EM fields lecture 18 - electric potential of charged sphere & charged disk and choosing V=0

Overview

Tim Gorringe derives the potential of a uniformly charged solid sphere by adding concentric spherical shells, then derives the on-axis potential of a uniformly charged disk by integrating cylindrical charge elements. The examples show how potential relates to electric field and why finite charge distributions conventionally use V = 0 at infinity, while infinite sheets require a chosen reference such as the sheet itself.

Key takeaways

A uniformly charged solid sphere can be assembled from shells with dq = 4πρr²dr, giving V = Q/(4πε₀z) outside and V = ρ(3a² − z²)/(6ε₀) inside.
The sphere’s potential is continuous at its surface, while its field changes from linear growth inside, E = ρz/(3ε₀), to inverse-square decay outside.
The on-axis potential of a uniformly charged disk is V(z) = σ[√(z² + R²) − |z|]/(2ε₀), obtained by integrating cylindrical area elements.
Far from a disk, its potential becomes that of a point charge with Q = σπR²; close to a large disk, its field approaches the constant infinite-sheet value σ/(2ε₀).
Adding a constant to potential does not change the electric field, so finite distributions can use V = 0 at infinity while infinite charge arrangements use a convenient reference surface or point.

Chapters

0:00 Potential Examples and the Choice of V = 0

The class builds on earlier results for an electric dipole and a thin spherical shell.
It compares potentials and fields for a solid sphere, a uniformly charged disk, and infinite charge arrangements.
For localized finite charge, the usual reference is V = 0 at infinity; infinite sheets or lines instead need a convenient reference point or surface.

2:35 Building a Uniform Solid Sphere from Spherical Shells

A sphere of radius a and volume charge density ρ is centered at the origin; the field point lies a distance z from the center.
The sphere is decomposed into thin concentric shells of radius r and thickness dr, each carrying dq = ρ·4πr²dr.
The shell results from the previous class provide the building blocks: outside a shell its potential is that of a point charge, while inside it is constant.

8:22 Outside the Sphere: Total Charge Acts Like a Point Charge

For z > a, the field point is outside every shell, so each shell contributes dq/(4πε₀z).
Superposition over r from 0 to a gives V(z) = Q/(4πε₀z), where Q = ρ(4πa³/3).
The result follows the convention V → 0 at infinity and matches the potential of a point charge with the sphere’s total charge.

13:45 Inside the Sphere: Split the Shell Integral at the Field Point

For z < a, shells with r < z contribute as external shells, while shells with r > z contribute their constant interior potential.
The two integrals run over radii 0 to z and z to a, respectively, using the shell charge element ρ·4πr²dr.
Combining the terms gives V(z) = ρ(3a² − z²)/(6ε₀) inside the sphere.

21:05 Sphere Potential and Field: Continuity at the Surface

At z = a, the interior and exterior potential expressions agree, so the potential is continuous at the sphere’s surface.
The interior potential is largest at the center and decreases quadratically with distance; outside it falls as 1/z.
Taking the negative gradient gives a field that grows linearly inside, E = ρz/(3ε₀), and decreases as 1/z² outside.

24:08 Setting Up the Disk Potential with Cylindrical Charge Elements

The disk has radius R and uniform surface charge density σ; the potential is evaluated on its central axis at height z.
A cylindrical area element is dA = r·dr·dφ, so its charge is dq = σr·dr·dφ.
Each element is a distance √(z² + r²) from the field point, and integrating its point-charge potential over the disk gives the total.

35:14 Disk Potential Integral and the Point-Charge Limit

Integrating over φ from 0 to 2π and then over r from 0 to R yields V(z) = σ[√(z² + R²) − |z|]/(2ε₀).
For |z| ≫ R, a binomial expansion reduces the expression to V ≈ σR²/(4ε₀|z|).
Since the disk’s total charge is Q = σπR², the far-field result is Q/(4πε₀|z|), as expected for a point charge.

42:36 Near the Disk: Infinite-Sheet Field and Reference Potential

For |z| ≪ R, the square-root term is nearly constant; after subtracting that constant, the potential change is approximately −σ|z|/(2ε₀).
The resulting electric field has constant magnitude σ/(2ε₀) near the disk, pointing away from the sheet for positive σ.
Equipotential planes are parallel to the sheet, and choosing V = 0 on the sheet is convenient because an infinite sheet cannot use a zero-at-infinity reference.

47:03 Finite Versus Infinite Charge and the Uncovered Line Example

For finite distributions such as the sphere, the potential reference is conventionally set to zero infinitely far away.
For an infinite sheet—or an infinite line—the reference is instead chosen at a convenient location; the sheet example sets V = 0 on the sheet.
The line-charge calculation is deferred, but it follows the same reference-choice principle as the infinite sheet.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, tim gorringe.