Forces from Charge Distributions

A 24 cm rod holding 8.0 nC carries 8.0 × 10⁻⁹ C ÷ 1.602 × 10⁻¹⁹ C = 5.0 × 10¹⁰ elementary charges. Spread along the rod, their mean spacing is about 5 × 10⁻¹² m. Probe the rod from a millimetre away and that staircase of lumps is indistinguishable from a smooth ramp, so we describe the object by a density instead: λ = dq/dl in C m⁻¹ for a wire, σ = dq/dA in C m⁻² for a sheet, ρ = dq/dV in C m⁻³ for a solid. Which one you use is set by the shape of the object, not by the problem. A density need not be uniform: λ(x) = λ₀ x/L is a rod whose charge grows towards one end, and only a uniform density may be pulled outside the integral as a constant.

The whole method is five lines, in this order. Choose one coordinate that labels position on the source — x along a rod, angle φ around a ring, radius s on a disc. Write the element in it: dq = λ dx, or dq = σ (2πs ds) for an annulus of a disc. Write the separation r from that element to the test charge as a function of that same coordinate. Write the direction of dF, which lies along the line from dq to q. Only then choose limits, which come from the geometry of the source. The commonest setup error is to treat r as a single fixed number, the distance to the rod. The test charge's position is what stays constant through the integration; r does not.

A ring of radius a and charge Q, with q on the axis at distance z, is the clean case. Every element lies the same distance r = √(z² + a²) away, and each dF tilts towards the ring by the same angle. Pair each element with the one diametrically opposite: their off-axis components are equal and opposite, so the whole perpendicular part cancels exactly. What survives is dF cos θ with cos θ = z/r, and because r and cos θ are the same for every element, both leave the integral and ∫dq = Q is all that remains: F = k q Q z/(z² + a²)³⁄². With a = 5.0 cm, Q = 30 nC, q = 2.0 nC and z = 12 cm this gives 29.5 μN, only 79% of the 37.5 μN you would get by lumping Q at the centre, because every element pulls at an angle.

Move the test charge onto a rod's own axis and the direction is constant but the distance is not. With the origin at q, the rod runs from x = d to x = d + L, each element contributes k q λ dx/x², and integrating dx/x² from x = d to x = d + L gives F = k q λ (1/d − 1/(d + L)) = k q Q/(d(d + L)) once λ = Q/L is used. Because the geometric mean √(d(d + L)) never exceeds the arithmetic mean d + L/2, the rod always exerts more force than the same charge parked at its midpoint would. Put the charge on the perpendicular bisector instead and both distance and angle vary: F = 2 k q λ ∫₀^(L/2) z dx/(x² + z²)³⁄², which the substitution x = z tan θ turns into a cosine integral, giving F = k q λ L/(z √(z² + L²/4)).

An integral is not finished until its limits are checked. Let d ≫ L in the rod result and d(d + L) → d², so F → k q Q/d²: far away, any bounded distribution looks like its total charge at a point. Let z ≫ a in the ring result and (z² + a²)³⁄² → z³, giving the same law. Now go the other way on the bisector result. Let z ≪ L, so √(z² + L²/4) → L/2 and F → 2 k q λ/z, an inverse-first-power law. That does not contradict Coulomb's law. Only the stretch of line lying within roughly z of the foot of the perpendicular contributes appreciably, so halving z brings each of those elements four times closer while halving how much of the line still counts, and 4 × ½ = 2. Dimensions confirm the result: [k q λ] = N m, so k q λ/z is a force.

Three failures are worth naming. The 2 k q λ/r line law diverges as r → 0, which is the idealisation admitting that it has no thickness; a real rod of radius R keeps the force finite, because at r < R the charge no longer sits all at one small distance. The density itself fails once r drops below the spacing of the actual charges, about 5 × 10⁻¹² m for the rod above, where the smooth ramp turns back into a staircase. And a fixed λ, σ or ρ assumes the source cannot respond. That holds for charge sprayed onto an insulator; on a conductor the approaching test charge redistributes the very charge you integrated over, and the problem stops being a single integral.