Problem statement

Problem 3.15. A thread carrying a uniform charge λ per unit length has the configurations shown in Figure a and b. Assuming a curvature radius R to be considerably less than the length of the thread, find the magnitude of the electric field strength at the point O.

Interpretation. In figure (a) a long straight part comes down vertically, turns through a quarter circle of radius R, and goes off to the right as a second long straight part. The two straight parts meet the arc without a corner. O is the centre of the arc. In figure (b) two long parallel straight parts, a distance 2R apart, are joined at their lower ends by a semicircle of radius R. O is the centre of the semicircle. The page has one tab for each figure.

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

Symbols: λ is the charge per unit length, R the radius of the curved part, Θ the angle the curved part turns through, S1 and S2 the two straight parts.

The big idea in plain words

Split the thread into its straight parts and its curved part. Add their fields at O. A long straight thread that ends at the foot of a radius, with O on the perpendicular through that end, always gives the same field: size √2 λ/(4πε₀R) at 45° to the thread (this is problem 3.14). The curved part gives a field along the middle of the arc. Then only the directions decide the sum.

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, a straight part. Take a straight part that ends where the arc starts. O is at distance R from the thread, on the perpendicular through its end. From problem 3.14, its field at O has two equal parts, each λ/(4πε₀R): one along the thread and one across it. Together E = √2 λ/(4πε₀R) = kV/m, at 45° to the thread and pointing away from it.

Step 2, the arc. An element of the arc of length R dθ carries λ R dθ. Its field at the centre has size dE = λ R dθ/(4πε₀R²) = λ dθ/(4πε₀R). Measure θ from the bisector of the arc. By symmetry the parts across the bisector cancel. The parts along it add: E = (λ/(4πε₀R)) ∫ cos θ dθ from −Θ/2 to +Θ/2 = (λ/(4πε₀R)) × 2 sin(Θ/2). It points away from the arc.

Step 3, figure (a). The two straight parts are at right angles. Take O at the origin. The vertical part is on the left, at x = −R, and its field at O points right and down at 45°. The horizontal part is below, at y = −R, and its field points left and up at 45°. They are equal and opposite, so they cancel. The arc has Θ = 90°, so 2 sin 45° = √2 and E = √2 λ/(4πε₀R) = kV/m (with the numbers on screen). It points toward the upper right, away from the arc, for positive λ.

Step 4, figure (b). The two straight parts are parallel, at x = −R and x = +R. Each gives √2 λ/(4πε₀R) at 45°. The parts across the threads point in opposite directions and cancel. The parts along the threads both point down and add: 2 × λ/(4πε₀R) = λ/(2πε₀R) = kV/m. The semicircle has Θ = 180°, so 2 sin 90° = 2 and its field is 2λ/(4πε₀R) = λ/(2πε₀R), pointing up. The two are equal and opposite: E = 0.

Step 5, the cases. The problem has two figures and each has its own result: (a) √2 λ/(4πε₀R), (b) zero. The assumption that R is much smaller than the length of the thread is what lets the straight parts count as infinitely long.

Five methods

1. Force method

Add the fields of the straight parts and the arc, as above.

2. Energy method

This does not shorten the work: the potential of a semi-infinite thread is not finite, and the field at O would still need the slope in two directions.

3. Graph method

The first graph shows each part against R: all are proportional to 1/R. The second graph shows the arc formula against the angle: 90° gives √2 and 180° gives 2. The third graph shows the arc pieces added up to the angle β, measured along the bisector.

4. Calculus method

The integral of cos θ over the arc in step 2, and the thread integrals of problem 3.14.

5. Check method

Units: λ/(ε₀R) is N/C. In figure (b), all of the downward pushes come from the straight parts and the upward push from the semicircle, which is the same size, so zero is a sensible result. A full circle (Θ = 360°) gives 2 sin 180° = 0, as it should.

Fastest: use the result of 3.14 and the arc formula. Most reliable: the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Charge per length λ	Yes	Every field, every arrow length and the colour of the thread
Radius R	Yes	Every field (as 1/R), the size of the drawing and every curve
Arc angle β counted	No	How much of the arc is counted: the partial Ex and Ey, their resultant, the net field so far and the dots on the graphs. The final answer does not change
Figure tabs	Yes	Switch between figure (a) and figure (b)
Given-answer mode	Sets λ	The page finds λ for a chosen field of the curved part, and warns if it needs more than 20 nC/m
Play and angle bar	No	Play counts more and more of the arc and holds 3 s at the middle, then repeats. The bar sets β

Parameters

Quantity	Symbol	Value used
Charge per length	λ	nC/m
Radius	R	cm
Each straight part	E	kV/m
Curved part (this figure)	E	kV/m

Exam traps

• Ignoring the straight parts. They matter: in (a) they cancel and in (b) they cancel the semicircle.
• Treating the straight parts as if the field were only along the thread. Each also has an across part, equal in size.
• Adding the arc elements as numbers. Only the parts along the bisector add.
• Using the full-line field λ/(2πε₀R) for one straight part. A thread with one end gives √2 λ/(4πε₀R).