EM fields lecture 20 - arrangements of charged conductors and coefficients of potential p_ij

Overview

The lecture analyzes a charged conducting sphere of radius A inside a concentric conducting shell with radii B and C, deriving the charge distribution, electric field, and potentials in each region. It then introduces geometry-dependent potential coefficients, using the sphere-shell results to obtain a symmetric 2×2 coefficient matrix that maps arbitrary conductor charges to their potentials.

Key takeaways

For a sphere of charge Q1 inside a concentric shell of net charge Q2, the shell's inner surface carries −Q1 and its outer surface carries Q1 + Q2.
The electric field is zero inside either conductor, equals Q1/(4πε0r²) in the gap, and equals (Q1 + Q2)/(4πε0r²) outside the shell.
With potential zero at infinity, the inner sphere's potential is φ1 = Q1/(4πε0)(1/A + 1/C − 1/B) + Q2/(4πε0C).
Potential coefficients encode geometry: φi = Σj PijQj, and electrostatic reciprocity makes the coefficient matrix symmetric, Pij = Pji.
For the concentric sphere-shell arrangement, the three independent coefficients are P11 = (1/A + 1/C − 1/B)/(4πε0) and P12 = P21 = P22 = 1/(4πε0C).
An isolated conducting sphere has self-capacitance 4πε0A, the reciprocal of its potential coefficient 1/(4πε0A).

Chapters

0:00 Define the Concentric Sphere–Shell Geometry

Conductor 1 is a conducting sphere of radius A carrying charge Q1.
Conductor 2 is a concentric conducting shell with inner radius B, outer radius C, and total charge Q2.
The analysis divides space into the sphere interior, the A-to-B gap, the shell metal, and the exterior.

3:45 Determine the Electric Field in All Four Regions

Electrostatic equilibrium makes the field zero inside the sphere and within the conducting shell.
Between radii A and B, the radial field is E = Q1/(4πε0r²), as if the sphere's charge were at the center.
Outside radius C, the field is E = (Q1 + Q2)/(4πε0r²), set by the total enclosed charge.

7:05 Use Gauss’s Law to Fix the Surface Charges

Spherical symmetry distributes the sphere's Q1 uniformly over its surface area 4πA².
The shell's inner surface carries −Q1 so the field within the shell metal remains zero.
The shell's outer surface carries Q1 + Q2, leaving the shell's net charge equal to Q2.

10:07 Integrate the Exterior Field to Find the Shell Potential

The zero of potential is chosen at infinity, the standard reference for a finite charge distribution.
Integrating the exterior field from infinity to r = C gives the shell potential φ2 = (Q1 + Q2)/(4πε0C).
Because the electric field is zero in the shell metal, its entire thickness is at the same potential.

18:11 Find the Sphere-to-Shell Potential Difference

The potential difference φ1 − φ2 comes from integrating the field through the gap from B to A.
Only Q1 contributes to the gap field, giving φ1 − φ2 = Q1/(4πε0)(1/A − 1/B).
The potential is constant inside each conductor and varies as 1/r in the charge-free radial regions.

24:23 Combine Contributions to Obtain the Inner-Sphere Potential

Adding φ2 to φ1 − φ2 gives φ1 = Q1/(4πε0)(1/A + 1/C − 1/B) + Q2/(4πε0C).
The potential on the sphere depends on both conductor charges, even when the sphere itself has no charge.
Setting Q1 or Q2 to zero isolates each conductor's contribution; setting both to zero gives zero potential relative to infinity.

29:33 Introduce Potential Coefficients for Coupled Conductors

For two conductors, potentials are linear in charge: φ1 = P11Q1 + P12Q2 and φ2 = P21Q1 + P22Q2.
The off-diagonal coefficients are equal, P12 = P21, leaving three independent coefficients for two conductors.
The coefficients depend on conductor geometry, not on the particular charges or resulting potentials.

36:23 Calculate the Self-Coefficient of an Isolated Sphere

An isolated sphere of radius A has potential φ1 = Q1/(4πε0A).
Comparing this with φ1 = P11Q1 gives P11 = 1/(4πε0A), the sphere's potential coefficient.
The corresponding self-capacitance is its reciprocal: Cself = 4πε0A.

40:32 Extract Sphere–Shell Coefficients from Two Special Cases

With Q1 = 0, both conductors have potential Q2/(4πε0C), yielding P12 = P21 = 1/(4πε0C) and P22 = 1/(4πε0C).
With Q2 = 0, the sphere potential gives P11 = 1/(4πε0)(1/A + 1/C − 1/B).
The two special charge configurations recover all three independent coefficients for this geometry.

48:40 Assemble the Symmetric Potential-Coefficient Matrix

The matrix is P = (1/(4πε0)) [[1/A + 1/C − 1/B, 1/C], [1/C, 1/C]].
Multiplying this matrix by the charge vector (Q1, Q2)ᵀ gives the potentials (φ1, φ2)ᵀ for any charge values.
Once the geometry-dependent coefficients are known, potentials can be calculated without re-solving the field for each new charge arrangement.

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.