Problem statement

Problem 3.11. A system consists of a thin charged wire ring of radius R and a very long uniformly charged thread oriented along the axis of the ring, with one of its ends coinciding with the centre of the ring. The total charge of the ring is equal to q. The charge of the thread (per unit length) is equal to λ. Find the interaction force between the ring and the thread.

Interpretation. A ring and a very long thread share one line, which is the axis of the ring. The thread starts at the centre of the ring and goes off to infinity. Both carry charge, so they push or pull on each other. We want the size of that force.

We assume q = 20 nC, λ = 5 nC/m and R = 10 cm, because the problem gives no numbers. All three are sliders. The constant 1/(4πε₀) is 8.988 × 10⁹ N m²/C².

Symbols: q is the charge of the ring, R its radius, λ the charge per metre of the thread, x the distance along the axis from the ring centre, E the field of the ring.

The big idea in plain words

Think of the ring as a source of an electric field, and think of the thread as many tiny charged pieces. Each piece feels the push of the ring field at its own spot. Add up all these small forces and you have the total. Far from the ring the field is weak, so only the part of the thread near the ring matters much.

The solution in each frame

Both charges are at rest, so only one frame is needed.

The solution step by step

Step 1, the field of the ring on its axis. Take a small bit of the ring with charge dq. Its field at a point on the axis, at distance x from the centre, has the size dq/(4πε₀(R² + x²)). The sideways parts of the fields from opposite bits cancel. The parts along the axis are multiplied by x/√(R² + x²). Adding over the whole ring gives E(x) = qx/(4πε₀(R² + x²)^3/2), pointing along the axis.

Step 2, the force on one piece. A piece of thread of length dx carries the charge λ dx. The force on it is dF = λ E dx = λq x dx/(4πε₀(R² + x²)^3/2). All these forces point along the axis, so they can be added as plain numbers.

Step 3, add up the whole thread. F = (λq/4πε₀) times the integral of x dx/(R² + x²)^3/2 from 0 to infinity. The integral equals [−1/√(R² + x²)] between the limits, which is 0 + 1/R. So F = λq/(4πε₀R) = 8.988 µN.

Step 4, the cases. The thread is long, so there is one case. The force on the ring is equal and opposite (Newton's third law). The force depends on 1/R: a smaller ring gives a bigger force. The part of the thread within a distance x of the ring gives the share 1 − R/√(R² + x²) of the total. For x = R this is 29.3%, for x = 10R it is 90.0%.

Five methods

1. Force method

This is the method used above: field of the ring, then the force on each piece, then a sum.

2. Energy method

The potential of the ring on its axis is φ(x) = q/(4πε₀√(R² + x²)). The work to bring the thread in from infinity to its place is λ times the integral of φ dx, which diverges, so we use the force on a displacement instead. Moving the thread by dx changes the energy by λ φ(0) dx − λ φ(∞) dx = λq dx/(4πε₀R). So the force is F = λq/(4πε₀R).

3. Graph method

The first graph shows how much of the force comes from the thread up to x. It climbs quickly and then flattens at 1. The second graph shows the ring field along the axis, which peaks at x = R/√2.

4. Calculus method

Substitute x = R tan θ. Then x dx/(R² + x²)^3/2 = sin θ dθ/R, and the limits are θ = 0 to π/2. The integral of sin θ is 1, so the result is 1/R.

5. Check method

Units: λq/(ε₀R) is (C/m)(C)/((C²/(N m²))(m)) = N. As R gets very small, the force grows as 1/R, since the thread then touches the charge of the ring. As R gets large, the force falls.

Fastest: the energy method. Most reliable: the calculus method with the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Ring charge q	Yes	The force, the ring field and the ring colour
Thread charge per metre λ	Yes	The force and its direction, the force on each piece and the thread colour
Ring radius R	Yes	The force (as 1/R), the ring size in the picture and the shape of every graph
Position of the shown piece x	No	The piece, its force arrow, the dots on the graphs and the live readings
Given-answer mode	Sets R	The page finds the ring radius for a chosen force, and warns if it falls outside 3 to 30 cm
Play and time bar	No	Play moves the shown piece out along the thread, holds 3 s at x = R, then runs to 100 cm and repeats. The bar sets x

Parameters

Quantity	Symbol	Value used
Ring charge	q	20 nC
Thread charge per metre	λ	5 nC/m
Ring radius	R	10 cm
Constant	1/(4πε0)	8.9876 × 10⁹ N m²/C²
Force	F	8.988 µN
Peak ring field	E at x = R/√2	6.919 kV/m

Exam traps

• Using the Coulomb force between point charges. The ring and the thread are not points, so the field must be integrated.
• Forgetting that the thread starts at the ring's centre, so the integral starts at x = 0 and not at x = R.
• Using a field that points sideways. On the axis, only the axis component remains.
• Giving the answer for the force on the ring with the wrong sign. It is equal and opposite to the force on the thread.
• Thinking the infinite thread gives an infinite force. The field falls fast enough with distance for the sum to be finite.