AP® Physics C: Electricity and Magnetism · Unit 8 · Topic 8.4

The Charged Disk and the Infinite Sheet Limit

The field on the axis of a charged disk comes from adding up the fields of many thin rings. As the disk's radius grows without bound, the formula collapses to a constant, sigma/(2 eps0), the field of an infinite sheet.

Start with a ring of charge, since you already know its axial field. Slice the disk into many concentric rings, each of radius r, thickness dr, and charge dq = sigma(2 π r dr). Each ring contributes a sliver of field along the axis, and because every ring's field points the same direction on the axis, you just add their magnitudes with an integral instead of worrying about vector components. Integrating from r = 0 to r = R gives E = (sigma / (2 eps0)) (1 - x / √(x2 + R2)), where x is the distance along the axis from the disk's center.

Look at what that formula does as R grows huge compared to x. The fraction x / √(x2 + R2) shrinks toward zero, since the R2 term swamps everything else, leaving E = sigma / (2 eps0). The distance x drops out completely. That is the mathematical reason a sheet's field does not weaken as you back away from it: at any finite distance from an infinite plane, the plane still looks infinite, so the geometry connecting you to the charge never changes.

This is also where Gauss's law earns its keep. Instead of grinding through the ring-by-ring integral, you can picture a pillbox straddling the sheet, note that flux only exits through the two flat end caps, and solve E(2A) = sigma A / eps0 in one line. Both routes land on sigma/(2 eps0): the disk calculation shows you why the number is true, and Gauss's law shows you how fast you can get there once the symmetry is good enough.

Key terms

Surface charge density (sigma)
Charge per unit area on a two-dimensional surface, measured in coulombs per square meter.
Axial field
The electric field evaluated at points lying along the central perpendicular axis of a symmetric charge distribution.
Infinite sheet limit
The approximation where a charged plane is treated as having no edges, making its field constant and independent of distance.
Gauss's law
A law stating that the electric flux through a closed surface equals the enclosed charge divided by epsilon0.
Pillbox surface
A cylindrical Gaussian surface straddling a charged sheet, used to extract the field from Gauss's law by symmetry.

Watch it explained

  1. How it worksWalk me through building a charged disk out of nested rings and integrating their fields.
  2. Worked exampleFind the field 2 cm along the axis of a 10 cm disk carrying 4 microcoulombs per square meter.
  3. Going furtherWhat happens to the disk's field formula when its radius grows without any bound?
  4. Ties togetherHow does the disk result connect to the sheet field Gauss's law gives you in one line?

Check yourself

Your turn1 of 2
A disk of radius 22 cm carries a uniform surface charge density of 13 microC per square meter. Find the magnitude of the electric field 14 cm along the disk's axis from its center. Use epsilon0 = 8.85x10−12 C2/(N m2).

End of topic.