1. When we are calculating the potential due to the sphere, why do we integrate from $r$ to $\infty$ instead of from $\infty$ to $r$? Or do we integrate from $r$ to $\infty$? Regardless of who does the work, the charge itself is moving from $\infty$ to $r$, correct? So, shouldn't the integral be from $\infty$ to $r$? I am a little confused as to how we define the potential. Is it negative of the work done by the electric field?

  2. Also in a sphere, since the potential remains constant from any point on the sphere to any point inside the sphere, is it true that once a charge is on the sphere, and we want to move it inside the sphere, no extra work is done?

  3. I could not understand why it is the case, that if we have a system of charges, then the potential energy is the same for the configuration which is one value, whereas the electric potential is different when you are closer to one charge as opposed to the other?


1 Answer 1



The order of integration really doesn't matter, as you can easily flip the order and pick up a negative sign. With that being said, there are usual conventions, and it depends on which force you want to look at as doing work.

The easiest thing for me to consider is thinking about moving a charge in from infinity (where usually the potential can be set to $0$) and thinking about the work done (per charge) by the electric force. We know that the work done by conservative forces is equal to the negative change in potential energy. Therefore, we have (usuaing some lose notation in the integral limits) $$\Delta V=V(\mathbf r)-V(\infty)=V(\mathbf r)=-\int_{\infty}^{\mathbf r}\mathbf E\cdot\text d\mathbf x=\int^{\infty}_{\mathbf r}\mathbf E\cdot\text d\mathbf x$$

However, sometimes you see potential defined as the work you do in order to make sure the charges move very slowly to their final position. In this case your force is equal and opposite to the electric force $\mathbf F=-q\mathbf E$ (so as to not introduce significant kinetic energy). Therefore, we have for the work per charge $$\frac{W_\text{you}}{q}=\int_{\infty}^{\mathbf r}\frac{\mathbf F}{q}\cdot\text d\mathbf x=-\int_{\infty}^{\mathbf r}\mathbf E\cdot\text d\mathbf x=\Delta V$$

and everything is consistent. Either the work done per charge by the electric force is $-\Delta V$, or the work per charge you do is equal to $\Delta V$. Either way of looking at potential is fine, and either order of integration limits is acceptable as long as you are careful with the signs. I personally prefer to think in terms of the first integral I have posted, but usually certain problems lend themselves to certain interpretations.

2) Yes, if you are talking about a spherical shell of charge. But this is not true for all spheres, of course. For example, an insulating sphere with a uniform charge distribution will not have a constant potential inside of the sphere.

3) You are mistaken. The potential of a configuration is essentially the same thing as the potential energy of the system, its just that certain terms are scaled by various charges when you add the energy or potential up. A configuration of charges has an associated potential energy or potential due to the interactions of the charges. However, if you want to think of $U(\mathbf r)$ or $V(\mathbf r)$ then really what we are doing is considering what another test charge would experience if it were at position $\mathbf r$.

Note that you do the same thing for forces and fields between two charges, you probably just don't think about it in the same way. You can either consider the interaction between two charges, or you can consider the force/field another test charge would experience in the presence of the original two charges. (Of course this analogy breaks down for more than two charges because it wouldn't make sense to have a "total force" or "total field" of the configuration, but the idea is still valid I believe).

  • $\begingroup$ I am a little confused about one thing. So when we consider the sign of potential energy, do we consider the work done by the electric field or the external force? also, when do we say that something is at a higher potential? $\endgroup$
    – Oishika
    Commented Oct 19, 2019 at 18:34
  • $\begingroup$ @OishikaChaudhury As you can see from my answer, you can consider either one. $$W_\text{electric}=-W_\text{external}=-\Delta U$$ Higher potential really just means a larger value. for example, $5\,\mathrm V$ is a higher potential than $3\,\mathrm V$ $\endgroup$ Commented Oct 19, 2019 at 18:39
  • $\begingroup$ So physically this would just mean that more work is needed to bring a unit test positive charge near the 5V than near the 3V? $\endgroup$
    – Oishika
    Commented Oct 19, 2019 at 18:42
  • $\begingroup$ @OishikaChaudhury Yes that is correct $\endgroup$ Commented Oct 19, 2019 at 18:46
  • $\begingroup$ Just to be clear, the potential energy/potential is related to the negative of the work done by the "conservative force"? So it is equal to the positive work done by the external force? $\endgroup$
    – Oishika
    Commented Oct 19, 2019 at 21:53

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.