A very long uniformly charged thread oriented along the axis of a circle of radius R rests on its centre with one of the ends. The charge of the thread per unit length is equal to λ. Find the flux of the vector E across the circle area.
The flux of E through the circle is Φ = λR/(2ε0). With λ = nC/m and R = cm, Φ = V m. The flux goes through the disc in the direction away from the thread when λ is positive.
Problem statement
Problem 3.19. A very long uniformly charged thread oriented along the axis of a circle of radius R rests on its centre with one of the ends. The charge of the thread per unit length is equal to λ. Find the flux of the vector E across the circle area.
Interpretation. The thread runs along the axis of the circle, perpendicular to the plane of the circle, and extends to infinity on one side. Its other end is exactly at the centre of the circle. The flux is that of E through the flat disc bounded by the circle.
We assume λ = 5 nC/m and R = 10 cm, because the problem gives no numbers. Both are sliders. The constant ε0 is 8.854 × 10−12 F/m. The unit of flux is N m²/C, which equals V m.
Symbols: λ is the charge per unit length, R the radius of the circle, r the distance of a point of the disc from its centre, Φ the flux.
The big idea in plain words
Every point of the disc sees the thread end on, from the same end point O. The field there, from problem 3.14, has one part along the thread and one part across it. Only the part along the thread goes through the disc. Add this up over thin rings of the disc.
The solution in each frame
The thread and the circle are at rest and nothing moves, so only one frame is needed.
The solution step by step
Step 1, the field on the disc. The circle is only an imaginary surface. All the charge is on the thread, which starts at O and runs up the axis. Take a point P of the circle at distance r from O. A bit of the thread of length dz at height z carries λ dz and is at distance ρ = √(r² + z²) from P. Its field at P points away from the bit and has size dE = λ dz/(4πε0ρ²). The part across the thread is dE r/ρ and the part along it is dE z/ρ. Adding all the bits from z = 0 to infinity, with the substitution z = r tan α, gives λ/(4πε0r) for each part. The along part does not cancel, because there is no thread below the circle. For a thread that passes right through the plane, the along parts of a bit above and a bit below would cancel and only the across part would be left. Here E = √2 λ/(4πε0r), at 45° to the thread.
Step 2, the part that gives flux. The disc is perpendicular to the thread. The normal to the disc points along the thread. So the flux density is the part along the thread: Ealong = λ/(4πε0r). The part across the thread lies in the plane of the disc and adds nothing. For positive λ the along part points away from the thread, so the flux goes through the disc in that direction.
Step 3, one ring. Take a thin ring of the disc between r and r + dr. Its area is 2πr dr. The flux through it is dΦ = Ealong 2πr dr = (λ/(4πε0r)) 2πr dr = λ dr/(2ε0). The r has cancelled: every ring of the same width carries the same flux.
Step 4, the whole disc. Φ = ∫ λ dr/(2ε0) from 0 to R = λR/(2ε0). With the numbers on screen, Φ = V m.
Step 5, the cases. The problem has one situation. If the sign of the area vector is taken the other way, the flux has the opposite sign: −λR/(2ε0). The size is the same. The flux grows in a straight line with R, even though the field grows large near the centre, because the area of a ring there is small.
A separate calculation adds the Coulomb fields of 3000 small pieces of the thread at each of 200 rings of the disc, then adds the flux through the rings. It compares the total with λR/(2ε0). The relative difference is .
Five methods
1. Force method
Find the field on the disc from the thread, then add the flux through the rings, as above.
2. Energy method
Not suited: flux is a surface integral of the field, not an energy.
3. Graph method
The first graph shows the field along the axis against r: a hyperbola. The second graph shows the flux of each ring per cm: a flat line. The third graph shows the flux up to r: a straight line, ending at λR/(2ε0).
4. Calculus method
The single integral in step 4.
5. Check method
Units: λR/ε0 is (C/m)(m)/(C²/(N m²)) = N m²/C, the unit of flux. Limit: as R shrinks to zero the flux goes to zero, and it doubles when R doubles.
Fastest: the ring integral with the cancelling r. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Charge per length λ | Yes | The field on the disc, the flux, the arrow colours and every curve |
| Radius R | Yes | The size of the disc, the total flux (proportional to R) and the range of every curve |
| Ring radius r | No | The highlighted ring, the field there, the flux so far and the dots on the graphs |
| Height z of the thread bit | No | Which bit of the thread is drawn in pane B, its distance to P and the two parts of its field. The total field does not change |
| Given-answer mode | Sets λ | The page finds λ for a chosen flux, and warns if it needs more than 20 nC/m |
| Play and ring-radius bar | No | Play grows the highlighted ring from the centre to the edge and holds 3 s at r = R/2, then repeats. The bar sets r |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Charge per length | λ | nC/m |
| Radius | R | cm |
| Flux per cm of ring width | λ/(2ε0) | V m per cm |
| Total flux | Φ | V m |
Exam traps
- Using the full size of E (the √2 result of 3.14). Only the part along the thread goes through the disc.
- Forgetting the area of the ring is 2πr dr. The r in it cancels the 1/r of the field.
- Thinking the flux is infinite because the field is large near the centre. The integral is finite.
- Dropping the sign. The flux is positive in the direction away from the thread when λ is positive, and negative if the area vector is chosen the other way.