Electric Fields of Charge Distributions: Practice Questions (Physics C: E&M 8.4)

Answer: (D). Every small piece dq has a partner directly opposite it, at the same distance a from the centre. Their fields at the centre are equal in size and opposite in direction, so they cancel. Adding all the pairs gives E = 0.

• (A) adds the magnitudes k dq/a² of all the pieces as if every dE pointed the same way. E is a vector, so you must add components.
• (B) is the field at the centre of a semicircle of radius a carrying Q. The other half of the full ring cancels it.
• (C) is the largest field on the ring’s axis, found at z = a/√2, not at the centre.

Answer: (C). Integrating k λ ds/(distance)² along the rod gives E = kQ/[d(d + L)] = (8.99 × 10⁹)(4.0 × 10⁻⁹) ÷ [(0.10)(0.30)] = 35.96 ÷ 0.030 = 1.20 × 10³ N/C.

• (A) puts all the charge at the far end, 0.30 m away: kQ/(0.30)². Every piece is closer than that, so this is too small.
• (B) puts all the charge at the centre, 0.20 m away: kQ/(0.20)² = 899 N/C. Because of the 1/r² rule, the near half adds more than the far half takes away, so the true field is larger.
• (D) puts all the charge at the near end, 0.10 m away: kQ/(0.10)². Every piece is further than that, so this is too large.

Answer: (B). For the semicircle, E₀ = 2kλ/R. Each quarter contributes kλ/R along the semicircle’s symmetry line and kλ/R across it; the two across-components cancel. The single quarter keeps both of its components, kλ/R and kλ/R, at right angles. Its field is √[(kλ/R)² + (kλ/R)²] = √2 kλ/R = E₀/√2 ≈ 0.71E₀.

• (A) assumes the field is proportional to the amount of charge. That ignores direction: the remaining quarter’s across-component is no longer cancelled.
• (C) assumes the removed quarter added nothing to E₀. It added kλ/R along the symmetry line.
• (D) assumes removing charge doubles the field. Only the across-component stops cancelling.

1. (a) Every piece of the ring is r = √(0.030² + 0.040²) = 0.050 m from P. Using E = kQz/(z² + a²)^(3/2) = (8.99 × 10⁹)(5.0 × 10⁻⁹)(0.030) ÷ (0.050)³ = 1.08 × 10⁴ N/C, along the axis, away from the ring.
2. (b) kQ/r² = 1.80 × 10⁴ N/C is what you get by adding the sizes of all the dE vectors. By symmetry only their axial components survive, each multiplied by cos θ = z/r = 0.60. So E = 0.60 × 1.80 × 10⁴ = 1.08 × 10⁴ N/C. The components across the axis cancel in pairs.
3. (c) F = qE = (2.0 × 10⁻⁹)(1.08 × 10⁴) = 2.2 × 10⁻⁵ N, directed along the axis towards the ring’s centre, because the charge is negative.

Suggested mark points (3): 1 for E = 1.08 × 10⁴ N/C with direction along the axis away from the ring; 1 for explaining the cancellation of the components across the axis (or the factor cos θ = z/r); 1 for the force 2.2 × 10⁻⁵ N with the correct (towards the centre) direction.

Common error: using kQ/z² = 5.0 × 10⁴ N/C, which treats the ring as a point charge at its centre. P is closer to the ring’s centre than its radius, so this model is badly wrong.

1. λ = Q/L = 3.0 × 10⁻⁹ ÷ 0.60 = 5.0 × 10⁻⁹ C/m, and kQ = 26.97 N·m²/C.
2. (a) Exact: √(0.040² + 0.30²) = 0.3027 m, so E = 26.97 ÷ (0.040 × 0.3027) = 2.23 × 10³ N/C. Infinite line: E = 2(8.99 × 10⁹)(5.0 × 10⁻⁹) ÷ 0.040 = 2.25 × 10³ N/C, less than 1% high. P is very close to a long rod, so the rod looks infinite from P.
3. (b) Exact: √(3.0² + 0.30²) = 3.015 m, so E = 26.97 ÷ (3.0 × 3.015) = 2.98 N/C. Point charge: E = 26.97 ÷ 9.0 = 3.00 N/C, only 0.5% high. P is far from a short rod, so the rod looks like a point.
4. Both directions are perpendicular to the rod, away from it.

Suggested mark points (4): 1 for the exact value at 0.040 m; 1 for the infinite-line value and a comment that the model fits because x is much smaller than L; 1 for the exact value at 3.0 m; 1 for the point-charge value and a comment that the model fits because x is much larger than L.

Common error: swapping the models. The infinite-line model at 3.0 m gives 30 N/C, about ten times too big; the point-charge model at 0.040 m gives 1.7 × 10⁴ N/C, more than seven times too big.

(a) The upper quarter has length πR/2, so λ₁ = Q/(πR/2) = 2Q/(πR). A piece at angle θ (0 to π/2) has dq = λ₁R dθ, is a distance R from O, and pushes a positive test charge at O in the direction (−cos θ, −sin θ). So: E_x = −(kλ₁/R) ∫₀^(π/2) cos θ dθ = −kλ₁/R and E_y = −(kλ₁/R) ∫₀^(π/2) sin θ dθ = −kλ₁/R. With λ₁ = 2Q/(πR), E_x = E_y = −2kQ/(πR²).

(b) The lower quarter is the mirror image of the upper quarter in the x-axis. Reflecting the charge reflects its field: E_x stays the same and E_y changes sign. With 3 times the charge, each component is 3 times larger: E_x = −6kQ/(πR²) and E_y = +6kQ/(πR²).

(c) Add the components: E_x = −8kQ/(πR²) and E_y = +4kQ/(πR²). The magnitude is √(8² + 4²) kQ/(πR²) = 4√5 kQ/(πR²) ≈ 2.85kQ/R². The direction is tan⁻¹(4/8) = 26.6° above the −x direction, tilted away from the more heavily charged lower quarter. Check: if both quarters carried +Q, the y-components would cancel and E = 4kQ/(πR²) along −x, which is the semicircle result 2kλ/R.

(d) kQ/(πR²) = (8.99 × 10⁹)(3.0 × 10⁻⁹) ÷ [π(0.15)²] = 381.5 N/C, so E = 4√5 × 381.5 = 3.4 × 10³ N/C (3413 N/C), at 26.6° above the −x direction.

Accept a direct integral over the lower quarter (θ from −π/2 to 0, λ₂ = 6Q/(πR)) in (b).

(a) Q = ∫₀ᴸ βx dx = βL²/2.

(b) A piece at x has dq = βx dx and is a distance (x + d) from P. Every piece is to the right of P and positive, so every dE points in the −x direction (away from the rod). So:

E = kβ ∫₀ᴸ x dx/(x + d)²

Substitute u = x + d: ∫(u − d)/u² du = ln u + d/u. Evaluate from u = d to u = L + d:

E = kβ[ln((L + d)/d) + d/(L + d) − 1] = kβ[ln((L + d)/d) − L/(L + d)]

(c) Let ε = L/d, which is small. Then ln(1 + ε) ≈ ε − ε²/2 + ε³/3 and L/(L + d) = ε/(1 + ε) ≈ ε − ε² + ε³. The bracket is about ε²/2 − 2ε³/3, so E ≈ kβL²/(2d²) = kQ/d², a point charge.

(d) Q = (4.0 × 10⁻⁸)(0.50)²/2 = 5.0 × 10⁻⁹ C. The bracket is ln 6 − 0.50/0.60 = 1.7918 − 0.8333 = 0.9584. E = (8.99 × 10⁹)(4.0 × 10⁻⁸)(0.9584) = 3.4 × 10² N/C (345 N/C), in the −x direction. A uniform rod with the same Q gives kQ/[d(d + L)] = 44.95 ÷ (0.10 × 0.60) = 7.5 × 10² N/C. The non-uniform rod’s field is less than half of that, because most of its charge sits near x = L, far from P, while the uniform rod has more charge near P.

Carry forward an error in (b) into (d) once. Accept the bracket in the equivalent form ln(1 + L/d) − L/(L + d).