I was thinking about the definition of this 2 concepts and I don't know if I understand exactly what they mean. Electric potential is just the work that must be done to bring a charge from infinity to a region in space with an electric field generated by any mean (for example a point charge) and the electric potential energy is the energy that charge has just because it yields in that electric field. Are this 2 definitions accurate?


5 Answers 5



When a charged particle moves in electric field, the field exerts a force that can do work on the particle. This work can be expressed in terms of potential energy. Just as gravitational energy depends on the height of the mass above the earth's surface, the electric potential energy depends on the position of the charged particle in the electric field when a force $\vec F$ acts on a particle that moves from point a to point b, the work $W_{a\rightarrow b}$ done by the force is given by,

$$W_{a\rightarrow b}=\int \vec F\cdot\vec{\mathrm{d}s}=\int_a^b F \cos\theta \mathrm{d}s$$

where $\vec{\mathrm{d}s}$ is an infinitesimal displacement along the particle's path and $\theta$ is the angle between $\vec F$ and $\vec{\mathrm{d}s}$ at each point along the path.

Second, if the force $\vec F$ is conservative, the work done can be expressed in terms on potential energy. When a particle moves from a point where the potential energy is $U_a$ to a point where it is $U_b$, the change in potential energy is, $\Delta U$=$U_b$- $U_a$. This is related by the work $ W_ {a\rightarrow b}$ as

$$ W_ {a\rightarrow b} = U_a- U_b = -\Delta U\tag{1}$$

Here $ W_{a\rightarrow b}$ is the work done in displacing the particle from a to b by the conservative (here electrostatic) force, not by us. Moreover you can see from eq (1) that if $ W_ {a\rightarrow b}$ is positive, $\Delta U$ is negative and the potential energy decreases. So whenever the work done by the conservative forces is positive, the potential energy of the system decreases and vice-versa. That's what happens when the particle is thrown upwards, the work done by the gravity is negative, and the potential energy increases.


Potential is defined as the potential energy per unit charge.

Electrical potential is the energy per unit charge gained or lost, when a unit positive charge is moved from some reference point, at which the potential is defined to be zero.



To summarise:

Electric potential energy is the energy the body has due to its position in an electric field (the capacity for doing work which arises from position or configuration if you want to get specific). I.e if you have say, an postive charge (+ charge) and you move it near another positive charge, it will want to deflect. If you push it closer to the positive charge, it will want to deflect more. The unit of electrical potential energy unit is the Joule.

Electric potential is the value that we get when we displace a unit charge from a reference point (having potential $0$) to a specific point. It's unit is Joule/charge.

Hope this helps you.

  • $\begingroup$ complete answer. I gave a thought to including some information about the gradient, but that would be getting into the calculus field $\endgroup$
    – JD_PM
    Commented Apr 25, 2018 at 13:06

Your definitions are not quite accurate.

You might think of potential energy as the energy needed to bring a particular charge in from infinity.

You might think of potential as the energy needed to bring a unit charge in from infinity.

Potential is more flexible in the sense that once it's obtained for a unit charge, it can be used to calculated the energy needed to bring a particular charge. This process is conducted by the familiar $$U=qV$$

  • $\begingroup$ Exactly so. The Potentials are a feature of the e-field itself, not a feature of the test-charge, or of the stored energy. Improved definition: the potential is the line-integral of the e-field intensity. (See, no charges involved. Pure e-field.) Or this: the volt is the energy/charge ratio, when the charge approaches zero. We use an infinitesimal test-charge, rather than a one-coulomb unit charge. The energy is zero, while the potentials involved can be immense, and unrelated to test-charge stored energy. $\endgroup$
    – wbeaty
    Commented Sep 14, 2023 at 22:57

Before defining the electric potential, let me describe the electric field. Any space region where a static charge is placed and a force is exerted on is regarded as an electric field. Imagine we have 3 different charges: $q_{1}, q_{2}$ and $q_{3}$ in a region. If a charge $q$ is placed in that region, the force exerted on $q$ is the sum: $F= F_{1}+F_{2}+F_{3}$

As the force that each $q_{i}$ exerts on $q$ is proportional to $q$, $F$ will be proportional to $q$. Therefore:


Note that if the charge $q$ is positive, then the electric field and the electric force have both the same direction. If the charge $q$ is negative, then both the electric field and the electric force have opposed directions. This fact explains why when an electric field is applied to an electrolyte both positive and negative ions have opposite directions.

If $q$ is placed in a region where the electric field is due to $q_{1}$, we have:

$$E= \frac{Kq_{1}}{r^2}\hat{r}$$

When there are many charges in a space region, the total electric field in that region is the sum of each and every single electric field produced by each charge in the region. That is:

$$E=\sum_{i} E_{i}=K\sum_{i} \frac{q_{i}}{r_{i}^2} \hat{r}$$

If in a region there are so many charges, it can be said there is a continuos charge distribution. Therefore the space region can be divided into infinitesimal charge elements $dq$. Besides, if charge density is defined as:

$$Q= \frac{dq}{dv}$$

Where dv is the infinitesimal volume occupied by dq.

Based on what has been described, electric field:

$$E=K\int\frac{Q}{r^2}dv \hat{r}$$

As the electric field is conserved, it can be assured there is a scalar function $V$ called electric potential, which is defined:

$$E=- \nabla V$$

The work done by the forces inside the field on the charge/s along a distance r (from an initial position to a final position) equals to the difference between potencial energy values at an initial position and final position. That is:

$$\int_{i}^{f} Fdr = U_{p}(i) - U_{p}(f)$$

Then, the electric potential can be defined at any point of the electric field as:



According to Wikipedia, an electric potential is "the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing any acceleration.Typically, the reference point is Earth or a point at Infinity, although any point beyond the influence of the electric field charge can be used."

There a couple of points (npi) in this definition worth mentioning. First, the potential is defined for a positive unit charge, not just a charge. Second, the potential at the reference point is defined as zero, which is just a convention. Taking those points into account, we can unambiguously determine an absolute potential, Φ and its sign at any point in space and an absolute potential energy and its sign of a charge q brought to that point, as U=qΦ.

So if the field is created by a positive charge, the work to bring a positive unit charge to that field is positive (i.e., actual work has to be performed against the field) and therefore the potential at any point in this field will be positive as reflected in the formula for the potential created by charge Q, Φ=kQ/r, which says that a positive charge will create a positive potential.

If a positive charge q is brought to that field, it will acquire positive potential energy U=qΦ.

If a negative charge, -q, is brought to such field ("minus" here is used explicitly for illustrative purposes), it'll want to move by itself (yield into the field), so, actually, a negative work would have to be done to keep it from accelerating and, therefore, by definition, its acquired potential energy would be negative, which will be reflected in the formula: U=-qΦ. Or we can say that the charge will lose some potential energy working against the force keeping it from accelerating.

The potential of an electric field created by a negative charge could be treated similarly.

This convention does not violate the principle that the system, left to itself, will try to reduce its potential energy. If a charge in the field, with a positive potential energy, is released, it'll move back to the reference point or to zero potential. If its potential energy is negative, it'll move toward the charge producing the field, making the potential energy even more negative.

So your both definitions are accurate, but I would be a little more careful with the phrase "yields into electric field", because, as we've seen, depending on the signs of the charges involved, the charge may either yield into electric field or be forced into electric field.


Electric potential is work done per unit charge to bring some charge from infinity to some region near a source of electric field. Important here is this part per unit charge. Electric potential energy is just the work done to bing some charge from infinity to a region near the source. So if q is charge, and U is work than potential is just U/q.


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