# How is electric potential related to potential energy and work in electrostatics?

In Section 2.3.1 of Griffiths' text on electrodynamics he defines the scalar potential $$V$$ as $$V(\mathbf{r})=-\int_{\mathcal{O}}^{\mathbf{r}}\mathbf{E}\cdot d\mathbf{l}.$$

In Section 2.3.2 he states

The word ''potential'' is a hideous misnomer because it inevitably reminds you of potential energy. This is particularly insidious, because there is a connection between ''potential'' and ''potential energy'', as you will see in Sect. 2.4. I'm sorry that it is impossible to escape this word. The best I can do is to insist once and for all that ''potential'' and "potential energy" are completely different terms and should, by all rights, have different names.

In Section 2.4 he shows that the work required to move a test charge $$Q$$ from $$\mathbf{a}$$ to $$\mathbf{b}$$ in the presence of a collection of source charges is given by $$W=\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{F}\cdot d\mathbf{l}=-Q\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{E}\cdot d\mathbf{l}=Q[V(\mathbf{b})-V(\mathbf{a})].$$

And this implies that $$V(\mathbf{b})-V(\mathbf{a})=\frac{W}{Q},$$ so that the scalar potential difference between $$\mathbf{a}$$ and $$\mathbf{b}$$ is equal to the work per unit charge to carry a particle from $$\mathbf{a}$$ to $$\mathbf{b}$$. In particular, if we wish to bring $$Q$$ in from far away and stick it at the point $$\mathbf{r}$$, the work we must do is $$W=Q[V(\mathbf{r})-V(\infty)],$$ so if the reference point is infinity $$W=QV(\mathbf{r}).$$ He then states

In this sense, potential is potential energy (the work it takes to create a system) per unit charge (just as the field is the force per unit charge).

I find this very confusing. I thought that energy was the ability to do work, and that work is defined as the line integral $$W=\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{F}\cdot d\mathbf{l}.$$ If these definitions are correct, then $$V(\mathbf{r})=W/Q$$ doesn't seem to have anything to do with energy.

If, on the other hand, energy is "the work it takes to create a system", then it is correct to say that the scalar potential is the potential energy per unit charge required to create the system. But if energy is "the work it takes to create a system", I don't see how this definition is related to "the ability to do work".

I apologize for the obtuse question. I am from a mathematical background, and am not familiar with energy and work beyond a very elementary level, and would really appreciate any help in understanding this.

I think my question can be phrased more concisely as: what is energy in the context of electrostatics, and how does it relate to work?

• Hey jackrodgers1554, if you think you can write a better title for your question (such as the phrasing you just edited in), please feel free to replace the title I added. Note that, when editing questions, we prefer you don't explicitly mark your changes with "Edit:"; instead, just change the question to make it look as if you'd written it that way all along. Jun 15, 2020 at 22:12
• @DavidZ Thank you very much for writing a better title, for some reason I was having trouble thinking of a better one. I post a better one next time. And thank you for letting me know the communities preference on edits, I have edited my post above accordingly. Jun 16, 2020 at 0:03

I would advise you to forget both of these supposed definitions:

I thought that energy was the ability to do work...

...energy is "the work it takes to create a system"...

These are things we tell people who don't know any math, to give them a vague conceptual inkling of what energy is. They are imprecise and you cannot use them with any benefit in calculations.

When you know math, energy is defined by equations. There are some very general definitions in more advanced physics, but at the level of elementary electrostatics,1 you're likely to find different definitions for different types of energy. One such definition is the electric potential energy, which can be defined (for a single particle at position $$\mathbf{r}$$ in a static electric field) as $$U_e(\mathbf{r}) = -\int_{\mathcal{O}}^{\mathbf{r}}\mathbf{F}_e(\mathbf{r})\cdot\mathrm{d}\mathbf{l}$$ Here $$\mathbf{F}_e(\mathbf{r})$$ is the electric force, $$\mathbf{F}_e(\mathbf{r}) = Q\mathbf{E}(\mathbf{r})$$. Note that this is physically and mathematically equal to the work required to put the charge at $$\mathbf{r}$$, given that $$\mathcal{O}$$ is a reference location (meaning, by definition, that a charge at that point contributes nothing to the potential energy).

From this it's straightforward to conclude that $$U_e(\mathbf{r}) = QV(\mathbf{r})$$ There's your link between energy and electric potential.

1 Griffiths' book is not exactly an elementary one, but what you're asking about doesn't involve any advanced concepts so I'll stick to the basics so as to avoid confusing readers of this answer more than necessary.

• Thank you so much for your clear and precise answer, I was driving myself nuts trying to understand vague notions like "ability to do work" and "the work it takes to create a system". I will promptly forget these "definitions". Is your definition of energy a special case of a more general definition? Can you please point me to where I might find more general definitions? They might be too advanced for me, but I am just curious. Jun 16, 2020 at 0:06
• @jackrodgers1554 The definition given here is for the electric potential energy. This definition generalizes straightaway to a definition for the potential energy of any conservative force $\mathbf{F}$ by replacing the electrostatic force $\mathbf{F}_e$ with $\mathbf{F}$. Note however that this is only a definition of the potential energy associated with a conservative force field, and not of energy in general.
– d_b
Jun 16, 2020 at 0:15
• @d_b Got it. Can one show that the definition of potential energy $\int_{\mathcal{O}}^{\mathbf{r}}\mathbf{F}\cdot d\mathbf{l}$ itself follows from a more general definition of energy in general? Jun 16, 2020 at 0:19
• @jackrodgers1554 That might actually make a pretty good followup question, if you flesh it out a bit. Or, if you start writing it up, the system may suggest some existing questions which might give you the answer you're looking for. Jun 16, 2020 at 0:27

work is defined as the amount of energy required to move a mass form point $$a$$ to $$b$$ against an opposing force.

the force acting on a charge $$q$$ in field $$\mathbf{E}$$ is simply $$\mathbf{f}=q\mathbf{E}$$. Hence the work done is: $$W = \int_a^b{q\mathbf{E}\cdot dl}$$ now $$q$$ is a test charge. Thus the potential is simply: $$\phi=\frac{W}{q}$$