Problem statement

Problem 3.12. A thin non-conducting ring of radius R has a linear charge density λ = λ₀ cos φ, where λ₀ is a constant, φ is the azimuthal angle. Find the magnitude of the electric field strength (a) at the centre of the ring; (b) on the axis of the ring as a function of the distance x from its centre. Investigate the obtained function at x >> R.

Interpretation. The ring is charged unevenly. The charge per length is largest and positive at φ = 0, zero at φ = 90°, largest and negative at φ = 180°. So the ring has a positive half and a negative half, and the total charge is zero. Part (a) asks for the field at the centre. Part (b) asks for the field at a point P on the axis through the centre, at a distance x from it, and how it behaves when x is much larger than R.

We assume λ₀ = 10 nC/m and R = 10 cm, because the problem gives no numbers. Both are sliders. The constant 1/(4ε₀) is π × 8.988 × 10⁹ N m²/C².

Symbols: λ₀ is the largest charge per length, φ the angle round the ring measured from the y axis, R the radius, x the distance of P from the centre along the axis, p the dipole moment.

The big idea in plain words

Cut the ring into small pieces. Each piece makes a tiny field at the point. Some of these tiny fields point along the axis and some point sideways. Because the positive half and the negative half balance, the parts along the axis cancel. The sideways parts, which all point the same way, add up. This is the same as the field of two opposite charges close together: a dipole.

The solution in each frame

The ring is at rest and nothing moves, so only one frame is needed.

The solution step by step

Step 1, one element. An element of the ring at angle φ, of length R dφ, carries the charge dq = λ₀ cos φ R dφ. It makes a field of size dq/(4πε₀r²), where r is its distance from the point.

Step 2, the centre (part a). At the centre r = R, and the field of every element lies along the line from the element to the centre. By symmetry, the part along z cancels (the integral of cos φ sin φ over a full turn is zero). The part along y is dE_y = −dq cos φ/(4πε₀R²) = −λ₀ cos²φ dφ/(4πε₀R). The average of cos²φ is 1/2, so its integral over 2π is π. Therefore E = λ₀π/(4πε₀R) = λ₀/(4ε₀R) = kV/m. It points toward −y, away from the positive half.

Step 3, the axis (part b). For P at distance x on the axis, every element is at distance r = √(R² + x²). The vector from the element to P has parts x (along the axis), −R cos φ (along y) and −R sin φ (along z). The part along the axis is proportional to ∫ cos φ dφ = 0. The part along z is proportional to ∫ cos φ sin φ dφ = 0. The part along y is dE_y = −λ₀R² cos²φ dφ/(4πε₀(R² + x²)^3/2), and the integral is π. So E = λ₀R²/(4ε₀(R² + x²)^3/2), perpendicular to the axis. For x = 0 this gives the result of part (a).

Step 4, far away. For x much larger than R, (R² + x²)^3/2 is about x³, so E = λ₀R²/(4ε₀x³). The dipole moment of the ring is p = the integral of y dq = ∫ R cos φ λ₀ cos φ R dφ = πR²λ₀. So E = p/(4πε₀x³), the field of a dipole at a point on its perpendicular bisector.

Step 5, the cases. The problem has one situation. The result in (b) covers all x, including the centre. The field falls as 1/x³, faster than the 1/x² of a single charge, because the total charge is zero.

Five methods

1. Force method

The field is added up element by element, as above.

2. Energy method

The potential on the axis is zero for every x, because the positive and negative halves are the same distance from the axis. So the field along the axis is zero, which matches the result. The field across the axis cannot be got from the axis potential alone, so this method only checks part of the answer.

3. Graph method

The first graph shows E along the axis. It is largest at the centre and falls. The second graph shows the exact result and the dipole formula against x/R: they join for x/R beyond about 5.

4. Calculus method

This is the integral over φ in steps 2 and 3, using ∫ cos²φ dφ = π over a full turn.

5. Check method

Units: λ₀/(ε₀R) is (C/m)/((C²/(N m²)) m) = N/C. At x = 0 the general formula gives part (a). At large x it matches p/(4πε₀x³).

Fastest: the integral of cos²φ. Most reliable: the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Density amplitude λ0	Yes	Both fields, the dipole moment, the field direction and the ring colours
Ring radius R	Yes	The field at the centre (as 1/R), the field on the axis, the ring size and every graph shape
Distance of P, x	No	The point P, the field arrow, the dots on the graphs and the live readings
Highlighted element φ	No	Which small element is drawn with its own field at P. The total does not depend on it
Given-answer mode	Sets λ0	The page finds λ0 for a chosen field at the centre, and warns if it needs more than 20 nC/m
Play and position bar	No	Play holds 3 s at the centre, moves P out along the axis, holds 3 s at x = 5R (far away), then runs to 60 cm and repeats. The bar sets x

Parameters

Quantity	Symbol	Value used
Density amplitude	λ0	nC/m
Ring radius	R	cm
Field at the centre	E	kV/m
Dipole moment	p = πR²λ0	pC m

Exam traps

• Saying the field on the axis is along the axis. For this ring the axis parts cancel and the field is across the axis.
• Forgetting that the average of cos²φ is 1/2, not 1 or 0. This gives the factor π.
• Using the total charge as q. The total is zero, so the field is not that of a point charge.
• Writing E ∝ 1/x² far away. It falls as 1/x³, like a dipole.
• Putting R instead of R² in the numerator of the field on the axis.