Spherical Capacitor with an air gap and a dielectric

PROBLEM:
Consider a spherical capacitor with conducting spheres of radii a and c, an air gap (ϵ = (Subscript[ϵ, o])) for a  < r < b, and a linear dielectric with relative permittivity (Subscript[ϵ, r]) for b < r < c.  Where appropriate, assume the inner sphere is charged positive.

Your mission:

(a) Derive an expression for the Capacitance of this system using the fields.
(b) Find P in the dielectric.
(c) Find the bound charges associated with the dielectric.
(d) From the definition of D, show that D in the air gap and the dielectric are equal.  (Explain why D should be equal.)
(e) Find the E field for r > c; justify your answer.
(f)  A voltage V is applied between the electrodes (sphere and shell).  Find the stored energy.

(a) Derive an expression for the Capacitance of this system using the fields.
	When appropriate, use M to carry out integration, etc.

(this is a text cell)
 C is the ratio (Q/(ΔV)) where ΔV is the potential difference between the inner and outer conductors and  Q is the charge on the capacitor.  It is understood that the charges on the inner conductor and the outer conductor have opposite signs.  Below I assume that the inner sphere carries the positive charge, +Q, and the outer sphere carries the negative charge,-Q .
 
To find ΔV we need the E fields.  

Call E in the air gap (a < r < b) (Subscript[E, I]);  E in the dielectric (b < r < c) (Subscript[E, II]), and E outside the outer sphere (r > c), (Subscript[E, III]).

We use D to find E using D = ϵ E.  D is due to free charge only.  The charge inside a dielectric is normally bound charge only.  So, given +Q on the inner sphere, Gauss’s Law for D gives the same equation for D in the air gap AND the dielectric:  i.e.,   D = (Q/(4 π (r^2))) (OverHat[r])  for a < r < c  (using the appropriate symmetry arguments).

Using the same subscripts, I, II, and III:     DI = (Q/(4 π (r^2))) (OverHat[r])  (a < r < b)    and DII = (Q/(4 π (r^2))) (OverHat[r])  (b < r < c)
In region I, ϵ = ϵo.  

Therefore, for a linear dielectric,  (Subscript[E, I]) = ((DI)/(ϵo))= (Q/(4 πϵo (r^2))) (OverHat[r])     (a < r < b)   
Again, for a linear dielectric, (Subscript[E, II]) = ((DII)/ϵ)= (Q/(4 πϵo ϵr (r^2))) (OverHat[r])     (b < r < c), where:  ϵ = ϵo ϵr     

To repeat, the free charges in the problem are ±Q on the inner and outer cylinders; we used the +Q at r = a to determine DI and DII.  We used D = ϵ E to determine EI and EII.

This gave us:
				(Subscript[E, I]) = (Q/(4 π (Subscript[ϵ, o]) (r^2))) (OverHat[r])  (in air gap, ϵ = (Subscript[ϵ, o]));    
				(Subscript[E, II]) = (Q/(4 π (Subscript[ϵ, o])(Subscript[ϵ, r])(r^2))) (OverHat[r])  (in dielectric,  ϵ=(Subscript[ϵ, o])(Subscript[ϵ, r]) ) 

FYI:    We anticipate that (Subscript[E, III]) = 0.

To find the potential difference between the two electrodes, we integrate the E field along a path connecting the inner and outer spheres.  Any path will do, but the easiest path is radial.  Likewise, we might as well use M.  

	Δ(Subscript[V, between r = a and r = c]) = (+(∫_a^b)(Subscript[E, I]) ·dr) + (+(∫_b^c)(Subscript[E, II]) ·dr)      
	
Integrals are + because we are integrating from the most + electrode to the most – electrode.

I will be using print statements to label the most important results.  It is easy to get lost in the calculations.

ClearAll["`*"]
$Assumptions = a > 0 && a ϵ Reals&& b > 0 && b ϵ Reals && c > 0 && c ϵ Reals && a < b < c; 
E1 = Q/(4*Pi*ϵo*r^2); 
E2 = Q/(4*Pi*ϵo*ϵr*r^2); 
V1 = Integrate[E1, {r, a, b}];
Print["V1 = ",V1]
V2 = Integrate[E2, {r, b, c}];
Print["V2 = ",V2]
DeltaV = Simplify[V1 + V2]; 
Print["The potential difference between the electrodes: DeltaV = V1 + V2 = ", DeltaV]

Using this DeltaV, we can find the capacitance, CC:      CC =  ((
Q)/(ΔV))

((
Q)/DeltaV);
CC = %/.{ϵo-> (Subscript[ϵ, o]), ϵr-> (Subscript[ϵ, r])}//Simplify;
Print["The capacitance C = ", CC]

The result is rather messy.  To check it, we can compare this result to our calculation of the capacitance of an air-filled spherical capacitor [ Cair gap = ((4 a b π (Subscript[ϵ, o]))/(b-a)) ] by assuming that c = b and (Subscript[ϵ, r])= 1.  (This fills the gap from r = a to r = b with AIR — vacuum.)

CC/.{c-> b,(Subscript[ϵ, r])-> 1}//Simplify;
Print["Cairgap = ", %]

Obviously, this result agrees with our expression for the capacitance of an air-filled spherical capacitor.  This is an important check on our results.

(b) Find P in the dielectric.  Again, assume that the inner sphere carries the positive charge Q.

For a linear Dielectric, P = ϵo χe E,   here E = (Subscript[E, II])   and  χe = (Subscript[ϵ, r]) - 1;  ∴  P = χe (Q/(4 π (Subscript[ϵ, r])(r^2))) (OverHat[r])

P will be in the + r direction ((OverHat[r])) because P is in the same direction as (Subscript[E, II]) = (Q/(4 π (Subscript[ϵ, o])(Subscript[ϵ, r])(r^2))) (OverHat[r]) .

P=(ϵr -1) (Q/(4 π ϵr (r^2))) ; (* r component; Pvector = {this P, 0, 0} *)
Print["The polarization in the dielectric P = ", P]
Print["For Q > 0, the vector P points in the +(OverHat[r]) direction"]

(c) Find the density of bound charge associated with the dielectric.

We first find the charge densities from the polarization vector, P.  In the interior (bulk) of the dielectric, the bound charge density is given by the divergence of P.  

Bulk:   ρbound = –Div[P]   (we’ll use M to compute the divergence in spherical coordinates)

rhobound =Div[{P, 0, 0}, {r, θ, ϕ}, "Spherical"]
Print["The bound charge density in the interior of the dielectric = ", rhobound]

Therefore, there is no bulk bound charge density.

This is expected:  The E-field from a point charge = ((Q )/(4 π ϵo (r^2) )) ~  ((1 )/( (r^2) ))
For r > 0, Div[E] = (ρ/(ϵo)).  Where there is no charge (ρ = 0), Div[E] = 0.

Basically, any vector field with a ((1 )/( (r^2) )) dependence will have zero divergence (for r ≠ 0).
SO:  we have no bulk bound charge density, ρbound = 0.

We know that E in the dielectric is lower than E in the air gap.  This change requires extra charge somewhere.  It must be bound surface charge.  What is the bound surface charge density?

Bound Surface Charge Density:      σbound = P· (OverHat[n])   
Inner surface at r = b, (OverHat[n]) = - (OverHat[r]) = {-1, 0, 0}  
Outer surface at r = c, (OverHat[n]) =  (OverHat[r]) = {1, 0, 0};        (b and c define the surfaces of the dielectric)

Note that the vector normal points out of the dielectric.  Thus the normal vector for the inner surface points inward, toward the center of the sphere, in the – (OverHat[r]) direction.  Likewise, the normal vector for the outer surface points outward, in the + (OverHat[r]) direction.

Again, using M:

σboundinner = {P, 0, 0}.{-1, 0, 0} /. r -> b ;
σboundouter = {P, 0, 0}.{1, 0, 0} /. r -> c ;
Print["Bound surface charge density on the inner conductor = ",σboundinner]
Print["Bound surface charge density on the outer conductor = ",σboundouter]

Note the (1/(4  π (b^2) )) and (1/(4  π (c^2) )) factors in these two Surface DENSITIES — we’re going to be using these factors to get the total surface charges.

Total bound surface charges = σbound * (Surface Area), because σbound is constant on both surfaces (no need to integrate).  The two (spherical) surface areas are 4 π (b^2)  and 4 π (c^2).

Qboundinner = σboundinner* 4 π (b^2)  ;
Qboundouter = σboundouter* 4 π (c^2)   ;
totalboundsurfacecharge = Qboundinner + Qboundouter;
Print["Total bound surface charge on the inner conductor = ", Qboundinner]
Print["Total bound surface charge on the inner conductor = ", Qboundinner]
Print["The total bound surface charge, Qbound = ",totalboundsurfacecharge]

So the total bound charge is zero!

(d) From the definition of D, show that D in the air gap and the dielectric are equal (explain why it should be).

(this is a text cell)
At every point, D = ϵo E + P.   
	(All the vectors are radial outward so we can simply deal with magnitudes.)
	
For the air gap:  E = E1; P = 0
For the dielectric E = E2;  P = P

Recall that Q is the magnitude of the free charge on the inner conducting sphere.

E1
E2
Dairgap = ϵo E + P /. {E -> E1, P -> 0}
Ddielectric = ϵo E + P /. {E -> E2, P -> P} // Simplify
(* Test Equality *)
Print["D in the airgap equals D in the dielectric—True or False??:  ",Dairgap == Ddielectric]  (*these are the magnitudes *)

Both magnitudes and directions ( (OverHat[r])) of the two D vectors are equal.

Justification:  Gauss’s Law for  D:  (∫_S)D.dS = (Subscript[Q, enclosed]^free);  For our geometry, the only free charge inside of a spherical Gaussian surface with radius a < r < c is on the conducting sphere of radius a.  Therefore D is the same in the gap as it is in the dielectric.

(e) Find the E field outside the outer sphere, that is, for  r > c  ((Subscript[E, III]));   justify your answer.

(this is a text cell)
We use Gauss’s Law and note that all the charges (both bound and free) inside a sphere of radius r ⩽ c are distributed in a spherically symmetric fashion (remember that E depends on both free AND bound charge).  For a Gaussian surface of r > c:

Qenclosed = Qtotal where  Qtotal = Qinnerconductor + Qbound + Qouterconductor

Qtotal = 0   (because Qinnerconductoer = –Qouterconductor  and  Qbound  = 0)

Therefore RHS of Gauss’s law = 0.

By symmetry, E take the form E[r] (OverHat[r])   so left hand side = E[r] * 4π (r^2) which we set = RHS = 0

r > 0, so Eoutside [r] = 0.

Additional justification:

Displacement D for r > c:  Again, Gauss’s Law for D says (∫_S)D·dS = (Subscript[Q, enclosed]^free)
By same symmetry arguments as made for E:  LHS = D[r] * 4π (r^2) .   

Because only free charge is on conductors AND  because Qinnerconductor = –Qouterconductor,
RHS =  (Subscript[Q, enclosed]^free) = 0)

Thus D[r] = 0 for r > c.   BUT since E[r] = ((D[r])/(ϵo)), E[r] = 0 for r > c  ⇒ the vector E[r] = 0 for r > c.

(f)  Find the energy stored in the structure when a potential difference V is applied between the inner and outer spheres.  How is the energy stored?

We only need the capacitance (CC) and the equation:   (Subscript[U, stored]) = (1/2) CC  (V^2)

Print["Energy stored in capacitor = ", (1/2) CC (V^2)]

i.e.,  (Subscript[U, stored]) = ((2 a b c π  (Subscript[ϵ, o]) (Subscript[ϵ, r]))/(a (-b+c)+(-a+b) c (Subscript[ϵ, r])))(V^2)

How is the Energy Stored?:  The energy is stored in the dielectric, i.e., in the polarized material.  It takes energy to polarize the dielectric, LIKE PULLING ON or STRETCHING A SPRING.  This is the same process that polarizes an atom:  the electron cloud is distorted so that its center no longer coincides with the atomic nucleus.  It takes energy to pull negative charge “away” from positive charge.  This is true even when the electron remains bound to the nucleus.