A very long straight uniformly charged thread carries a charge λ per unit length. Find the magnitude and direction of the electric field strength at a point which is at a distance y from the thread and lies on the perpendicular passing through one of the thread’s ends.
The field at P has size E = √2 λ/(4πε0y) and makes an angle of 45° with the thread. Its two parts, along the thread and across it, are equal: each is λ/(4πε0y). With λ = nC/m and y = cm, each part is kV/m and E = kV/m.
Problem statement
Problem 3.14. A very long straight uniformly charged thread carries a charge λ per unit length. Find the magnitude and direction of the electric field strength at a point which is at a distance y from the thread and lies on the perpendicular passing through one of the thread’s ends.
Interpretation. The thread starts at an end point O and runs to infinity in one direction. P is at distance y from the thread, on the line that crosses the thread at right angles at O. So P is straight above the end of the thread.
We assume λ = 5 nC/m and y = 10 cm, because the problem gives no numbers. Both are sliders, and λ is positive in the first picture. The constant 1/(4πε0) is 8.988 × 109 N m²/C².
Symbols: λ is the charge per unit length, y the distance of P from the thread, x the distance of a small element from the end O, α the angle between the perpendicular OP and the line from P to the element.
The big idea in plain words
Cut the thread into small elements. Each makes a small field at P, pointing away from it. Every element pushes P a little along the thread (away from the thread) and a little across it. Add the two kinds of push separately. Both sums turn out to be the same number, so the total field leans exactly halfway, at 45°.
The solution in each frame
The thread is at rest and nothing moves, so only one frame is needed.
The solution step by step
Step 1, one element. An element of length dx at distance x from O carries the charge λ dx. P is at distance r = √(x² + y²) from it. The field has size dE = λ dx/(4πε0r²), and points along the line from the element to P. Call α the angle between this line and the perpendicular OP. Then tan α = x/y.
Step 2, change the variable. Put x = y tan α. Then dx = y dα/cos²α and r = y/cos α, so dx/r² = dα/y. Therefore dE = λ dα/(4πε0y). As the element goes from x = 0 to infinity, α goes from 0 to π/2.
Step 3, the part along the thread. The part of dE along the thread is dE sin α. So Ex = (λ/(4πε0y)) ∫ sin α dα from 0 to π/2 = λ/(4πε0y). For positive λ it points back past the end O, away from the charge.
Step 4, the part across the thread. The part across is dE cos α. So Ey = (λ/(4πε0y)) ∫ cos α dα from 0 to π/2 = λ/(4πε0y). It points away from the thread, along OP.
Step 5, add the parts. E = √(Ex² + Ey²) = √2 λ/(4πε0y) = kV/m. Because Ex = Ey, tan θ = 1 and the field makes 45° with the thread (and 45° with the perpendicular OP). For negative λ the size is the same and every direction is reversed.
Step 6, the cases. The problem has one situation. For comparison, a thread that runs to infinity on both sides gives only the across part, λ/(2πε0y), which is twice Ey here: the second half of the thread cancels the along part and doubles the across part.
A separate calculation adds the Coulomb fields of 20 000 small pieces of the thread, out to 10 000 times y. It compares both parts with the closed form for that length. The largest relative difference is .
Five methods
1. Force method
Add the field of every element, as above.
2. Energy method
The potential of a semi-infinite thread is not finite, so the potential itself is not useful. The difference of potential between two points is, but the field across the thread would still need its slope in the y direction and along x. This method does not shorten the work.
3. Graph method
The first graph shows the parts of the field made by elements out to X. Both curves climb to 1 as X grows: the part across climbs faster. The third graph shows the angle of the partial field. It starts at 90° (only the nearest elements, which push straight across) and settles at 45°.
4. Calculus method
The two integrals of sin α and cos α from 0 to π/2 in steps 3 and 4. For elements out to X the limits are 0 and arctan(X/y), which gives Ex = (λ/(4πε0y))(1 − 1/√(1 + X²/y²)) and Ey = (λ/(4πε0y)) (X/y)/√(1 + X²/y²).
5. Check method
Units: λ/(ε0y) is (C/m)/((C²/(N m²)) m) = N/C. Symmetry: for a thread on both sides, the along parts cancel and the across part doubles, matching the known result λ/(2πε0y).
Fastest: the angle substitution. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Charge per length λ | Yes | The field size, both parts, the arrow colours and every curve |
| Distance y | Yes | The height of P, the field (as 1/y) and every reading. The angle stays 45° |
| Horizontal distance X from O | No | How much of the thread is counted: the partial Ex and Ey arrows, their resultant, its angle and the dots on the graphs. The final field stays as a faint ghost |
| Given-answer mode | Sets λ | The page finds λ for a chosen field at the chosen y, and warns if it needs more than 20 nC/m |
| Play and distance bar | No | Play counts more and more of the thread as X grows, holds 3 s at X = y and 3 s at X = 5y (when they fit), then repeats. The bar sets X |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Charge per length | λ | nC/m |
| Distance | y | cm |
| Each part of the field | Ex, Ey | kV/m |
| Total field | E | kV/m |
Exam traps
- Using the formula of a thread that is infinite on both sides, λ/(2πε0y). Here the thread has an end, so the along part does not cancel.
- Forgetting that sin α and cos α both integrate to 1 between 0 and π/2, which makes the two parts equal.
- Giving only the size. The problem also asks for the direction: 45° with the thread, pointing away from it.
- Mixing up which part is which. The part along the thread comes from sin α, the part across from cos α.