Consider a conducting wire bent in a circle (alternatively, a perfectly smooth metal ring) with a positive (or negative) electric charge on it. Technically, this shape constitutes a torus. Assume that the radius of the wire is much smaller than the radius of the circle.

Find the tension in the wire due to the repulsive electrostatic force (in Newtons).

For this problem, here are example physical values, (may or may not be necessary to solve the problem) charge, $Q = 1 ~\mathrm{C}$, circle radius, $R = 1~\mathrm{m}$, wire radius, $r = 1 ~\mathrm{mm}$. Use the most straightforward notations possible for Coulomb's law, $F=\frac{k q_1 q_2}{r^2}$. This is a question about the mathematics of a set of inverse-square problems, not about theoretical E&M. Other notations that will be of useful to this discussion:

  • $\lambda = \frac{Q }{ 2 \pi R } $ = linear charge density
  • $\sigma = \frac{\lambda }{ 2 \pi r }$ = surface charge density

What happens to the tension in the wire as the wire radius goes to zero with a constant circle radius, assuming:

  • A constant charge on the entire thing (also constant linear charge density)
  • A constant surface charge density
  • A constant voltage
  • If it really bothers you, consider the tensile stress for the above as opposed to the wire tension, although my question is for the latter

The following are some helping equations. Any of them could be wrong.

Electric field from an infinite straight line charge $$F(r) = \frac{ 2 \lambda k }{ r }$$

Electric field face-on for a ring of charge, R is radius of ring and d is distance away $$F(d) = \frac{2 \pi \lambda R k d }{ (d^2+r^2)^{3/2} }$$

Electric potential in vicinity of ring of charge, $\rho$ is distance from z-axis with torus in xy-plane, centered at origin. Electric field is simple knowing this.

  • $l = \sqrt{ (\rho+R)^2 + z^2 }$ = distance to furthest point on ring
  • $K(x) =$ Elliptic integral of the 2nd kind

$$E(\rho,z) = \frac{ 4 R k \lambda }{l} K\left( \frac{ 2 \sqrt{ R \rho } }{ l } \right) $$

Now, I think that a valid approach would be to find the net effective field that a differential part of the ring experiences. If you found this, I believe the tension could be written as follows, with T being tension, R the circle radius, and F the electric field. $$T = \lambda R F$$

I have my own thoughts on the answer (and more equations than you would ever want), and I'll return to offer them later, but I certainly don't have a complete picture, and I want to see what approaches people here use. Maybe this could be answered very simply and maybe some people will know intuitively which cases limit to infinity.

  • $\begingroup$ A very interesting question.. +1 $\endgroup$
    – qftme
    May 13, 2011 at 21:34
  • $\begingroup$ Is this homework? I tagged it as such, then removed it since I wasn't sure. Sentences like, "Use the most straightforward notations possible for Coulomb's law" are commands, which you would typically find in a homework assignment. On reading the problem, though, it seems like maybe you simply chose wording similar to what I'd expect on a homework problem. $\endgroup$ May 13, 2011 at 21:53
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    $\begingroup$ It's not homework and it wouldn't feasible as homework. My thinking is that this problem is categorically never given or discussed in physics classes because it's too involved for low level problems, and as a simple 1/r^2 integral it's not novel enough for higher classes. The field face-on for a ring and disk is the closest I've seen. FYI, the equation for the electric potential is entirely my creation. I wanted the Coulomb's law like that so it could be applied to other 1/r^2 problems, like Newtonian gravity. $\endgroup$ May 14, 2011 at 19:23
  • $\begingroup$ I have an answer for this now, but it won't let me post it here. Any advice? It's too long for the comment section here. $\endgroup$ May 18, 2011 at 2:41
  • $\begingroup$ just add a new answer, there is nothing wrong with answering your own questions. In fact, i think its a great way to document the solution to a interesting problem $\endgroup$
    – lurscher
    May 18, 2011 at 22:42

2 Answers 2


J.M. Hure addressed parts of the problem in Solutions of the axi-symmetric Poisson equation from elliptic integrals (2005), giving a form of the electric field in the vicinity of a loop charge. I will post his published result here with his notation.

These equations are for a mass distributed in a loop with linear mass $\lambda$, radius $a$, and located elevated to a height of $z$ (as far as I'm concerned $z=0$). Then the coordinates used are $R$ for distance to vertical axis and $Z$ for the vertical position. Note that this notation is very different from what I use. Two intermediary values are introduced which are $k$ and $k'$. The answer is for the gravitational field using the case of Newtonian gravity (standard use of $G$) and is denoted $< \kappa_R, \kappa_\Phi, \kappa_Z>$. The equations follow.

$$k^2 = \frac{ 4 a R }{ (a+R)^2 + (z-Z)^2 }$$ $$k'^2 = 1-k^2$$

$$\kappa_R = \frac{G}{R} \sqrt{\frac{a}{R} } k \left( E(k)-K(k)+\frac{(a-R) k^2 E(k)}{2 a k'^2 } \right)$$ $$\kappa_\Phi = 0$$ $$\kappa_Z = \frac{G (z-Z)}{ 2 R \sqrt{a R} } \frac{k^3 E(k) }{k'^2}$$

And yes, this is what I had. I gave the potential in my question because it was the most simple expression that could convey the mathematics for a loop geometry. Hure also noted a paper by Durand in 1964 in regarding these equations, Electrostatique. I. Les distributions, which may have the first published form for loop geometry. Hure also discusses self-gravitation of the loop geometry and introduces the concept of a loop singularity. I think he means to say that a loop singularity implies infinite self-gravitation but I think his wording does not explicitly say that. Nonetheless, he avoids the problem by writing an integral for self gravitation using a mass density, $\rho$, and not linear mass density, $\lambda$, (similar to my $F_{effective}$ below) indicating that the issue of infinite self-gravitation was well understood. Hure's paper then goes on to offer empirical methods for the self-gravitation as well as some other problems.

Before I move on to my own solution, I should note one difference about the gravitational problem. This question is about the electric wire force, in which case the charge is a surface charge. In the case of gravitation, the entire volume contains a mass density, meaning that ultimately I address a surface integral here but the gravitational problem requires a volume integral. The volume integral is not easy with the $r<<R$ method I apply here. I've tried.

I will write my own equation for the field here which is mathematically equivalent to the above one. I will make one new definition, which is the closest distance to the loop from a point.

these should really be negative, since I've written them for electric field, but I'll keep them positive for consistency with the above equations, which are for gravitational field

$$r = \sqrt{ (\rho-R)^2 + z^2 }$$

$$< F_\rho, F_\theta, F_z> = \nabla E(\rho,z)$$

$$F_\rho = \frac{2 k R \lambda }{ l \rho } \left( K \left( 2 \frac{ \sqrt{ R \rho } }{l} \right) + \frac{ \rho^2-R^2-z^2}{r^2 } E\left( 2 \frac{ \sqrt{ R \rho } }{l} \right) \right)$$ $$F_\theta = 0$$ $$F_z = \frac{ 4 k R \lambda z }{r^2 l} E\left( 2 \frac{ \sqrt{ R \rho } }{l} \right)$$

Starting from here I will directly address the question I asked. Recall the equation for the field around an infinite line charge above. The potential then follows a $ln(r)$ form, but keep in mind that the potential is a relative measure, but also, since the potential around a ring charge given above was found by integrating the $1/r$ differential potentials, it will have the same offset.

$$E(r) = - 2 \lambda k ln(r) = 2 k \lambda ln \left( \frac{C}{r} \right) $$

Objectively, we will find a simplified approximation of the previously given $E(\rho,z)$ as $r$ goes to zero and a part of it will be identified as being equivalent to $E(r)$ above, along with another, yet unknown, component. I did just that, and it was no easy task. Here is what I obtained. The two parts are identified and written separately to make the result crystal clear.

$$E(\rho,z) \approx 2 k \lambda \left( ln( \frac{8 R}{r} ) + \frac{\rho-R}{R} \left( 1-ln( \frac{8 R}{r} ) \right) \right)$$ $$E_{line}(\rho,z) = 2 k \lambda \left( ln( \frac{8 R}{r} ) \right)$$ $$E_{loop}(\rho,z) = 2 k \lambda \left( \frac{\rho-R}{R} \left( 1-ln( \frac{8 R}{r} ) \right) \right)$$

Again, this is an approximation of the previous equation for $E(\rho,z)$ as $r$ goes to zero. I found it to be a good approximation for $r/R<0.1$ or so, and the approximation becomes more correct as $r$ decreases further. It is almost exact for wire radius of $cm$ or $mm$ with a loop radius of $m$. Next, the electric field due to the global loop geometry, $E_{loop}$, is used to answer the problem. The gradient of this quantity gives the field, $\vec{F}_{loop}$, which can then be used to obtain the force. The problem is that this field acts very non-uniformly at a certain $r$. Note that $E_{line}$ acts uniformly and both the potential and field for the infinite line approximation will dominate in magnitude compared to the global loop effects, but results in no net force on the wire while $E_{loop}$ does, this means that a uniform surface charge distribution is a valid approximation. The average value of $\vec{F}_{loop}$ at some $r$ is obtained by integrating, and the vector nature is dropped while it is noted that the field is directed outward, or in the positive $\rho$ direction.

$$\rho(r,\psi) = R+r \cos{\psi}$$ $$z(r,\psi) = r \sin{\psi}$$

$$F_{effective} = \frac{1}{2 \pi} \int_0^{2 \pi} \left( \frac{d}{d\rho} E_{loop}(\rho,z) |_{(\rho,z)=\rho(r,\psi),z(r,\psi)} \right) d\psi = \frac{\lambda k}{R} ( ln( \frac{8 R}{r} ) - \frac{3}{2} ) $$

This point is very close to the final answer. The math for finding the tension in the wire is returned to and an expression for written for it in terms of the desired quantities. I also answer the problem for the previously stated values.

$$T = \lambda R F_{effective} = \lambda^2 k ( ln( \frac{8 R}{r} ) - \frac{3}{2} ) = 1.7 GN$$

Now, the other part of the question was about what happens as the wire thickness goes to zero. For this case (constant charge) the force goes to infinity. Well well all knew that (otherwise why would I have asked the question). But what about the case of constant voltage and constant surface charge density? The previous is written using $V$ for the wire voltage and $\sigma$ for the surface charge density. Note that the voltage of the wire of radius $r$ can validly be found by the infinite line charge approximation as per my prior arguments.

$$T = \frac{V^2}{16 k} \frac{ ln( \frac{8 R}{r} ) - \frac{3}{2} }{ln( \frac{8 R}{r})^2} = 4 \sigma^2 \pi^2 r^2 k ( ln( \frac{8 R}{r} ) - \frac{3}{2} )$$

Now the limits can be addressed. Here is what happens to the force in the wire as the radius goes to infinity with the quantities kept constant. For completeness, I address the same things in terms of tensile stress.

  • Constant charge --> infinity, infinity
  • Constant voltage --> zero, infinity
  • Constant surface charge density --> zero, infinity

This is an interesting result. I should note that due to the tensile stress result this means that no curved wire with an extraordinarily small radius, a curve, and a charge can exist without tearing itself apart. I welcome disagreement on that point.


I will try to answer by using simple physical arguments.

In the first, the approximate formula for the capacitance of a ring is:

$$C=\frac{\pi R}{ln (\frac{8R}{r})}(CGS-system)$$ where R is a radius of the ring and r is a radius of the wire. ($R>>r$)

Let $q$ be the total charge(distributed uniformly) of the ring. Then the electrostatic energy of the ring: $$W_q=\frac{q^2}{2C}$$ Now, the radial force (due to the electrostatic repulsions), applied to the ring, can be obtained from a simple differentiation: $$F_R=-\frac{dW_q}{dR}$$ Thus $$F_R=\frac{q^2}{2\pi}\frac{ln(\frac{8R}{r})}{R^2}$$ The tension $T$ in the wire: $$T=\frac{F_R}{2\pi}= \left(\frac{q}{2\pi}\right)^2\frac{ln(\frac{8R}{r})}{R^2}$$ Tensile stress of the wire: $$\sigma=\frac{T}{S}=\frac{T}{\pi r^2} =\frac{1}{\pi}\left(\frac{q}{2\pi}\right)^2\frac{ln(\frac{8R}{r})}{(rR)^2}$$

  • $\begingroup$ If the first equation for approximate capacitance is in CGS units, I believe it should have either an $\epsilon$ on top or a $k$ on the bottom. With that revision, you are consistent with my result, since I believe the $3/2$ factor I had is likely to come from the set of assumptions employed. Obtaining the capacitance is mathematically equivalent to the electric potential calc, and possibly easier. Great work! $\endgroup$ May 23, 2011 at 14:12

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