Half the energy within radius √(ab)

Setup: a cylindrical capacitor of length L has an inner conductor of radius a and an outer cylinder of radius b. The charge per unit length is λ. We show that half the stored energy lies within radius r = √(ab).

   outer cylinder (radius b)
   +--------------------+
   |   +------------+   |
   |   |  inner a   |   |
   |   +------------+   |
   +--------------------+
   field region: a < r < b

Derivation

1. By Gauss' law with a coaxial cylinder of radius r (a < r < b): E (2πrL) = λL/ε₀, so E = λ / (2πε₀ r). The field is zero outside.
2. Energy density: u = ½ ε₀ E² = λ² / (8π² ε₀ r²).
3. Take a thin cylindrical shell of radius r, thickness dr, volume dV = 2πrL dr. The energy in it is dU = u dV = [λ² / (8π²ε₀r²)] · 2πrL dr = (λ² L / 4πε₀) dr / r.
4. Total energy stored (a to b): U = (λ²L / 4πε₀) ln(b/a).
5. Energy stored between a and some radius r: U(r) = (λ²L / 4πε₀) ln(r/a).
6. Require U(r) = U/2: ln(r/a) = ½ ln(b/a) = ln √(b/a) r/a = √(b/a)

r = √(ab)

So half the energy is stored within the cylinder of radius √(ab) (the geometric mean of a and b), and half outside it. Check with the capacitor formulas: U = q²/2C with C = 2πε₀L/ln(b/a), which gives the same total.