Lessons · circuits

The field of a charged disk: between a point and a sheet

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 disk of radius $R = 0.3$ m carries a uniform surface charge density $\sigma = 100$ nC/m². We want the electric field a distance $z = 0.1$ m out along its axis. Algebra hands you two extremes — a point charge $kQ/z^2$ and an infinite sheet $\sigma/2\varepsilon_0$ — but a real disk is neither, and there is no algebra formula for one. Calculus is the only way in: slice the disk into concentric rings $dq = \sigma\,2\pi r\,dr$ , each contributing $k\,dq\,z/(z^2+r^2)^{3/2}$ on the axis, and integrate. The shaded area under those ring contributions is the field $E = \tfrac{\sigma}{2\varepsilon_0}(1 - z/\sqrt{z^2+R^2})$ — which collapses to the point charge far away and to the infinite sheet for a huge disk.

Total field

E = \dfrac{\sigma}{2\varepsilon_0}\!\left(1 - \dfrac{z}{\sqrt{z^2+R^2}}\right)

3862 N/C

Point-charge approx

kQ/z^2

(exact only far away; overestimates up close)

25420 N/C

Infinite-sheet limit

\sigma/2\varepsilon_0

(overestimate)

5649 N/C

Slice the disk into rings and add their on-axis fields

dE = \frac{k\,dq\,z}{(z^2+r^2)^{3/2}}, \qquad dq = \sigma\,(2\pi r\,dr)

Cut the disk into concentric rings of radius $r$ and width $dr$ , each carrying charge $dq = \sigma\,2\pi r\,dr$ . By symmetry only the axial part of each ring's field survives, $k\,dq\,z/(z^2+r^2)^{3/2}$ . Summing over the rings is an integral — the only way to handle a continuous distribution.

check $dE/dr = 2\pi k\sigma z\,r/(z^2+r^2)^{3/2}$ ; check $E = \int_0^r f\,dr'$ ; far limit $kQ/z^2$ ( $z\to\infty$ ); large-disk limit $\sigma/2\varepsilon_0$ ( $R\to\infty$ )

✓ The accumulated field's slope is the ring contribution: $\dfrac{dE}{dr} = \dfrac{2\pi k\sigma z\,r}{(z^2+r^2)^{3/2}}$ . [structural]
✓ The field is the area under the ring contributions: $E(r) = \int_0^r \dfrac{2\pi k\sigma z\,r'}{(z^2+r'^2)^{3/2}}\,dr'$ . [structural]
✓ Far away the disk looks like a point charge: $E \to kQ/z^2$ with $Q = \sigma\pi R^2$ as $z \to \infty$ . [structural]
✓ A very large disk is an infinite sheet: $E \to \sigma/2\varepsilon_0 = 2\pi k\sigma$ , independent of $z$ , as $R \to \infty$ . [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A charged disk is basically a flat sheet, so its field is just

\sigma/2\varepsilon_0

— or, if you prefer, lump all its charge at the centre and use

kQ/z^2

.”

Both are limits, not the answer. $\sigma/2\varepsilon_0$ is the field of an infinite sheet (or the disk in the limit $z \to 0$ ); for a finite disk at finite distance it is an overestimate, because the missing charge beyond radius $R$ would only have added more field. $kQ/z^2$ is the far-field limit ( $z \gg R$ ); up close it overestimates badly, because the inverse-square point model piles all the charge at one distance instead of spreading it from $r=0$ out to $R$ . The true field is the integral $E = \tfrac{\sigma}{2\varepsilon_0}(1 - z/\sqrt{z^2+R^2})$ , which lies below both extremes and only matches one of them in its own regime. Drag the cursor to sweep how much of the disk is included; the area under the ring-contribution curve is the field.

Modeling assumptions — author-asserted, disclosed not discharged

A flat disk with uniform surface charge density $\sigma$ , and the field evaluated on its central axis — so by symmetry the off-axis components of each ring cancel and only the axial field survives.

The dE/dr–r graph, fully annotated

A static rendering (Matplotlib): the shaded area under dE/dr is the accumulated integral E, and the slope of E is dE/dr. 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):