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When calculating electric potential using : $$\ V(r) = \frac{q}{4\pi\epsilon_0 r}$$ (r pointing any point).

It's implied we are referring to this full formula : $$\ V(r) - V(\infty)=\frac{q}{4\pi\epsilon_0 r}-\frac{q}{4\pi\epsilon_0 \infty}$$ (which comes from the definite integral of an electric field).

I used in an exercise :$\ V(r) = \frac{q}{4\pi\epsilon_0r}$ , to calculate the potential difference between a point $K$ located on a uniform ring of charge (so $q$ would be a line integral), radius $R$, and a point $P$ on the ring axes ($z\perp R$).
For what i said above i can't do this, my result coincides with the book only because i wrongly considered $\ V(r_K)$ as the reference potential and therefore it's contribuition to the last part of the equation equal to zero.
Now, from what i understood, if i want to consider $\ V(r_K)$ my reference potential i should have calculated $\ V(r_P)$ like this:
$$ \Delta V_{PK}=V(r_P)=\frac{q}{4\pi\epsilon_0 r_P}-\frac{q}{4\pi\epsilon_0r_K}$$
With :$\ V(r_K)=0$ (reference potential); $r_k=0$ ($K$ is located on the ring).

2.Edit: I wrongly said that $r_K=R$ so i changed it to $0$ since as @Quasihorse pointed out, the distance used i those formulas is the one from the ring of charge (not from the origin as i previously wrote)

BUT this is also probably wrong because there is an inconsistency in the equation, it can probably be solved by saying that if i want to calculate "the raw potential of a point" by making an indefinite integral i'll end up with : $$\ V(r_K)=\frac{q}{4\pi\epsilon_0r_K}+V_0$$ so by equating $\ V(r_K)$ to $0$ we don't necessarily say that $\frac{q}{4\pi\epsilon_0r_K}$ is $0$.

  1. Edit in this case $\frac{q}{4\pi\epsilon_0r_K}$ is undefined ($r_K=0$) so we cant say much about it, but if $\frac{q}{4\pi\epsilon_0r_K}$ was a finite number $V_0$ would be -$\frac{q}{4\pi\epsilon_0r_K}$ to balance off the equation or i totally went bonobo on this one?

I don't know if any of what i wrote makes sense so feel free to harrass me ;), i'd like if someone confirms (or not) what i said and clears up my mind which is permanently confused.
(read the edit for clarification)

  1. Edit: to clarify the question; the problem asks to calculate the potential of every point of the ring axe, this problem is solved by simply using $$\ V(r) = \frac{q}{4\pi\epsilon_0 r}$$ with $r$ as the distance between a point on the ring and a point on the axis.
    My question is: what if instead of using $\infty$ as my zero potential point i use $K$ (located on the ring) as my reference? What are the implications?
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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Apr 19 '20 at 23:19
  • $\begingroup$ @David Z they were clarification comments, can you please put them back? $\endgroup$ Apr 19 '20 at 23:21
  • $\begingroup$ @Philip Wood, when I say ring I mean a linear distribution of charge, positioned on a ring. The formula I used should work also on continuous charge distribution, like this one, making the right adjustment q "becomes" the line integral of a linear density of charge. I have a demonstration on my notes that starts from using this formula for a point charge, then for a discrete distribution of charge and then for a continuous distribution of charge, so it should be" legal", also because any distribution of charge "watched from an infinite distance" looks like a point. I'm not sure if I'm right $\endgroup$ Apr 19 '20 at 23:28
  • $\begingroup$ Also since a linear distribution of charge is basically a series of charges that form a line, due to the conservative field hypotesis, i can use the summation of effects and evaluate every point charge contribution and sum it up. (this is more of an explanation dont consider what i said before of things collapsing to a point charge when observed form far away) $\endgroup$ Apr 19 '20 at 23:28
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Setup your $dV=-\vec E\cdot d\vec l$, indefinitely integrate it and choose your constant however you like depending on $\vec r$.

About the implications of choosing one reference point over another, this should address it: $$\int_a^bf(x)\,dx=\left[F(b)+C\right]-\left[F(a)+C\right]=F(b)-F(a)$$ independently of $C$.

If $\underset{x\to-\infty}{\lim}F(x)=0$ then: $$\int_{-\infty}^bf(x)\,dx=\left[F(b)+C\right]-\left[\lim_{t\to-\infty}F(t)+C\right]=F(b)-\lim_{t\to-\infty}F(t)$$ the same as the first integral.

If I want my potential to be zero at $a$, for example, then I'd set $C=-F(a)$.

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    $\begingroup$ Ok, I didn't quite understand, if $\ F(x)$ is my potential, in the first case you chose $C=\infty$, chosing $C$ this way translates to: I say my potential is infinite at an infinite distance from the ring (but not necessarily $0$ on the ring) right? Can you please explain more... bear with me ahah $\endgroup$ Apr 20 '20 at 9:53
  • $\begingroup$ That was a mistake, I meant the limit to be $0$. Thank you! $\endgroup$
    – Luyw
    Apr 20 '20 at 10:28
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If the potential is known for a given choice of "zero potential value" reference point, such as infinity, then the potential difference between two points may be found simply by subtracting the potential values at them — there is no need to change the zero potential reference point.


edit: Responding to the edit in the question: in this case, to calculate a potential $\Phi’(\mathbf{x})$ with reference point on the ring, it is sufficient to take

$$\Phi’(\mathbf{x}) = \Phi(\mathbf{x}) - \Phi(K)$$

where $\Phi(\mathbf{x})$ is the potential calculated using an infinite reference point. One sees immediately that $\Phi’(K) = 0$ as required. The problem with doing this for $K$ on the ring is that $\Phi$ is undefined/divergent on the ring itself (see the rest of the answer) so $\Phi(K)$ does not exist. This is due to the field strength being arbitrarily large as you approach the ring, similar to what happens if you try to calculate the potential of a point charge at the point charge. For your question, this means that the ring can not be taken as a meaningful reference point — speaking loosely, you would then have $\Phi’(\mathbf{x}) = \Phi(\mathbf{x}) -\Phi(K) = \Phi(\mathbf{x}) - \infty = -\infty $ at any point on the axis — this means the potential with this reference point is undefined and is useless.


edit 2: Responding to the further edit of the question, I will assume that $r_K$ is the Euclidian distance from the ring to the the point $K$, and $r_P$ is the distance from the ring to $P$, with both $K$ and $P$ on the ring's axis. In this case, we have (with the reference point at $\infty$)

$$\Phi_\infty(r) = \frac{q}{4\pi\epsilon_0 r}$$

so that your formula is correct: the potential at $P$ with $K$ as the reference point is

$$\Phi_K(r_P) = \Phi_\infty(r_P) - \Phi_\infty(r_K) = \frac{q}{4\pi\epsilon_0 r_P} - \frac{q}{4\pi\epsilon_0 r_K}$$

which we can see is right since $\Phi_K(r_K) = 0$.

To eliminate a seeming point of confusion, please note that your expression $\Phi_\infty(r)$ (what you call $V(r)$ at the start) is only valid for points on the ring's axis and with $r > R$ (consider the diagram in my first link, which shows how the expression is obtained and why it is only valid in these cases). This means that you cannot set $r_K = 0$ in your expression to show the divergence, since your formula is not valid at that point — you would need to do something else, such as find a formula valid in the whole plane of the ring or make an argument based on approximating the ring locally as a straight wire and arguing the electric field diverges as you approach the wire.


Considering the specific problem at hand, the potential due a uniformly charged ring on the ring's axis is a standard calculation, presented for instance at http://www.phys.uri.edu/gerhard/PHY204/tsl81.pdf. If the point $K$ is on the axis at the ring's centre, the formula derived in the link should be sufficient to find the potential difference between it and a point $P$ further along the axis.

If $K$ is genuinely on the ring, we might try to calculate the potential in the plane of the ring. The potential $\Phi(\mathbf{x})$ at a point $\mathbf{x}$ due to a continuous charge distribution $\rho(\mathbf{r})$ is (in SI units)

$$\int_\text{all space} d\mathbf{r}\, \frac{\rho(\mathbf{r})}{4\pi\epsilon_0|\mathbf{x} - \mathbf{r}|}. $$

For a linear charge density $\lambda$ on a ring of radius R in the $x–y$ plane, $\rho(\mathbf{r}) = \lambda\,\delta(z)2R\delta(x^2 + y^2 - R^2)$. This means that if $\mathbf{x}$ is in the plane of the ring, i.e. $\mathbf{x}_z = 0$, then the $z$ integral will simply eliminate the $\delta(z)$ Dirac delta and the remaining integral will be in the $x–y$ plane.

A natural parametrisation of a circle in 2D is $\mathbf{r}(t) = R(cos(t/R), sin(t/R))$ (giving a line element $dl = dt$), so that we may rewrite the above integral in the plane of the ring as

$$\Phi(\mathbf{x}) = \int_0^{2\pi R} dt\, \frac{\lambda}{4 \pi \epsilon_0 \sqrt{(\mathbf{x}_x - R cos(t/R))^2 + (\mathbf{x}_y - Rsin(t/R))^2}}$$

Setting aside whether this integral possesses a closed form (Mathematica does not yield anything helpful), we see that there is a problem if $K$ lies on the ring. This is because at $t$ such that $\mathbf{r}(t) = \mathbf{x}$, where $\mathbf{x}$ are the coordinates of $K$, the denominator of the integrand goes to zero. The integrand is thus not bounded and the Riemann integral is not defined (the improper integral also diverges). Thus, in realm of classical electrodynamics, the potential of a uniformly charged ring is not defined on the ring. A similar issue is covered in this answer about the potential energy of such a ring: Potential energy of a charged ring.

In light of the above problem, it seems likely that your book considers $K$ to be located at the ring's centre rather than actually on the ring.

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  • $\begingroup$ Thank you for the answer, i noticed i didn't actually say what the problem was asking to calculate, i'll add an edit... anyway my question is: what if i want to calculate the potential of a general point on the axis of this ring of charge using a general point instead of $\infty$ as my reference potential. $\endgroup$ Apr 20 '20 at 8:15
  • $\begingroup$ @GiuseppeReitano I have an added an edit to my answer that may help clarify the situation — I believe the original answer answered the question, but was too brief at the beginning. $\endgroup$
    – Quasihorse
    Apr 20 '20 at 9:44
  • $\begingroup$ Thank you very much, can you please have a bit more patience and see if what i say in edit .3 in my question is right? $\endgroup$ Apr 20 '20 at 10:32
  • $\begingroup$ @GiuseppeReitano I have made a further edit, showing that your $\Delta V_{PK}$ is in fact correct, provided that both $P$ and $K$ are on the ring's axis. $\endgroup$
    – Quasihorse
    Apr 20 '20 at 11:06

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