I have spent a lot of time trying to understand potentials (gravitational and electrical) and I can't grasp it. I understand that it is the work in bringing a point mass (or charge) from infinity to a point in the gravitational (or electrical) field and I understand that I must integrate the force with respect to the displacement. But everytime I do it I seem to be coming out with the wrong sign. Boiled down, my questions are:

  1. For a gravitational field, is the work done in moving a point mass from infinity to a point at radius $R$ given by:

    $ -\displaystyle\int_\infty^R\dfrac{-GMm}{r^2}~dr$

  2. Why is it a point mass or point charge, is this because if it was larger, it would have an affect on the other planet or charge?

  3. Between positive charges, is the electrostatic potential energy?: $+\displaystyle\int_\infty^R \dfrac{kQq}{r^2}dr$

  4. If so how would this change for positive and negative charges, or negative and negative?

  5. Is positive work if I have to "struggle" against something and negative work when I "give in", effectively going along with whatever the gravitational or electric field wants?

If you can help me out with anyone of these I'd appreciate it. If you close the question at least tell me what's wrong with it specifically in the comments so I don't make the same 'mistake' again.

  • $\begingroup$ You haven't shown us how you've performed the integration in the first part, likely you've forgotten to parameterize the path with a negative sign since it starts at infinity and moves towards the origin. $\endgroup$
    – Triatticus
    Commented Jan 29, 2023 at 22:01

3 Answers 3


First $0$ altitude is an arbitrary height set by convention. Typically is it set at mean sea level.

Potential energy is similar. Often we say the gravitational potential energy of an object is when it is at altitude $0$.

This means an object has negative potential energy when in Death Valley a couple hundred feet below sea level. There is nothing stopping us from redefining $0$ to suitable Death Valley levels.

A common physics problem is a universe containing a point mass, and perhaps a light test object. A common physics convention is to set $0$ altitude at the center of the point mass, and the $0$ for potential energy at infinity.

The usual conservation of energy law applies. The test object at rest at infinity has $0$ potential energy + $0$ kinetic energy = $0$ total energy. The point mass attracts the test mass. It falls. $\vec F \cdot \vec d > 0$. $$\int_{\infty}^{R}\vec F \cdot d\vec s > 0$$ Work is positive. Kinetic energy becomes positive. Potential energy becomes negative.

The gravitational potential at a point is the potential energy of a test object with mass $m=1$.

This much should allow you to work out all of your gravitational sign questions.

Electrical signs are similar. The potential of a point charge is the potential energy of a test charge with $q = +1$. Given a positive point charges, the force is repulsive and the potential is positive. Given a negative point charge, the force is attractive and the potential is negative.

Any mass or charge distribution can be broken down into a set of point masses/charges. If you add up the potential from each point, you get the potential for the distribution.


The reason that it only really makes sense to talk about "point charges" is that the definition is using a line integral, about a specific path of a one dimensional line.

If the object has volume, how would I assign a one dimensional path to this object, when each part takes a different path?

The idea of potential (V) is more usefull, since it is independant of any specific object, meaning that given a Conservative vector field, we can assign potential to a point in space.

This means that if I want to find the total amount of energy it takes to move an object through a vector field, I can first find the potential at a point, and then sum up the contribution that each part of an object imparts to the total potential energy. Breaking up each part of an object as $\rho dv$ ( which itself moves through a 1d line).

$$U = \int_{R^3} V [\rho dv]$$

This method allows us to sensibly talk about the potential energy of an object with volume.

  • $\begingroup$ I would like to add that you probably could formulate the definition of potential energy to a macroscopic object using a distance field rather than a single line, but that is not convenient, nor would it be useful and would be harder to construct a quantity that is object independant $\endgroup$ Commented Jan 30, 2023 at 1:16

You're having a hard time because you're starting from a very rigourous point of view. Your knowledge is correct, but let me try a different approach.

You should think of potential as a scalar field.

For a electric field, for example, you can think that the potential created by a point charge is $$V_E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r}$$

and that's true. However, think it this way: the electric potential is the potential energy per unit charge $$V_E=\frac{E_p}{q}$$

This means:

  1. In an empty space, nothing happens (this doesn't exist btw)
  2. If you place a charge $Q$, for the sole reason of being there, it assigns one number to every single point in space. That number is the potential $V_E$

This number is different in each point. The farther you are from $Q$, the smaller the number is (if $Q>0$).

  1. However, if you place a second charge, $q$, it will immediately acquire a potential energy, given by $E_p=q\cdot V_E$

This is a way to explain how particles can "get energy" from other particles without contact. It's simply that the energy is the $q$ itself times the potential that "was already there".

Since particles "want" to have the least energy, a steady possitive charge $q$ will "flow" from the higher potential to the lower potential (and the other way around if it is negative). This is how you should regard potential. It's somehow a gradient in which particles tend to move.

If $Q$ is possitive, the potential is possitive around it (supposing $V_\infty =0$). A possitive $q$ will get the lowest potential at infinity (repulsion). A negative $q$ will get the lowest potential as close to $Q$ as possible (atraction), and so on.

Extra balls:

  1. We do it for a point charge because it is linear. If you want more charges, just add up the individual potentials.
  2. Instead of working with these $1/r^2$ forces, consider first trying a simpler example: a constant force is also conservative and thus has a potential energy. Think for example about a constant gravitational field $\vec{g}$, where the potential energy is $mgh$.
  3. If you have more questions about potential energy, rather than potential itself, you should ask a separate question.

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.