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Equipotential lines

I immediately understood that this question is about how work done to get a test charge from one equipotential line to another doesn’t depend on the path taken but this diagram seems a little strange to me. I only understand the part with a and b where there is a dashed circle with I and II by it. I’m assuming that path I represents the shorter looking path between a and b while path II represents the longer looking path (please correct me if I’m wrong).

What I don’t understand is why ra and rb are included; I’m assuming ra represents the distance between the source charge and any point (like point a?) on the first equipotential line aka the smaller circle and rb represents the distance between the source charge and any point (like point b?) on the second equipotential line aka the larger circle. They seem irrelevant but I could be mistaken.

Also what’s with path III? If I’m not mistaken there are 2 equipotential lines here: the smaller dashed circle and the larger dashed circle. Going from any point on the smaller circle to the larger circle or vice versa will result in the same magnitude of potential difference. So why specify two points?

Also I’m confused about the explanation about electric potential and electric potential energy. Why does electric potential depend on the source charge and EPE depend on the test charge?

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  • $\begingroup$ electric potential energy of a SYSTEM is the energy stored inside the configuration of a system of 2 or more charges that interact with eachother through the conervative electric force. electric potential at a point in an electric field is the electric potential energy per unit positive charge at that point. Electric potential $\endgroup$ Feb 9, 2021 at 8:00

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It seems to me that you understood the excercise correctly. This is the key:

Going from any point on the smaller circle to the larger circle or vice versa will result in the same magnitude of potential difference.

Actually, going from $r_{A}$ to $r_{B}$ differs from going from $r_{B}$ to $r_{A}$ in a minus sign. $$\Delta\phi_{AB} = - \Delta\phi_{BA}$$

What I don’t understand is why ra and rb are included

They just seem to be included to show what path III looks like, but again it's not relevant how the path look. As you said the work needed only depends on the points $A$ and $B$ (in this particular case, it just depend on $r_{A}$ and $r_{B}$).

Also I’m confused about the explanation about electric potential and electric potential energy. Why does electric potential depend on the source charge and EPE depend on the test charge?

The electric potential $\phi$ is the function where the difference of potential between 2 points $\Delta\phi = \phi_{B} - \phi_{A}$ is the amount of work per unit of charge you have to do to move a charge from $A$ to $B$. It's defined per unit of charge becuase the electric field is defined as force per unit of charge. The electric potential energy of a charge is just the work that was done to bring that particular charge from the reference point (ussually set in infinity, if there is no charge distribution at infinity) to where it is. Note that the electron is not a test charge: it's a ordinary charge with charge $-e$.

Formally "test charge" means you take a infinitesimal charge or $q\to 0$. With that definition, it wasn't strictly a test charge as the electron has a finite charge $-e$. Howerver, the definition your book used for "test charge" is something like you can consier as a test charge any particle whose propieties are negligible wrt the system propieties (except for the propiety you are studying). In this case, the electron won't apply a considerable electric force on the charge $+q$ (just like considering the mass of the earth is negligable compared to the sun when studying earth-sun gravitational interaction). Much less a considerable gravitational force. In conclusion, it's equivalent to say the positive charge $+q$ is fixed. If you think the charge $q$ is fixed, all this concepts apply to any charge.

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  • $\begingroup$ Thanks for your answer. I understood the part about ra to rb and rb to ra, that’s what I meant by same magnitude. Also your explanation about electric potential and electric potential energy is great. Only thing is the book says “the test charge is negative” and there is no other charge besides the electron. I’ve also never heard of a “simple charge” nor can I find it online $\endgroup$
    – Ibby
    Feb 9, 2021 at 9:29
  • $\begingroup$ With "simple charge" I meant just "charge". I had updated my answer to answer your concern about "test charge". $\endgroup$
    – Gilgamesh
    Feb 9, 2021 at 9:42
  • $\begingroup$ I see what you’re talking about now. Thanks again $\endgroup$
    – Ibby
    Feb 9, 2021 at 16:18
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In the exercise, we propose 3 possible paths: I: between a and b according to the small end of a circle in dotted green II: along a larger end of a circle in black dotted lines III: passing through a part of conductor a, which passes through the perpendicular between the two conductors then which returns to b along conductor a.

As the work of the force depends only on the final and initial state, the 3 paths are identical for the work of the force: whether you calculate according to the path I, II or III the numerical value will be the same.

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  • $\begingroup$ Welcome to PSE. The circles shown in the Figure are neither circles nor conductors. They are ideal equipotential surfaces (spheres). $\endgroup$
    – Frobenius
    Feb 9, 2021 at 7:41
  • $\begingroup$ the work done by a CONSERVATIVE force only depends on the inital and final positions. The work done by non conservative forces like friction does depend on the path taken $\endgroup$ Feb 9, 2021 at 7:57
  • $\begingroup$ @Frobenius thanks for pointing that out, I thought I was missing something and putting together your correction with Nicholas’s answer I think I understand this now $\endgroup$
    – Ibby
    Feb 9, 2021 at 8:59
  • $\begingroup$ Thanks. This doesn’t involve conductors but I think your explanation for path III makes sense otherwise. I guess I could think of this like an atom because there is a positive source charge (like a nucleus) and this problem involves an electron. Atoms generate an electric field and like @Frobenius said this is an ideal equipotential surface so one path that could be taken is the path of electric field lines which are perpendicular to the equipotential lines $\endgroup$
    – Ibby
    Feb 9, 2021 at 9:18

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