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The ring and the disc are the two continuous charge distributions that JEE problems return to again and again. Once you can derive them, you can handle half-rings, rotating rings and discs with holes. This post builds both results step by step, checks them in limiting cases, and works through numbers you can test yourself on.
Why are these derivations in the syllabus?
The NCERT Class 12 chapter Electric Charges and Fields introduces continuous charge distributions in section 1.12, defining linear, surface and volume charge densities by taking a small element that is tiny on a large scale but contains many charges. The Syllabus for JEE (Main) 2026 lists "superposition principle and continuous charge distribution" under Electrostatics. Check the syllabus for your exam year.
Why do students find these integrals hard?
Two studies explain where the trouble lies. Nguyen and Rebello (2011) interviewed 15 students in a calculus-based course and found that most recognised a problem needed an integral but failed to set it up, which the authors linked to not understanding the small term dq or the idea of adding up small contributions. Meredith and Marrongelle (2008) found that students do bring useful ideas, such as "parts of a whole", to electrostatics problems, but that naive ideas about limits can stop them seeing an integral as a sum.
So the fix is to say in words what each small piece is before you integrate.
How do you derive the field of a ring?
A thin ring of radius R carries charge Q spread evenly. Find the field at point P on the axis, a distance x from the centre.
Small piece. Take a tiny piece of the ring with charge dq. Every piece is the same distance from P: r = √(x² + R²).
Its field. dE = k dq / (x² + R²), pointing from the piece to P.
Symmetry. The piece on the opposite side of the ring has an equal field whose component perpendicular to the axis cancels. Only the axial component survives: dE_x = dE × cos α, where cos α = x / √(x² + R²).
Add. Every piece contributes the same dE_x, so summing dq gives Q:
E = kQx / (x² + R²)^(3/2), along the axis.
If this looks familiar, it should. NCERT's Moving Charges and Magnetism (section 4.5) uses the same symmetry argument for the magnetic field on the axis of a current loop, and gets the same (x² + R²)^(3/2) in the denominator.
Where is the ring's field strongest?
Since E is zero at the centre and falls to zero far away, it has a maximum. Setting dE/dx = 0 gives x = R/√2, and the maximum value is E_max = 2kQ / (3√3 R²).
Worked example. A ring of radius 0.10 m carries 10 nC. Take k = 9.0 × 10⁹ N m²/C².
At x = 0.10 m: x² + R² = 0.020 m², and (0.020)^(3/2) ≈ 2.83 × 10⁻³ m³. E = (9.0 × 10⁹ × 1.0 × 10⁻⁸ × 0.10) / (2.83 × 10⁻³) ≈ 3.2 × 10³ N/C.
The point-charge estimate kQ/x² would give 9.0 × 10³ N/C, almost three times too large. At this distance the ring is not a point.
The maximum is at x = 0.10/√2 ≈ 0.071 m, where E_max = 2 × 90 / (3√3 × 0.010) ≈ 3.5 × 10³ N/C.
How do you build the disc from rings?
A disc of radius R has surface charge density σ. Split it into thin rings of radius r and width dr.
Small piece. A ring of area 2πr dr has charge dq = σ 2πr dr.
Its field. Using the ring result: dE = k x σ 2πr dr / (x² + r²)^(3/2).
Limits. r runs from 0 to R.
Integrate. With the substitution u = x² + r², ∫ r dr / (x² + r²)^(3/2) = 1/x − 1/√(x² + R²). So
E = 2πkσ [1 − x / √(x² + R²)] = (σ / 2ε₀) [1 − x / √(x² + R²)].
Limits. As x → 0, or for a very large disc, E → σ/2ε₀, the field of an infinite plane sheet that our guide to Gauss's law problems derives in one line. Far away, a binomial expansion gives E ≈ kQ/x² with Q = σπR².
Worked example. σ = 1.0 × 10⁻⁶ C/m², R = 0.10 m, x = 0.050 m. Then σ/2ε₀ = 1.0 × 10⁻⁶ / (2 × 8.85 × 10⁻¹²) ≈ 5.65 × 10⁴ N/C. The bracket is 1 − 0.050/0.112 ≈ 0.553. So E ≈ 3.1 × 10⁴ N/C.
Which mistakes cost marks?
Forgetting the cos α. Without it, you add whole field vectors that point in different directions.
Using R instead of √(x² + R²) as the distance from the piece to P.
Integrating the ring formula over the disc with R inside. In the disc integral, the ring radius r is the variable; the disc radius R is a limit.
Treating a nearby ring as a point charge. Check whether x ≫ R before using kQ/x².
Losing the direction. For positive charge the field points away from the ring along the axis on both sides.
Our post on electrostatics mistakes covers more sign and direction traps.
A practice routine
Derive the ring result from a blank page twice this week.
Derive the disc result from the ring, then check both limits.
Sketch E against x for the ring and mark the peak at R/√2.
Try one variant: a disc with a hole of radius a (limits a to R).
Field patterns around rings and discs are easier to see than to read. The Fermi Simulation Lab has interactive simulations for ideas easier to see than read, a useful next step once you have sketched the field along the axis yourself.
Before your next practice set, derive E for a ring without notes and check that it gives zero at the centre.
