__Gauss' law, spherical symmetry__

A spherical capacitor with conducting surfaces of radii R_{1}
and R_{2} has a material of dielectric constant

ε(r) = ε_{0}(R_{1}/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**: D4πr^{2 }= Q_{free inside}in SI units.Free charges reside only on the surfaces of the capacitor. Between the surfaces we have

D = Q_{free}/( 4πr^{2}), D(r) = ε(r)E(r),

V = ∫_{R1}^{R2}E(r)dr = ∫_{R1}^{R2}dr D(r)/ε(r) = ∫_{R1}^{R2}dr Q_{free}/(4πε_{0}R_{1}^{2}).

V = Q_{free}(R_{2}- R_{1})/(4πε_{0}R_{1}^{2}).

C = Q_{free}/V = 4πε_{0}R_{1}^{2}/(R_{2 }- R_{1}) is the capacitance of the capacitor.

(b) U = ½Q^{2}/C = Q^{2}(R_{2 }- R_{1})/(8πε_{0}R_{1}^{2}) is the total energy stored in the capacitor.

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/(4πr^{2}),**D**points radially away from the origin, if the center of the conducting sphere is placed at the origin.

**E**=**D**/ε =**D**/(ε_{0}K) =**D**/(ε_{0 }+ ε_{0}n(r - a)) for a < r < b.

E = [Q/(4πε_{0}r^{2})](1/(1 + n(r - a))) = [Q/(4πε_{0})]/(nr^{3}+ (1 - na))r^{2}).

**E**=**D**/ε_{0}for r > b. E = Q/(4πε_{0}r^{2}).

(b) To find the surface charge distributions we use (SI units).

Surface at r - a: σ = [Q/(4πa^{2})](1/(1 + n(a - a))) = Q/(4πa^{2}).

σ_{bound}= σ - Q/(4πa^{2}) = 0.

Surface at r - b: σ_{bound}= Q/(4πb^{2}) - [Q/(4πb^{2})](1/(1 + n(b - a)))

= [Q/(4πb^{2})](n(b - a)/(1 + n(b - a)).

To find the volume charge distribution we use ρ_{bound}= -**∇∙P**,**P**= ε_{0}(K - 1)**E**.

For a < r < b we have P = ε_{0}n(r-a)E = [nQ/(4πr^{2})](r - a)/(nr + 1- na).**∇∙P**= (1/r^{2})(∂/∂r)(r^{2}P)

ρ_{bound}= -[nQ/(4πr^{2})][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 isW = ½∫

_{all space}**E∙D**dV.W = ½4π∫

_{a}^{b}r^{2}dr [Q/(4πε_{0}r^{2})](1 + n(r - a))^{-1}[Q/(4πr^{2})]

+ ½4π∫_{b}^{∞}r^{2}dr [Q/(4πε_{0}r^{2})][Q/(4πr^{2})].

W = [Q^{2}/(8πε_{0})]∫_{a}^{b}dr [r^{2}(1 + n(r - a))]^{-1}+ [Q^{2}/(8πε_{0}b)]

= [Q^{2}/(8πε_{0})][-1/((1 - na)r) + (n/(1 - na)^{2})ln((1 - na + nr)/r)]_{a}^{b}+ [Q^{2}/(8πε_{0}b)].

W = [Q^{2}/(8πε_{0})][1/b + (b-a)/((1 - na)ab) + (n/(1 - na)^{2})ln((1 - na + nb)a/b)].

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)4πr^{2}= 4πr^{3}ρ/3, D(r) = rρ/3. E(r) = D(r)/(ε_{0}k_{e}) = ρr/(3ε_{0}k_{e}) = Q_{tot}r/(4πε_{0}k_{e}a^{3})

r > a: D(r)4πr^{2}= 4πa^{3}ρ/3, D(r) = a^{3}ρ/(3r^{2}) = Q_{tot}/(4πr^{2}). E(r) = a^{3}ρ/(3ε_{0}r^{2}) = Q_{tot}/(4πε_{0}r^{2}).

(b) The energy U of the system is U = ∫_{all space}(ε/2)E^{2}dV.

U = 2πε∫_{0}^{a}E^{2}r^{2}dr + 2πε_{0}∫_{a}^{∞}E^{2}r^{2}dr = [Q_{tot}^{2}/(8πε_{0})][1/(5ak_{e}) + 1/a]

= [2ρ^{2}π/(9ε_{0})][a^{5}/(5k_{e}) + a^{5}].

__Gauss' law, cylindrical symmetry__

A
capacitor is made of two concentric cylinders of radii r_{1}
and r_{2} (r_{1 }< r_{2}) and
length L >> r_{2}.

The region between r_{1}
and r_{3} = (r_{1}r_{2})^{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 (r_{1 }< r < r_{3})?
Assume a potential difference V between r_{1} and
r_{2}.

(c) How much mechanical work must be done to remove the dielectric
cylinder while maintaining this constant potential difference between r_{1}
and r_{2}?

Solution:

- Concepts:

Gauss' law, relationship between**E**,**P**, and**D**, capacitance, U = ½CV^{2}. - 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) (r_{1}< r < r_{2}) 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)/(ε_{0}k) = ∫_{r1}^{r2}dr D(r)/ε_{0 }= (λ_{f}/(2πε_{0}k))[ln(r_{3}/r_{1}) + k ln(r_{2}/r_{3})]

= (λ_{f}/(2πε_{0}k))[ln√(r_{2}/r_{1}) + k ln(√r_{2}/r_{3})]

= (λ_{f}(1 + k)/(2πε_{0}k))ln√(r_{2}/r_{1}).

C = Q_{f}/ΔV = λ_{f}L/ΔV = [4πε_{0}kL/(k + 1)][1/ln(r_{2}/r_{1})].

(b)**D**= (λ_{f}/(2πr))(**r**/r),**E**= (λ_{f}/(2πε_{0}kr)) (**r**/r),

**P**= ε_{0}χ_{e}**E**= ε_{0}(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 = ½CV^{2}.

Without the dielectric the capacitance is C_{out}= 2πε_{0}L/ln(r_{2}/r_{1}) < C.

The energy stored in the capacitor decreases by ΔU = ½(C_{out}- C)V^{2}.

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)V^{2}is done by the battery.

The energy stored in the capacitor decreases only by ½(C_{out}- C)V^{2}.

The mechanical work done must therefore equal W = ½(C - C_{out})V^{2}.

W = [2πε_{0}L/ln(r_{2}/r_{1})]V^{2}(k - 1)/(k + 1).

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 V_{0}be positive. Then the direction of**E**is radially outward.

E(r) = A/r, with A positive. Then V_{0}= -∫_{b/2}^{a/2}E(r)dr = A∫_{a/2}^{b/2}(1/r)dr

= A(lnb/2 - lna/2) = Aln(b/a). A = V_{0}/ln(b/a),

with a = 5 mm and b = 9 mm.

E(r) = V_{0}/(r ln(b/a)).

If E_{max}= E(a/2) = 2V_{0}/(a ln(b/a)) = 60 MV/m, then

V_{0}= (6*10^{7}V/m)*(2.5*10^{-3}m)*ln(9/5) = 8.82*10^{4}V is the maximum voltage which can be applied.

Energy stored in the cable:

W = (ε/2)∫E^{2}dV = (εL/2)2π∫_{a/2}^{b/2}E^{2}rdr = επL∫_{a/2}^{b/2}(V_{0}/(rln(b/a)))^{2}rdr.

W = [επLV_{0}^{2}/ln(b/a)] = [2πε_{0}(500 m)*(8.82*10^{4}V)^{2}/ln(9/5)] = 368 J.

(b) The cable is a capacitor. W= U = ½CV^{2}, 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 V_{1}to V_{2}, the energy stored in the capacitor decreases by an amount ΔU = ½C(V_{2}^{2}- V_{1}^{2}).

V = V_{1}exp(-t/τ). V_{2}/V_{1}= exp(-t'/τ) where t' is time it takes to reach V_{2}.

ln(V_{1}/V_{2}) = t'/τ.

The power dissipated in the resistor is P = V^{2}/R = (V_{1}^{2}/R)exp(-2t/τ).

The total energy dissipated is E_{diss}= (V_{1}^{2}/R)∫_{0}^{t'}exp(-2t/τ)dt = (V_{1}^{2}τ/(2R))(1 - exp(-2t'/τ)).

E_{diss}= ½V_{1}^{2}C(1 - exp(-2ln(V_{1}/V_{2}))) = ½V_{1}^{2}C(1 - (V_{2}/V_{1})^{2}) = ½C(V_{1}^{2}- V_{2}^{2}) ~ 368 J.

E_{diss}+ ΔU = 0.

__Gauss' law, planar symmetry__

The dielectric of a parallel plate capacitor
has a permittivity that varies as ε_{1 }+ 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 ε_{1 }+ ax varies from
ε_{1} to 2ε_{1}. 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_{free}ΔA, D = σ_{free}, D = εE.

V = ∫_{0}^{s}Edx = ∫_{0}^{s}dx/(ε_{1 }+ ax) = (σ_{free}/a)ln(1 + as/ε_{1})).

C = Q/V = σ_{free}A/((σ_{free}/a)ln(1 + as/ε_{1})) = aA/ln(1 + as/ε_{1}).

If a << ε_{1}then ln(1 + as/ε_{1}) ≈ as/ε_{1}- ½(as/ε_{1})^{2}+ ... .

C ≈ ε_{1}A/s.

(b) 2ε_{1}= ε_{1}+ as. a = ε_{1}/s. ε = ε_{1}+ (ε_{1}/s)x.**D**= σ_{free}**i**,**E**= σ_{free}**i**/(ε_{1}+ (ε_{1}/s)x),**P**=**D**- ε_{0}**E**= σ_{free}**i**(1 - [ε_{0}/(ε_{1}+ (ε_{1}/s)x)]

(c) ρ_{p}=**-∇∙P**= -∂P/∂x = -σ_{free}ε_{0}(ε_{1}/s)/(ε_{1}+ (ε_{1}/s)x)^{2}= -σ_{free}s(ε_{0}/ε_{1})/(s + x)^{2}.^{}σ_{p}=**P∙n**. At x = 0 σ_{p}= -σ_{free}(1 - ε_{0}/ε_{1}). At x = s σ_{p}= σ_{free}(1 - ε_{0}/2ε_{1}).

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 = ε_{0} + ax fills the region between the plates from x =
0 to x = d/2, and dielectric 2 with permittivity ε = ε_{0} + a(d - x) fills
the region between the plates from x = d/2 to x = d. Let a = ε_{0}/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**/(ε_{0}+ ax), 0 < x < d/2,**E**= σ**i**/(ε_{0}+ a(d - x)), d/2 < x < d.

(b) V =_{0}^{d/2}dx (1/(ε_{0}+ ax)) + σ∫_{d/2}^{d}dx (1/(ε_{0}+ a(d - x)))

= (σ/a)∫_{ε0}^{ ε0 +ad/2}dx' (1/x') - (σ/a)∫_{ ε0+ad/2}^{ ε0 }dx' (1/x')

= 2(σd/ε_{0})ln(1 + ½).

V = 2(σd/ε_{0})ln(3/2).

(c)**P**= (ε - ε_{0})**E**.**P**= (ε_{0}x/d)**E**, 0 < x < d/2,**P**= (ε_{0}(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.