Exercise 3: Worked solutions

7 · Gauss’s law in the two regions

1. Spherical symmetry makes the field radial and equally strong everywhere on a concentric Gaussian sphere. Its area is 4πr², so the flux law reduces to:

   ∯ D · dA = D_r(r) 4πr² = Q_enc(r)

   Since D = ε₀E, solving this gives E_r once we know the enclosed charge.

2. For 0 < r ≤ a, the Gaussian sphere contains only a smaller volume of the uniform density. The spherical volume is 4πr³/3:

   Q_enc(r) = ρ₀4πr³3

3. Substitute that enclosed charge into Gauss’s law and divide by 4πε₀r²:

   E_r(r) = Q_enc4πε₀r² = ρ₀r3ε₀,   0 ≤ r ≤ a

4. For r > a, the Gaussian sphere encloses all the charge, whose volume is the full sphere of radius a:

   Q_total = ρ₀4πa³3;   E_r(r) = ρ₀a³3ε₀r²,   r > a

5. Result: for the positive density shown, the vector field points radially outward. Inside:

   E(r) = ρ₀r3ε₀e_r,   0 ≤ r ≤ a

   Outside, all charge is enclosed:

   E(r) = ρ₀a³3ε₀r²e_r,   r > a

   At the origin, E = 0; negative ρ₀ reverses the radial direction.

   Check: both formulas give E(a) = ρ₀a/(3ε₀), so the field is continuous at the surface. Far away it falls as 1/r², as if the total charge were concentrated at the centre. With SI inputs, E is in N/C.

8a · One voltage, two regions

1. The field is uniform and axial in each region. Moving from the lower plate at z = 0 to the upper plate at z = h gives the upper-minus-lower potential:

   U₁₀ = V₁ − V₀ = −∫₀^h E_z dz

2. The radial boundary at ρ = a₁ is parallel to the z-directed field. Its tangential field component is continuous, so the assumed axial components agree:

   E₁ = E₂ = E_z

3. Because each field is uniform over the gap, the integral is E_zh. Solve for the common component:

   E₁ = E₂ = −U₁₀h

4. Result: in the requested gap (not inside the metal), the field is the same everywhere:

   E = −U₁₀h e_z,   0 ≤ ρ ≤ a₂,   0 < z < h

   The same field applies on both sides of ρ = a₁ under the uniform-field approximation. The material changes D, not this tangential E.

   Check: E has units V/m. For positive upper-minus-lower voltage, potential decreases in the +z direction, so E points downward.

8b · Add the two material patches

1. The outward normal from the lower face of the upper metal plate into the gap is −e_z. The conductor boundary relation gives free surface charge density σ_f = D · n. Since E = −(U₁₀/h)e_z, each patch has positive σ for positive U₁₀.

2. Over the dielectric circle, D = ε₁E; its area is πa₁²:

   Q_core = σ₁πa₁² = ε₁U₁₀hπa₁²

3. Over the air annulus, D = ε₀E; its area is the whole plate disk minus the core disk:

   Q_air = σ₂π(a₂² − a₁²) = ε₀U₁₀hπ(a₂² − a₁²)

4. Add the two disjoint patches on the same metal face:

   Q = Q_core + Q_air = πU₁₀h[ε₁a₁² + ε₀(a₂² − a₁²)]

   Result: this is the charge on the underside of the upper plate. For positive U₁₀ it is positive; the lower plate carries the opposite charge.

   Check: each term has permittivity × voltage × area / length, which is coulombs. If a₁ = 0 the full plate is air; if a₁ = a₂ it is fully filled by the dielectric.

8c · Charge-to-voltage ratio

1. Capacitance is defined by C = Q/U. Divide the expression from 8b by U₁₀:

   C = πh[ε₁a₁² + ε₀(a₂² − a₁²)]

2. Result: this is the capacitance under the stated uniform-field model. The two regions behave like capacitors in parallel because they span the same voltage and their charges add.

   Check: it equals the sum of area/permittivity contributions divided by h; capacitance has units farads.

8d · Fixed-voltage energy change

1. Initially the capacitance is the value from 8c. After removal, the whole plate area is air:

   C_i = πh[ε₁a₁² + ε₀(a₂² − a₁²)];   C_f = ε₀πa₂²h

2. With voltage held fixed, stored field energy is W = ½C U². Subtract the initial energy from the final energy:

   ΔW_field = ½(C_f − C_i)U₁₀²

3. The air-region terms cancel, leaving the contribution from the removed core:

   ΔW_field = −πa₁²(ε₁ − ε₀)2hU₁₀²

4. Follow the energy that leaves. With U₁₀ fixed, the plate charge changes by ΔQ = (C_f − C_i)U₁₀ < 0, so charge flows back into the source. The energy the source delivers is W_source = U₁₀ΔQ = 2ΔW_field, a negative number: the source takes energy back. Whoever pulls the core out supplies the difference:

   W_mech = ΔW_field − W_source = −ΔW_field = πa₁²(ε₁ − ε₀)2hU₁₀²

5. Result: if ε₁ > ε₀, the stored field energy decreases by this amount (ΔW_field < 0), measured in joules. Positive mechanical work of the same size is needed to pull the core out: at fixed voltage the field draws the dielectric into the gap, and the source takes back twice the field-energy drop.

   Check: if the core is indistinguishable from air, ε₁ = ε₀, the change is zero. The magnitude scales with the removed area and with voltage squared.