Problem statement

A thin wire ring of radius r carries an electric charge q. A point charge q0 is placed at the centre of the ring. Find the increase in the force that stretches the wire.

We assume the charge q is spread evenly along the wire. This means every small piece of the ring has the same charge per metre, and the pull of q0 is the same on every piece.

We assume the wire is thin, so its thickness is much smaller than r. This means the ring is a line, and every part of it is the same distance r from the centre.

We assume all charges are at rest in vacuum, with k = 1/(4π ε0) = 8.988 × 10^9 N m²/C² and ε0 = 8.854 × 10^-12 C²/(N m²). This means only Coulomb's law acts between q0 and the ring.

We assume the question asks only for the increase caused by q0. The ring already had some tension from its own charge q before q0 arrived. We do not count that part.

Symbols: q is the charge on the ring, q0 is the charge at the centre, r is the radius of the ring, and T is the extra tension in the wire (the force with which one part of the wire pulls on the next part).

The big idea in plain words

The charge q0 pushes every piece of the ring away from the centre, if q0 and q have the same sign. The ring does not fly apart because the wire pulls back. Cut the ring into tiny pieces. For each piece, the push outwards must be balanced by the pulls of the two neighbouring parts of the wire. Those pulls are the tension. Because the ring is a circle, the two pulls slant slightly inwards, and that is how they can cancel an outward push.

The solution in each frame

Lab frame

Everything is at rest, so this is the only frame we need, and the full solution is in the next section.

Charge rest frame

This frame does not apply. A frame that moves with a charge is the same as the lab frame here, because nothing moves. Every observer sees the same forces and the same tension.

The solution step by step

First we find the charge on one small piece, because the push of q0 on it depends on that charge. The piece is seen from the centre at a small angle dθ (in radians). A full circle is 2π radians, so the piece holds the fraction dθ/(2π) of the total charge.

Next we find the push of q0 on this piece, because this is the force the wire must hold back. Coulomb's law says two small charges push each other with a force k times the product of the charges over the distance squared. The piece is a distance r from q0, so the push dFe is along the radius, outwards.

Now we balance the forces on the piece, because the piece does not move. The wire pulls on each end of the piece with the tension T, along the tangent (the line that just touches the circle). The two tangents at the ends make an angle dθ with each other. Each tangent makes an angle dθ/2 with the line at right angles to the radius. So each pull has an inward part T sin(dθ/2). The two pulls together give 2 T sin(dθ/2) towards the centre. For a small angle, sin(dθ/2) is almost dθ/2, so this is T dθ. The outward push must equal it.

Last we put in dFe and cancel dθ, which appears on both sides.

We used k = 1/(4π ε0), so k/(2π) = 1/(8π² ε0). The result does not depend on dθ, so every piece of the ring has the same tension. The tension is also the same number as the push per radian of angle.

Five methods

1. Force method

The free-body diagram of one piece has three forces: the push dFe outwards, and two pulls T along the tangents. Newton's second law along the radius, with the piece at rest, gives T dθ = dFe. The steps above give T = N.

2. Energy method

We use the work-energy idea with a tiny imagined stretch. Let the ring grow by a small amount δ in radius. The wire gets longer by 2π δ, and the tension does work T × 2π δ against the stretch. The electric force on the ring does work too. The potential energy of the pair is U = k q0 q / r, so the outward force is −dU/dr = k q0 q / r², and its work is k q0 q δ / r². The two works are equal.

3. Graph method

The first graph shows T against q0. It is a straight line through zero, and its slope is k q / (2π r²). Read the value at your q0 to get T. The second graph shows log T against log r, a straight line with slope −2. The third graph shows the stress in the wire against its diameter, and shades the region where the wire gives way.

4. Calculus method

We cut the ring in half and balance the whole half. By symmetry the two cut ends pull with 2T along the axis of the half ring. A piece at angle φ from the cut line has the push dFe = k q0 q dφ / (2π r²), and its part along the axis is dFe sin φ. We add all pieces from φ = 0 to φ = π.

5. Check method

Units: k q q0 / r² is in newtons, and dividing by the pure number 2π keeps newtons. Limits: if q0 is zero, T is zero, as it should be. If r grows very large, T goes to zero. If q or q0 doubles, T doubles. Swapping q and q0 does not change T, which fits Coulomb's law. If the signs are opposite, T is negative: the ring is pulled inwards and a thin wire cannot push back, so it goes slack.

Fastest: the force method. Most reliable: the calculus method on the half ring, because it needs no small-angle step and gives the same result.

What each slider does

Parameters

Exam traps

• Adding up the pushes on all pieces. They point in different directions and sum to zero. Balance one piece, or one half ring.
• Forgetting the angle. The tension T does not equal dFe. It equals dFe divided by dθ.
• Using the full charge q in Coulomb's law. A piece holds only the fraction dθ/(2π) of q.
• Writing 4π instead of 8π² when using ε0. The 2π from the ring is on top of the 4π in k.
• Counting the ring's own repulsion. The question asks only for the increase from q0.