Gauss' law for D

Gauss' law, spherical symmetry

Problem:

A spherical capacitor with conducting surfaces of radii R₁ and R₂ has a material of dielectric constant
ε(r) = ε₀(R₁/r)²
between the spheres.
(a)  Find the capacitance C of the capacitor.
(b)  If the charge on the capacitor is Q find the total energy stored in the capacitor.

Solution:

• Concepts:
  Gauss' law for D.
• Reasoning:
  The problem has spherical symmetry.  We can find the field between the surfaces in terms of the charge Q_free on the surfaces using Gauss' law for D.
• Details of the calculation:
  (a)  Symmetry:
  D = D (r/r),  E = E (r/r).  
  if the origin is located at the center of the spheres.
  Gauss' law for D:  D4πr² = Q_{free inside} in SI units.

  Free charges reside only on the surfaces of the capacitor.  Between the surfaces we have
  D = Q_free/( 4πr²),  D(r) = ε(r)E(r),
  V = ∫_R1^R2 E(r)dr = ∫_R1^R2 dr D(r)/ε(r) = ∫_R1^R2 dr Q_free/(4πε₀R₁²).
  V = Q_free(R₂ - R₁)/(4πε₀R₁²).
  C = Q_free/V = 4πε₀R₁²/(R₂ - R₁) is the capacitance of the capacitor.
  (b)  U = ½Q²/C = Q²(R₂ - R₁)/(8πε₀R₁²) is the total energy stored in the capacitor.

Problem:

A spherical conductor of radius a is surrounded with a dielectric shell of outer radius b.  The dielectric constant varies with radius as K = 1 + n(r - a), where n is a constant.  A charge Q is placed on the conductor.
(a)  Find the electric displacement and the electric field at all points in space.
(b)  Find the distribution of bound charges in the interior of and on the surface of the dielectric shell.
(c)  Find the total energy stored in the system.

Solution:

• Concepts:
  Gauss' law for D,  E = D/ε,
• Reasoning:
  The problem has spherical symmetry.  Gauss' law for D alone can be used to find D everywhere.
• Details of the calculation:
  (a)  Inside the conductor D and E are zero. 
  Outside the conductor (r > a) we have D = Q/(4πr²), D points radially away from the origin, if the center of the conducting sphere is placed at the origin.
  E = D/ε = D/(ε₀K) = D/(ε₀ + ε₀n(r - a)) for a < r < b. 
  E = [Q/(4πε₀r²)](1/(1 + n(r - a))) = [Q/(4πε₀)]/(nr³ + (1 - na))r²).
  E = D/ε₀ for r > b.  E = Q/(4πε₀r²).
  (b)  To find the surface charge distributions we use   (SI units).
  Surface at r - a:  σ = [Q/(4πa²)](1/(1 + n(a - a))) = Q/(4πa²).
  σ_bound = σ - Q/(4πa²) = 0.
  Surface at r - b:  σ_bound = Q/(4πb²) - [Q/(4πb²)](1/(1 + n(b - a))) 
  = [Q/(4πb²)](n(b - a)/(1 + n(b - a)).
  To find the volume charge distribution we use ρ_bound = -∇∙P,  P = ε₀(K - 1)E.
  For a < r < b we have P = ε₀n(r-a)E = [nQ/(4πr²)](r - a)/(nr + 1- na).
  ∇∙P = (1/r²)(∂/∂r)(r²P)
  ρ_bound = -[nQ/(4πr²)][1/(nr + 1 - na)²].
  (c)  In the presence of a dielectric, the total work done in assembling the free charges into a charge distribution is

  W = ½∫_{all space} E∙D dV.

  W = ½4π∫_a^br²dr [Q/(4πε₀r²)](1 + n(r - a))^-1 [Q/(4πr²)]
  + ½4π∫_b^∞r²dr [Q/(4πε₀r²)][Q/(4πr²)].
  W = [Q²/(8πε₀)]∫_a^bdr [r²(1 + n(r - a))]^-1 + [Q²/(8πε₀b)]
  = [Q²/(8πε₀)][-1/((1 - na)r) + (n/(1 - na)²)ln((1 - na + nr)/r)]_a^b + [Q²/(8πε₀b)].
  W = [Q²/(8πε₀)][1/b + (b-a)/((1 - na)ab) + (n/(1 - na)²)ln((1 - na + nb)a/b)].

Problem:

A linear dielectric sphere of radius a and dielectric constant k_e carries a uniform charge density ρ, surrounded by vacuum.
(a)  Find E and D inside and outside the sphere.
(b)  Find the energy U of the system.

Solution:

• Concepts:
  Gauss' law, relationship between E, and D, energy in dielectrics
• Reasoning:
  We find D from Gauss' law for D.  We then use D to find E and U.
• Details of the calculation:
  (a)  E and D are radial.
  r < a:  D(r)4πr² = 4πr³ρ/3, D(r) = rρ/3.  E(r) = D(r)/(ε₀k_e) = ρr/(3ε₀k_e) = Q_totr/(4πε₀k_ea³)
  r > a:  D(r)4πr² = 4πa³ρ/3,  D(r) = a³ρ/(3r²) = Q_tot/(4πr²).   E(r) = a³ρ/(3ε₀r²) = Q_tot/(4πε₀r²).
  (b)  The energy U of the system is U = ∫_{all space}(ε/2)E²dV.
  U = 2πε∫₀^aE²r²dr + 2πε₀∫_a^∞E²r²dr =  [Q_tot²/(8πε₀)][1/(5ak_e)  + 1/a]
  = [2ρ²π/(9ε₀)][a⁵/(5k_e) + a⁵].

Gauss' law, cylindrical symmetry

Problem:

A capacitor is made of two concentric cylinders of radii r₁ and r₂ (r₁ < r₂) and length L >> r₂.
The region between r₁ and r₃ = (r₁r₂)^½ is filled with a circular cylinder of length L and dielectric constant k.
The remaining volume is an air gap.
(a)  What is the capacitance?
(b)  What are the values of E, P, and D at a radius r inside the dielectric (r₁ < r < r₃)?  Assume a potential difference V  between r₁ and r₂.
(c)   How much mechanical work must be done to remove the dielectric cylinder while maintaining this constant potential difference between r₁ and r₂?

Solution:

• Concepts:
  Gauss' law, relationship between E, P, and D, capacitance, U = ½CV².
• Reasoning:
  For a given charge on the cylinder we find D from Gauss' law for D.  We then use D to find E, P, V, and C.
• Details of the calculation:
  (a)  D = (λ_f/(2πr)) (r/r) (r₁ < r < r₂) from Gauss' law for D.  Here r is a cylindrical coordinate and we use SI units.
  ΔV = ∫_r1^r2 E(r)dr = ∫_r1^r3 dr D(r)/(ε₀k) = ∫_r1^r2 dr D(r)/ε₀
  =  (λ_f/(2πε₀k))[ln(r₃/r₁) + k ln(r₂/r₃)]
  =  (λ_f/(2πε₀k))[ln√(r₂/r₁) + k ln(√r₂/r₃)]
  = (λ_f(1 + k)/(2πε₀k))ln√(r₂/r₁).
  C = Q_f/ΔV = λ_fL/ΔV = [4πε₀kL/(k + 1)][1/ln(r₂/r₁)].
  (b)  D = (λ_f/(2πr))(r/r), E = (λ_f/(2πε₀kr)) (r/r),
  P = ε₀χ_eE = ε₀(k - 1)E = (λ_f/(2πr))[(k - 1)/k] (r/r).
  We can write this in terms of ΔV using the result of part (a).
  (c)  The energy stored in the the capacitor is U = ½CV².
  Without the dielectric the capacitance is C_out = 2πε₀L/ln(r₂/r₁) < C.
  The energy stored in the capacitor decreases by ΔU = ½(C_out - C)V².
  The charge on the capacitor is Q = CV.
  When the dielectric is removed, ΔQ = (C_out - C)V is removed from the cylinders.
  An amount of (negative) work W = ΔQ V = (C_out - C)V² is done by the battery.
  The energy stored in the capacitor decreases only by ½(C_out - C)V².
  The mechanical work done must therefore equal W = ½(C - C_out)V².
  W =  [2πε₀L/ln(r₂/r₁)]V²(k - 1)/(k + 1).

Problem:

A 500 m length of high-voltage cable is undergoing electrical testing.  The cable consists of two coaxial conductors, the inner of 5 mm diameter and the outer of 9 mm internal diameter.  The space between the conductors is filled with polythene which has a relative permittivity of 2 and which can withstand electric field strength of 60 MVm^-1.
(a)  Find the maximum voltage which can be applied between the conductors and the energy stored in the cable at this voltage.
(b)  If the cable is to be discharged to a safe level of 50 V in 1 minute, what value of resistance must be connected across it?  What is the maximum power and the total energy dissipated in the resistance?

Solution:

• Concepts:
  Gauss' law, the cylindrical capacitor
• Reasoning:
  The problem has enough symmetry to find D(r) and E(r) from Gauss' law alone.
• Details of the calculation:
  (a)  From Gauss' law: D(r), E(r ) ∝  1/r between the cylinders.
  Let the outer cylinder be grounded and V₀ be positive.  Then the direction of E is radially outward. 
  E(r) = A/r, with A positive.  Then V₀ = -∫_b/2^a/2E(r)dr = A∫_a/2^b/2(1/r)dr
  = A(lnb/2 - lna/2) = Aln(b/a).  A = V₀/ln(b/a),
  with a = 5 mm and b = 9 mm.
  E(r) = V₀/(r ln(b/a)).  
  If E_max = E(a/2) = 2V₀/(a ln(b/a)) = 60 MV/m, then
  V₀ = (6*10⁷ V/m)*(2.5*10^-3 m)*ln(9/5) = 8.82*10⁴ V is the maximum voltage which can be applied.
  Energy stored in the cable: 
  W = (ε/2)∫E²dV = (εL/2)2π∫_a/2^b/2E²rdr = επL∫_a/2^b/2(V₀/(rln(b/a)))²rdr.
  W = [επLV₀²/ln(b/a)] = [2πε₀(500 m)*(8.82*10⁴ V)²/ln(9/5)] = 368 J.
  (b)  The cable is a capacitor.  W= U = ½CV², C =  2επL/ln(b/a).
  Discharging the cable through a resistor R amounts to amounts to discharging a RC circuit wit time constant τ = RC.  If the voltage decreases from V₁ to V₂ , the energy stored in the capacitor decreases by an amount ΔU = ½C(V₂² - V₁²). 
  V = V₁exp(-t/τ).  V₂/V₁ = exp(-t'/τ) where t' is time it takes to reach V₂.
  ln(V₁/V₂) = t'/τ.
  The power dissipated in the resistor is P = V²/R = (V₁²/R)exp(-2t/τ).
  The total energy dissipated is E_diss = (V₁²/R)∫₀^t'exp(-2t/τ)dt = (V₁²τ/(2R))(1 - exp(-2t'/τ)).
  E_diss = ½V₁²C(1 - exp(-2ln(V₁/V₂))) = ½V₁²C(1 - (V₂/V₁)²) = ½C(V₁² - V₂²) ~ 368 J.
  E_diss + ΔU = 0.

Gauss' law, planar symmetry

Problem:

The dielectric of a parallel plate capacitor has a permittivity that varies as ε₁ + ax, where x is the distance from one plate.  The area of each plate is A and their spacing is s.
Assume a surface charge density ±σ_free for the plates.
(a)  Find the capacitance.
(b)  Assume ε₁ + ax varies from ε₁ to 2ε₁.  Find P from D and E for that case.
(c)  Find the polarization charge density ρ_p.

Solution:

• Concepts:
  Gauss' law, relationship between E, P, and D
• Reasoning:
  Assuming a surface charge density, Gauss' law for D can be used to find D, then E, then V.  We then can find C = Q/V.
• Details of the calculation:
  (a)  Gauss' law for D yields DΔA = σ_freeΔA,  D = σ_free, D = εE.
  V = ∫₀^sEdx =  ∫₀^sdx/(ε₁ + ax) = (σ_free/a)ln(1 + as/ε₁)).
  C = Q/V = σ_freeA/((σ_free/a)ln(1 + as/ε₁)) = aA/ln(1 + as/ε₁).
  If a << ε₁ then  ln(1 + as/ε₁) ≈ as/ε₁ - ½(as/ε₁)² + ... .
  C ≈ ε₁A/s.
  (b)  2ε₁ = ε₁ + as.  a = ε₁/s.  ε = ε₁ + (ε₁/s)x.
  D = σ_freei,  E = σ_freei/(ε₁ + (ε₁/s)x),  P = D - ε₀E = σ_freei(1 - [ε₀/(ε₁ + (ε₁/s)x)]
  (c)  ρ_p = -∇∙P = -∂P/∂x = -σ_freeε₀(ε₁/s)/(ε₁ + (ε₁/s)x)² = -σ_frees(ε₀/ε₁)/(s + x)².
  σ_p = P∙n.  At x = 0  σ_p = -σ_free(1 - ε₀/ε₁).  At x = s  σ_p = σ_free(1 - ε₀/2ε₁).

Problem:

The plates of an isolated parallel-plate capacitor have area A and are separated by a distance d << A^½.  Let one plate be located in the y-z plane at x = 0 and the other plate at x = d.  The plate at x = 0 carries a charge +Q and the plate at x = d carries a charge -Q.  Dielectric 1 with permittivity ε = ε₀ + ax fills the region between the plates from x = 0 to x = d/2, and dielectric 2 with permittivity ε = ε₀ + a(d - x) fills the region between the plates from x = d/2 to x = d.  Let a = ε₀/d.

(a)  Find the electric displacement and the electric field between the plates.
(b)  Find the voltage across the plates.
(c)  Find the distribution of bound charges in the interior of and on the surfaces of both dielectrics.

Solution:

• Concepts:
  Gauss' law for D
• Reasoning:
  Gauss' law can be used to find D.  Given D, we find E, P, V, and the charge densities.
• Details of the calculation:
  (a)  Gauss' law for D:  D = σ i = (Q/A)i.  E = D/ε.
  E = σ i /(ε₀ + ax),  0 < x < d/2,   E = σ i /(ε₀ + a(d - x)),  d/2 < x < d.
  (b)  V = σ∫₀^d/2dx (1/(ε₀ + ax)) + σ∫_d/2^ddx (1/(ε₀ + a(d - x)))
  =   (σ/a)∫_ε0 ^{ε0 +ad/2}dx' (1/x') - (σ/a)∫ _ε0+ad/2 ^ε0 dx' (1/x')
  = 2(σd/ε₀)ln(1 + ½).
  V  = 2(σd/ε₀)ln(3/2).
  (c)  P = (ε - ε₀)E.  P = (ε₀x/d)E,   0 < x < d/2,   P = (ε₀(d - x)/d)E,   d/2 < x < d.
  P = (xσ/(d + x))i,   0 < x < d/2,   P = (σ(d-x)/(2d - x))i,   d/2 < x < d.
  ρ_p = -∇∙P = -σd/(d + x)², 0 < x < d/2,  is the volume charge density in dielectric 1.
  ρ_p = -∇∙P = σd/(2d - x)², d/2 < x < d, is the volume charge density in dielectric 2.
  σ_p = P∙n is the expression for the surface charge density of bound charges.
  σ_p = 0 on the surfaces of the dielectrics at x = 0 and x = d.
  σ_p = σ/3 on the surface of dielectric 1 at x = d/2.
  σ_p = -σ/3 on the surface of dielectric 2 at x = d/2.