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

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.

Answer

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