Two ways of calculating energy
(a) Let the shells have radii r₁ and r₂ (with r₁ < r₂). The electric field between them is E = Q/4π𝜀₀r², and it is zero elsewhere. The first method of calculating the energy of the system therefore gives
U = (𝜀₀/2) ∫ E² dv = (𝜀₀/2) ∫ᵣ₁ʳ² (Q²/4π𝜀₀r²)² 4πr² dr
= (Q²/8π𝜀₀) ∫ᵣ₁ʳ² dr/r² = (Q²/8π𝜀₀)(1/r₁ - 1/r₂). (12.189)
The magnitude of the potential difference between the shells is
𝜑 = ∫ᵣ₁ʳ² E dr = ∫ᵣ₁ʳ² (Q/4π𝜀₀r²) dr = (Q/4π𝜀₀)(1/r₁ - 1/r₂), (12.190)
which is the familiar result. We then quickly see that the second method of calculating the energy via Q𝜑/2 also yields the result in Eq. (12.189), as desired.
(b) The electric field is zero both inside the inner conductor (by the uniqueness theorem) and outside the outer conductor (see the discussion in the example in Section 3.2), so the energy is confined to the volume between the conductors. If we integrate the given identity over this volume, we obtain
∫ᵥ ∇·(𝜑∇𝜑) dv = ∫ᵥ (∇𝜑)² dv + ∫ᵥ 𝜑∇²𝜑 dv. (12.191)
In the first term we can use the divergence theorem to write the integral as a surface integral. In the second term (and in the first) we can use E = -∇𝜑. And the third term equals zero due to Poisson's equation, ∇²𝜑 = -ρ/𝜀₀, since there is no charge between the conductors. So we have
-∫ₛ 𝜑E·da = ∫ᵥ E² dv, (12.192)
where the surface S is the boundary of the volume V between the conductors. S consists of two parts: a boundary S₁ just outside the inner conductor, and a boundary S₂ just inside the outer conductor. The potential 𝜑 takes on a constant value over each of these boundaries, because they coincide with conductors. Let these two constant values be 𝜑₁ and 𝜑₂. Then the left-hand side of Eq. (12.192) becomes
-∫ₛ 𝜑E·da = -𝜑₁∫ₛ₁ E·da - 𝜑₂∫ₛ₂ E·da. (12.193)
Now, the surface S₁ encloses the charge Q on the inner conductor, so Gauss's law tells us that ∫ₛ₁ E·da = -Q/𝜀₀. (The minus sign comes from the fact that da was defined to point outward from our
660 Solutions to the problems
volume V, which means inward toward the inner conductor. This is opposite to the direction of da if we were considering S₁ to be a surface surrounding the inner conductor.) Likewise, the surface S₂ also encloses the charge Q on the inner conductor (this time with the standard orientation), so Gauss's law tells us that ∫ₛ₂ E·da = Q/𝜀₀.
Equation (12.193) therefore gives
-∫ₛ 𝜑E·da = (Q/𝜀₀)(𝜑₁ - 𝜑₂) ≡ Q𝜑/𝜀₀, (12.194)
where 𝜑 is defined to be the potential difference between the conductors. Plugging this result into Eq. (12.192) yields
Q𝜑/𝜀₀ = ∫ᵥ E² dv ⇒ (1/2)Q𝜑 = (𝜀₀/2)∫ᵥ E² dv, (12.195)
as desired. If we instead have two conductors (with charges Q₁, Q₂ and potentials 𝜑₁, 𝜑₂) inside a third conductor (with charge -Q₁ - Q₂ and potential 𝜑₃), then you can show that the above procedure leads to
(1/2)Q₁(𝜑₁ - 𝜑₃) + (1/2)Q₂(𝜑₂ - 𝜑₃) = (𝜀₀/2)∫ᵥ E² dv. (12.196)
You should convince yourself that the left-hand side is the energy required to transfer charge from the outer conductor to the two inner conductors, in the correct proportion. If Q₁ = -Q₂ we have a standard two-conductor capacitor, with (1/2)QΔ𝜑 on the left.