Tension in a curved charged wire (electrostatic force) - does wire thickness matter?

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: $$\sigma=\frac{T}{S}=\frac{T}{\pi r^2} =\frac{1}{\pi}\left(\frac{q}{2\pi}\right)^2\frac{ln(\frac{8R}{r})}{(rR)^2}$$