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The field of a point charge is one line: kq/r². The field of a charged rod is not, because every bit of the rod is at a different distance and pulls in a different direction. The JEE and NEET syllabuses both list continuous charge distributions, and the finite rod is the standard test. This post sets up the integral for a point beside the rod, solves it with an angle substitution, and checks the answer against cases you already know.
Why is this in the syllabus?
The JEE (Main) 2026 syllabus lists the "superposition principle and continuous charge distribution" in its electrostatics unit, and the NEET (UG) 2026 syllabus uses almost the same words. The NCERT Class 12 chapter on Electric Charges and Fields introduces linear charge density, λ = charge per unit length, for exactly this kind of problem.
Why do students find the set-up hard?
Doughty, McLoughlin and van Kampen (2014) studied cohorts of about 50 students in an intermediate electromagnetism course. They found students saw integration mainly as "evaluation", a button to press, rather than as adding up small pieces. Technical slips meant almost no student reached a fully correct answer to a typical integration problem. The fix they describe is to treat each integral as a sum you have built. That is what the four steps below do.
How do you set up the integral?
Place the rod along the x-axis. Point P is at perpendicular distance d from the rod's line, and the foot of the perpendicular is the origin. The rod carries uniform charge density λ.
Pick a small piece. Take a length dx at position x. Its charge is dq = λ dx.
Write its field. Its distance from P is r = √(d² + x²), so dE = kλ dx / r², pointing along the line from the piece to P.
Split into components. Let φ be the angle between that line and the perpendicular. The perpendicular part is dE cos φ, and the part along the rod is dE sin φ.
Change variable. Use x = d tan φ, so dx = d sec²φ dφ and r² = d² sec²φ. Then dE cos φ = (kλ/d) cos φ dφ. Everything except φ has cancelled.
Integrating from −θ₁ to θ₂, where θ₁ and θ₂ are the angles from the perpendicular to the two ends:
E⊥ = (kλ/d)(sin θ₁ + sin θ₂)
E∥ = (kλ/d)(cos θ₁ − cos θ₂), where θ₂ is the larger angle. It points along the rod, away from the longer part.
Worked example 1: on the perpendicular bisector
A rod of length 0.30 m carries 6.0 nC spread evenly. Find the field at a point 0.20 m from its centre on the perpendicular bisector. Take k = 9 × 10⁹ N m²/C².
λ = 6.0 × 10⁻⁹ / 0.30 = 2.0 × 10⁻⁸ C/m, so kλ = 180 N m/C.
Each half is 0.15 m long. The distance to an end is √(0.15² + 0.20²) = 0.25 m, so sin θ = 0.15/0.25 = 0.6.
By symmetry the components along the rod cancel.
E = (180/0.20)(0.6 + 0.6) = 900 × 1.2 = 1080 N/C, pointing directly away from the rod.
Worked example 2: opposite one end
Same rod, but P is 0.40 m from one end, on the perpendicular through that end. Now θ₁ = 0 and tan θ₂ = 0.30/0.40, so sin θ₂ = 0.6 and cos θ₂ = 0.8.
kλ/d = 180/0.40 = 450 N/C.
E⊥ = 450 × (0 + 0.6) = 270 N/C.
E∥ = 450 × (1 − 0.8) = 90 N/C, pointing away from the rod along its length.
|E| = √(270² + 90²) ≈ 285 N/C.
Students often forget the second component here, because in example 1 it vanished.
How do you check the answer?
Push the formula to cases you already know.
Far away. In example 1, if d is much larger than the rod, sin θ ≈ (L/2)/d and E → kλL/d² = kQ/d². Even at 0.20 m, a point charge would give 1350 N/C, close to 1080 N/C and larger, as it should be, because the real charge is spread further away.
Very long rod. As θ₁ and θ₂ both approach 90°, E → 2kλ/d, the infinite-line result from Gauss's law. For example 1 that would be 1800 N/C. Our post on Gauss's law problems derives it the other way.
Units. kλ/d has units of (N m²/C²)(C/m)/m = N/C.
The point on the rod's own axis is a separate case with no angles; it is worked in our guide to the calculus you need for JEE physics.
Which mistakes cost marks?
Integrating the magnitude dE without splitting it. Components must be added, not magnitudes.
Wrong limits. Measure each angle from the perpendicular, with sign. A point beyond the end of the rod has both ends on the same side, so the perpendicular part becomes (kλ/d)(sin θ₂ − sin θ₁).
Using d as the distance to every piece. Only the nearest piece is at distance d.
Dropping λ. Write dq = λ dx, then put in λ = Q/L at the end.
A practice routine
Re-derive E⊥ and E∥ from a blank page, using the four steps, twice this week.
Solve three cases: P on the bisector, P opposite one end, and P beyond one end.
For each, check the far-away and long-rod limits.
Finish with a ring and a disc, from our post on the field of a ring and a disc, using the same four steps.
If you want to see how the field arrows change as P moves along the rod, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Tonight, redo example 2 with P at 0.30 m instead of 0.40 m, and check that |E| grows.
