What is the difference between the electrostatic potential energy of a charge immersed in an electrostatic field and the potential energy between two charges?

Is there a connection between the two? Are they the same thing expressed in different ways? What does change?

I mean, the two expressions are quite different, one is the potential energy for a single body, the other is a potential energy between two bodies. Moreover, one is zero at no distance between two objects, the other is zero at infinity (always assuming a certain constant $C$ is $0$).

What is an intuitive explanation for this? What do these energies represent?

Could the same be applied to gravitational potential energy as well?

  • $\begingroup$ Why do you think they are different? $\endgroup$ – sammy gerbil Jan 12 at 10:00
  • $\begingroup$ Because they have different expressions and one is $0$ at no distance between two objects, while the other is $0$ at infinity $\endgroup$ – Shootforthemoon Jan 12 at 10:02
  • $\begingroup$ @sammygerbil They are the same? $\endgroup$ – Shootforthemoon Jan 12 at 10:23

(a) "What is the difference between the electrostatic potential energy of a charge immersed in an electrostatic field and the potential energy between two charges?"

They are essentially the same, assuming that the field arises from one other charge. It's quite useful, by the way, to call this other charge "the source charge", and the charge that is immersed in the source charge's field, the "test charge". This is, though, an artificial distinction; the potential energy "belongs to" both charges, though we sometimes loosely speak of "the test charge's potential energy" (in the same way that we talk of the increase of our gravitational potential energy when we climb a mountain).

(b) Nature doesn't tell us the separation of charged bodies at which the potential energy is zero. It is an arbitrary choice. It's only changes in potential energy that have any physical significance. We could say that your gravitational potential energy (strictly the potential energy of the Earth–you system) is zero when you're standing on the downstairs floor of your house, but we could equally well choose to call it zero when you are on the subsoil below your house, or on the roof...

With two point charges, we choose the zero of their potential energy to be when they are so far away from each other as not to experience forces from each other, that is infinitely far away. This choice is made because then the potential energy formula for the two charges becomes very simple, as you know. We couldn't sensibly take the zero of potential energy to be when the separation is zero, because the potential energy at any other separation would then be infinite!

"Could the same be applied to gravitational potential energy as well?" Yes. An important case is when the source mass is a spherically symmetric body of radius $r_0$. Then for a test mass, $m$ outside the body, the source mass behaves as if all its mass, $M$, is concentrated at its geometrical centre, and $$\text {GPE of}\ M, m\ \text{system} =-\frac{GMm}{r}$$ Now suppose that m is at height $h$ above the surface of $M$, so that $$r=r_0+h.\ \ \ \ (h<<r_0)$$ We can substitute $(r_0+h)$ for $r$ and expand $(1+\frac{h}{r_0})^{-1}$ to first order.. $$\text{GPE} =-\frac{GMm}{(r_0+h)}=-\frac{GMm}{r_0(1+\frac{h}{r_0})}=-\frac{GMm}{r_0}\left(1-\frac{h}{r_0}\right)=-\frac{GMm}{r_0}+\frac{GMm}{r_0^2}h$$ Now if we're interested only in the region $h<<r$ (as dwellers on Earth, perhaps), it's convenient to abandon the convention of taking the zero of PE at $r=\infty$ and put it instead at $r=r_0$, that is $h=0$. The first term then drops out and we're left with the familiar elementary formula$$\text{GPE}\ = mgh.$$

  • $\begingroup$ Thanks! now I see it better. But if we take the gravitational potential energy for an object and the Earth: we know that locally $mgr=-\frac{GMm}{r}$. But how do we pass from $r$ to $h$ (or $y$)? $\endgroup$ – Shootforthemoon Jan 12 at 11:26
  • $\begingroup$ Thanks very much! $\endgroup$ – Shootforthemoon Jan 12 at 12:03
  • 1
    $\begingroup$ Yes, by integration as 𝑚 goes from infinity away from 𝑀 to 𝑟 from the centre of 𝑀(assumed spherically symmetric) we get $$\text{GPE}=-\frac{GMm}{r}=-\frac{GMm}{r^2}\times r = -mgr$$ This is the grav PE taking zero PE at infinite separation. Note that $-mgr$ is just a neat way of writing $-\frac{GMm}{r}$; it ISN'T where $-\frac{GMm}{r}$ comes from. This comment replaces my previous comment, which was poorly argued. I've added the correct treatment of "passing from $r$ to $h$" to my answer (above). $\endgroup$ – Philip Wood Jan 12 at 19:23
  • 1
    $\begingroup$ The difference in potential energy between the charges at separations $r_a$ and $r_b$ isn't what you've written. It is $$\Delta (\text {PE})=\frac{q q_0}{4 \pi \epsilon_0} \left(\frac{1}{r_a} - \frac{1}{r_b} \right).$$ You can indeed arrange this to get $(r_b - r_a)$ on the top. Give it a go! $\endgroup$ – Philip Wood Jan 12 at 21:17
  • 1
    $\begingroup$ Well done! My $\Delta \text{PE}$ was the amount by which the PE at $r_a$ is greater than that at $r_b$, so and (for charges of the same sign) is positive if $r_a<r_b$, that is $r_b>r_a$. And indeed a positive amount of work is done by the electrostatic force as the separation increases from $r_a$ to $r_b$. $\endgroup$ – Philip Wood Jan 12 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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