Lessons · circuits

The field of a charged rod: where algebra runs out

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 1 modeling assumptions (author-asserted)

A rod of length $L = 0.3$ m with linear charge density $\lambda = 100$ nC/m lies along the axis, its near end $a = 0.1$ m from the point where we want the field. Algebra gives the field of a point charge, $kq/r^2$ — but a rod is not a point, and there is no algebra formula for charge spread along a length. Calculus is the only way in: slice the rod into point charges $dq = \lambda\,dx$ , each contributing $k\,dq/x^2$ , and integrate. The shaded area under those contributions is the field $E = k\lambda(1/a - 1/(a+L))$ , which collapses to the point-charge $kQ/a^2$ when the rod is short.

Total field

E = \dfrac{kQ}{a(a+L)}

6742 N/C

Point-charge approx

kQ/a^2

(overestimate)

26970 N/C

Field from the near half of the rod

5394 N/C

Slice the rod into point charges and add their fields

dE = \frac{k\,dq}{x^2} = \frac{k\lambda\,dx}{x^2}

Cut the rod into slices of length $dx$ , each carrying charge $dq = \lambda\,dx$ and acting like a point charge at distance $x$ . Its field is $k\,dq/x^2$ . Summing over the rod is an integral — the only way to handle a continuous distribution.

check $dE/dx = k\lambda/x^2$ ; check $E = \int_a^x k\lambda/x'^2\,dx'$ ; total $kQ/(a(a+L))$ ; short-rod limit $kQ/a^2$

✓ The accumulated field's slope is the contribution density: $\dfrac{dE}{dx} = \dfrac{k\lambda}{x^2}$ . [structural]
✓ The field is the area under the contributions: $E(x) = \int_a^x \dfrac{k\lambda}{x'^2}\,dx'$ . [structural]
✓ The whole rod gives $E = k\lambda\left(\dfrac1a - \dfrac1{a+L}\right) = \dfrac{kQ}{a(a+L)}$ with $Q = \lambda L$ . [structural]
✓ For a short rod the field reduces to a point charge: $E/Q \to k/a^2$ as $L \to 0$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “To get the rod's field, just treat all its charge as sitting at the rod's centre (or near end) and use

kQ/r^2

.”

Lumping the charge at one point gives the wrong answer, because the field is non-linear in distance: the near slices ( $x$ small) contribute far more than the far ones ( $k/x^2$ falls off steeply), so you cannot replace the spread-out charge with a single point at the average position. Treating the rod as a point at its near end overestimates the field; the true field is the integral $E = k\lambda(1/a - 1/(a+L)) = kQ/(a(a+L))$ , which is genuinely smaller. Only when the rod is much shorter than its distance ( $L \ll a$ ) do the slices all sit at nearly the same distance and the integral collapses back to the point charge $kQ/a^2$ . The cursor sweeps how much of the rod is included; the area under the contribution curve is the field.

Modeling assumptions — author-asserted, disclosed not discharged

A thin rod with uniform linear charge density $\lambda$ , and the field evaluated at a point on the rod's axis beyond its near end (so every slice lies along the same line and the contributions add as scalars).

The dE/dx–x graph, fully annotated

A static rendering (Matplotlib): the shaded area under dE/dx is the accumulated integral E, and the slope of E is dE/dx. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):