is the same thing.

Often people hide the idea that potential or potential energy are about the difference between two points. They pick a point and choose the potential energy or potential to be $0$ at that point. Then they talk about the potential energy or potential at any other point. This is much like choosing gravitational potential energy to be the $0$ at $0$ altitude, where $0$ altitude might be a table top, sea level, the center of the Earth, or some other convenient point. For a circuit, this point is often the negative terminal of a battery. For charges in a field, an infinitely distant point often turns out to be convenient.

Once you do this, you have a potential scalar field whose gradient gives the vector E field.


Potential energy arises when you have a conservative force field. This is a region of space where you know the force on an object if you know where the object is. In particular, it means you don't need to know velocity. This excludes forces like friction.

In this case, kinetic energy grows and shrinks as the forces push the object around. If an object is pushed one way between two points, the kinetic energy grows by the amount of work done. If the object travels the other way, the same amount of kinetic energy is lost. You see this all the time with gravity.

$$\Delta KE = \int_a^b \vec F \cdot d \vec x$$

Potential energy is defined so that energy is conserved. The amount of kinetic energy gained is the same as the potential energy lost and vice verse.

$\endgroup$
4
  • $\begingroup$ This doesn't really explain what potential energy is ,if potential is scalar field such that its gradient is the original electric vector field ,then what does potential energy represent? $\endgroup$
    – Anuj
    Jan 9 at 17:20
  • $\begingroup$ I added to the answer. But for more on what energy is, see How is energy "stored in an electric field"? $\endgroup$
    – mmesser314
    Jan 9 at 18:27
  • $\begingroup$ "Potential energy arises when you have a conservative force field. This is a region of space where you know the force on an object if you know where the object is. In particular, it means you don't need to know velocity" It still doesn't explain how it differs to potential in a way that both are scalar fields and gradient gives us the original electric field? $\endgroup$
    – Anuj
    Jan 10 at 1:56
  • $\begingroup$ "Potential energy is defined so that energy is conserved. The amount of kinetic energy gained is the same as the potential energy lost." Energy isn't a thing or an object. It isn't real. It is something we define to characterize a system. It is useful because it is conserved - the number doesn't change when a system changes. For more on this, see the Feynman Lectures. Conservation of Energy $\endgroup$
    – mmesser314
    Jan 10 at 2:50
1
$\begingroup$

What does Electric Potential energy represent ,then ?

In electrostatics the electric force is conservative and so the potential energy $U$ represents the same thing that it always represents. Namely, a function of space from which the force can be derived by taking the negative of the derivative: $$ \vec F = -\vec \nabla U\;. $$

Or, to put it another way, in electrostatics the potential energy is something that, when combined with the kinetic energy $T$, is constant. That is, the total energy $T+U$ is a constant of the motion.


We have $$ \vec F = -\vec \nabla U $$ where $\vec F$ is the force due to the external electric field (the electric field sourced by all the charges that are not the test charge $q$): $$ \vec F = q\vec E\;, $$ where $\vec E$ is the external electric field (the electric field sourced by all the charges that are not the test charge $q$) and $q$ is the charge of the particle in the external electric field.

I.e., $$ q\vec E = -\vec \nabla U\;. $$

We write: $$ U = q\Phi $$ such that $$ \vec E = -\vec\nabla\Phi\;. $$


For example, if this conservative external electric field is the only force on the particle, then it is also the net force and we can write: $$ W_{net} = \int \vec F\cdot \vec {dx} = -\Delta U $$ but, since in this example the electric force is the net force, then by the work-energy theorem, the above expression is also equal to the change in kinetic energy $$ W_{net} = \Delta T $$ and thus $$ \Delta (T+U) = 0\;, $$ which says that the total energy $T+U$ of a particle in a conservative force field is constant.


For example, if there is also a conservative gravitational force acting on the particle, described by a potential $U_g$ then, by a similar calculation we have: $$ \Delta(T+U+U_g) = 0\;. $$

So, to put it yet another way, $U$ is the usual potential energy due to the external electric field, which should be additively combined with any other potential energy functions due to any other external forces, and the kinetic energy, in order to get a constant of the motion (the total energy).


Update:

To expand this answer a little more, it might be helpful to remember where the concept of a field came from. All that experiment tells us is that if $\vec r_0$ and $\vec r_1$ are two fixed points in space where two charges, $q_0$ and $q_1$, respectively, are located then the force that $q_0$ feels due to $q_1$ is: $$ \vec F_0 = \frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3}\;. $$

Similarly, if there are charges $q_0$, $q_1$, $q_2$, and $q_3$, the force felt by $q_0$ is: $$ \vec F_0 = \frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3} +\frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3} +\frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3} +\frac{q_0q_1(\vec r_0 - \vec r_1)}{|\vec r_0 - \vec r_1|^3} $$ $$ =q_0\sum_{j=1}^3\frac{q_j(\vec r_0 - \vec r_j)}{|\vec r_0 - \vec r_j|^3}\;, $$ where, we are considering the force on $q_0$, so here I will call $q_0$ the "test charge."

Similarly, if there are $N+1$ charges $q_0$, $q_1$, $q_2$, ..., $q_N$, the force felt by $q_0$ is: $$ \vec F_0=q_0\sum_{j=1}^N\frac{q_j(\vec r_0 - \vec r_j)}{|\vec r_0 - \vec r_j|^3}\;, $$

Clearly, the force on the test charge can be written as a vector field evaluated at $\vec r_0$: $$ \vec F_0 = \vec F(\vec x = \vec r_0)\;, $$ where $$ \vec F(\vec x) = q_0\sum_{j=1}^N\frac{q_j(\vec x - \vec r_j)}{|\vec x - \vec r_j|^3}\;.\tag{A} $$ This seems like a minor change, but the introduction of the field (which is what physicists call a function of space) is important conceptually.

But now our Eq. (A) looks a little funny... it has lost its dependence on $\vec r_0$, but it still depends on $q_0$. This might motivate us to consider a different field, the field in Eq. (A) divided by the test charge. We call this field the "electric field." $$ \vec E(\vec x) = \sum_{j=1}^N\frac{q_j(\vec x - \vec r_j)}{|\vec x - \vec r_j|^3}\;, $$ which doesn't depend on the test charge ($q_0$) or position ($\vec r_0$) explicitly, but does depend on the distribution of all the other "external" charge in the world ($q_1$, $q_2$, ..., $q_N$).

In terms of potentials, we can write the force in Eq. (A) as: $$ F(\vec x) = -\vec \nabla U\;, $$ where $$ U(\vec x) = q_0\sum_{j=1}^N\frac{q_j}{|\vec x - \vec r_j|}+C\;, $$ where $C$ is an arbitrary constant, which we here choose to be $C=0$.

This means that the Work done by the source charges on the test charge $q_0$ as it moves from infinity (infinitely far away from the source charges) to $\vec r_0$ is: $$ W = \int \vec F\cdot\vec{dx} = -\Delta U = -q_0\sum_{j=1}^N\frac{q_j}{|\vec r_0 - \vec r_j|} = -U(\vec r_0)\;. $$ We see that, as usual, the work done by the external force due to the source charges is the negative of the change in potential energy $U$ due to the source charges. Here the test charge $q_0$ again only comes in as a single overall multiple, which perhaps motivates the definition of the electric potential $\Phi$ as: $$ \Phi(\vec x) = \frac{U}{q_0} = \sum_{j=1}^N\frac{q_j}{|\vec x - \vec r_j|}\;. $$ Or, again in terms of the field: $$ \vec F_0 = q_0\vec E_0\;, $$ where $$ E_0 = E(\vec x=\vec r_0) = \sum_{j=1}^N\frac{q_j(\vec r_0 - \vec r_j)}{|\vec x - \vec r_j|^3}\;. $$ Here, we see that the force acting on the test charge $q_0$ is equal to the electric field of all the other charges in the world (not including $q_0$ itself) multiplied by the test charge value $q_0$.

[Question in comments:] "the gradient of electric potential energy... tells us how much force a charge Q, a non unit charge , would experience... what happens to electric field of charge Q itself ,is it considered ?"

The charge $Q$ itself is not considered as one of the sources that created the electric field $\vec E$ acting on $Q$. It is only considered in the sense that you still have to multiply the electric field by a single overall factor of $Q$ to get the force on $Q$.

$\endgroup$
10
  • $\begingroup$ Its just like what i said ,electric potential gradient gives us original electric vector field ,which is force per unit charge , and electric potential energy gradient gives the force between two charges ,is that it ? $\endgroup$
    – Anuj
    Jan 10 at 2:00
  • 1
    $\begingroup$ Yes, that is right. $\endgroup$
    – mmesser314
    Jan 10 at 2:54
  • $\begingroup$ @Anuj, yes, that is pretty much right, except for the sign convention. (The electric potential gradient gives the negative of the electric vector field, by convention. The electric potential energy gradient gives the negative of the force on the test charge, which by the third law, is the force on the source charge, if there is one source charge that created the field.) $\endgroup$
    – hft
    Jan 10 at 3:04
  • $\begingroup$ Can you expand on this ? Like isn't electric field just a vector field that tell us how much force a unit charge would experience at certain distance ,so is electric potential energy like a electric field but w.r.t a non-unit charge . $\endgroup$
    – Anuj
    Jan 10 at 3:41
  • $\begingroup$ @Anuj I will update the answer to expand $\endgroup$
    – hft
    Jan 10 at 4:40
0
$\begingroup$

here is my interpretation: electrostatic potential is "is defined as the amount of work energy needed per unit of electric charge to move the charge from a reference point to a specific point in an electric field" is right, and keep in mind that it is a property of a single point.
electrostatic potential energy is a property of a "system", so it is interpreted as "resulting from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system." or: "the enrgy needed to move the charge Q inside an electric field", because the electric field formula is:
$E=−∇V$ and then $ΔU=qΔV$
so it is not exactly equal to a Electric potential scalar field, but it is related and can be resulted from it.

$\endgroup$
0
$\begingroup$

The potential gives the potential energy divided by the charge. The potential is handy because it only depends on its source, where the potential energy depends on both interacting charges.

Forget ‘work’, it reminds of ‘horsepower’ and other pre mksi stuff.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.