Questions: 1 2 3 4 5

PHYS 2420 Electric Fields in Materials Solutions

1. A thin dielectric rod of cross section A extends along the x-axis from x = 0 to x = L. The polarization of the rod is along its length, and is given by P_x = ax² + b. Find the volume density of polarization charge and the surface polarization charge on each end. Show explicitly that the total bound charge vanishes in this case.




We are given that the polarization is P = (ax² + b)i. The volume polarization density of charge is given by ρ_P = -Ñ ·P. Here we find

ρP	= -Ñ·P
	= − ¶Px/¶x − ¶Py/¶y − ¶Pz/¶z
	= − d(ax2 + b)/dx − 0 − 0
	= − 2ax

The surface polarization charge is given by Q_P = P·nA, where A is area of the surface in question, n is the outward looking normal to the surface, and P is evaluated at the point in question. We have three surfaces here. For surface A₁ n₁ = − i, that is the surface normal points to the left. For surface A₂ the surface normal points right, n₂ = +i. For A₃ n₃ = r, where r is the cylindrical radial unit vector. Also A₃ = 2πrL.

Q₁ = (ax² + b)i · −iA|_x=0 = − bA ,

Q₂ = (ax² + b)i · +iA|_x=L = (aL² + b)A ,

and,

Q₃ = (ax² + b)i · r2πrL = 0 .

There is no surface charge on the side of the cylinder since the polarization is at right angles to the normal.

The charge in the interior of the cylinder is found by integrating the polarization density over the volume of the cylinder.

Qinterior	= ò ρP dV
	= ò ò ρP dA dx
	= ò dA ò ρP dx
	= A ò 0L (− 2ax) dx
	= − aA x2 |0L
	= − aL2A

The total of the interior and surface charges is thus

QTotal	= Q1 + Q2 + Q3 + Qinterior
	= − bA + (aL2 + b) A + 0 − aL2A
	= 0

The total charge is zero as expected.



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2. A dielectric cube of side L has a radial polarization given by P = Ar, where A is a constant, and r = ix + jy + kz. The origin of the coordinates is at the centre of the cube. Find all bound charge densities, and show explicitly that the total bound charge vanishes.


We are given that the polarization is P = Ar. The volume polarization density of charge is given by ρ_P = -Ñ·P. Here we find

ρP	= -Ñ·P
	= -¶(Ax)/¶x − ¶(Ay)/¶y − ¶(Az)/¶z
	= − A − A − A
	= − 3A

The surface polarization charge density is given by Σ_P = Q_P/A = P·n, where n is the outward looking normal to the surface, and P is evaluated at the point in question. We have six surfaces here, one for each side of the cube. If we look at the xy plane of the cube as shown below, we find



Similarly for the other faces, n₅ = +k at z = ½L and n₆ = − k at z = − ½L. We find

Σ₁ = P · n₁|_x=½L = A(ix + jy + kz) · i|_x=½L = ½AL ,

Σ₂ = P · n₂|_x=-½L = A(ix + jy + kz) · − i|_{x=− ½L} = ½AL ,

Σ₃ = P · n₃|_y=½L = A(ix + jy + kz) · j|_y=½L = ½AL ,

Σ₄ = P · n₄|_{y=− ½L} = A(ix + jy + kz) · − j|_{y=− ½L} = ½AL ,

Σ₅ = P · n₅|_z=½L = A(ix + jy + kz) · k|_z=½L = ½AL ,

and

Σ₆ = P · n₆|_{z=− ½L} = A(ix + jy + kz) · − k|_{z=− ½L} = ½AL .

The charge on each face is thus Q_face = ΣA_face = ΣL² = ½AL³.

The charge in the interior of the cylinder is found by integrating the polarization density over the volume of the cylinder.

Qinterior	= ò ρP dV
	= ò ò ò ρP dxdydz
	= −3A ò ò ò ρP dxdydz
	= −3AL3

The total of the interior and surface charges is thus

QTotal	= Q1 + Q2 + Q3 + Q4 + Q5 + Q6 + Qinterior
	= 6(½AL3) − 3AL3
	= 0

The total charge is zero as expected.



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3. A dielectric rod in the shape of a right circular cylinder of length L and radius R is polarized in the direction of its length. The polarization is uniform and of magnitude P. Calculate the electric field resulting from the polarization at a point on the axis of the rod outside the cylinder.




We are told that the polarization is P = Pi, where P is a constant. The electric field due to the polarization is the electric field due to the polarization charge on each surface an in the material. Thus these need to be calculated first.

The volume charge density is given by ρ_P = -Ñ·P. However we are told that the polarization P is uniform, hence its derivatives are zero. Thus ρ_P = 0.

The surface polarization charge density is given by Σ_P = Q_P/A = P·n, where n is the outward looking normal to the surface, and P is evaluated at the point in question. We have three surfaces here. For surface A₁ n₁ = − i, that is the surface normal points to the left. For surface A₂ the surface normal points right, n₂ = +i. For A₃ n₃ = r, where r is the cylindrical radial unit vector. Thus the charge densities are

Σ₁ = P · n₁|_x=0 = Pi · i|_x=o = P ,

Σ₂ = P · n₂|_x=-L = Pi · −i|_x=L = -P ,

and

Σ₃ = P · n₃|_r=R = Pi · r|_r=R = 0 .

There is no surface charge on the side of the cylinder since the polarization is at right angles to the normal.

What we have are two oppositely charge plates separated by a distance L, a capacitor. Various texts tell us that the well-known result for the electric field due to a single plate located at x = 0,

.

We have two plates, one at x = 0 and one at x = L, so we find the net electric field to be





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4. Two semi-infinite blocks of dielectric are placed almost in contact so that there is a narrow gap of constant separation between them. The polarization P is constant throughout all the dielectric material, and it makes an angle θ with the normal to the planes bounding the gap. Determine the electric field in the gap.


The situation described looks like



where I have included a coordinate system. In this coordinate system, the polarization is

P = iPcos(θ) + jPsin(θ)

where P is a constant. The electric field in the gap is due to the polarization and this electric field arises from the polarization charge on each surface an in the material. Thus these need to be calculated first.

The volume charge density is given by ρ_P = -Ñ·P. However we are told that the polarization P is uniform, hence its derivatives are zero. Thus ρ_P = 0.

The surface polarization charge density is given by Σ_P = Q_P/A = P·n, where n is the outward looking normal to the surface, and P is evaluated at the point in question. We have two surfaces here. For the right surface n_right = +i, that is the surface normal points to the right. For the left surface the surface normal points left, n_left = -i.

Σ_left = P · n_left|_x=0 = [iPcos(θ) + jPsin(θ)] · i|_x=o = Pcos(θ) ,

and

Σ_right = P · n_right|_x=d = [iPcos(θ) + jPsin(θ)] · − i|_x=L = −Pcos(θ) .

We do not need to worry about the other sides of the dielectric as they are an infinite distance away and will not contribute to the net electric field in the gap. We have two oppositely charged sides. Thus we have a capacitor situation and we know





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5. A metal sphere of radius a is surrounded by a thick dielectric shell of inner radius a, outer radius b, and dielectric constant ε . The metal sphere carries a free charge Q. There is no free charge on or in the dielectric. Determine the electric field everywhere. Determine the polarization of the dielectric and determine the bound charge distributions. Analyze this system thoroughly, determining the potential of the metal sphere and the distribution of bound charge.


When dealing with dielectrics, we deal with the electric displacement, D. Since we have spherical symmetry, we can use the integral form of Gauss' Law

ò D•ndS = ò ρ_CdV ,

where ò ρ_CdV = Q_enclosed, the enclosed free charge. Free charge is the charge not bound into matter, the charge that was added to the neutral bodies. Symmetry demands that the displacement be radial, D = D(r)r . The outward looking normal, n, is also radial here. Thus at r = R

ò D•ndS = ò D(r)r •rdS = D(R)4πR² .

Thus Gauss' Law reduces to

D(R)4πR² = Q_enclosed .

We know that the charge Q must be uniformly distributed on the surface of metal sphere. Thus the problem is naturally broken into two regions, inside the shell and outside the shell.

1. 0 £ R < a

Since there is no charge inside the shell, D = 0.

2. a £ R

Since the enclosed charge is Q,

D = Q/4πR² .

The electric field is related to the electric displacement by E = D/ε , where ε is the dielectric constant where you wish to determine the field. The electric field will have the same radial symmetry as D. However, there are three regions of different dielectric here, inside the metal, inside the dielectric, outside the dielectric.

1. a £ R

   The dielectric constant of a metal is infinite, but that doesn't matter since D = 0. Hence E = 0 inside the metal sphere. This is the usual expected result.

2. a £ R £ b

   The dielectric constant is ε, so

   E = D/ε = Q/4peR² .

3. b £ R

Presumably there is just air outside the dielectric and air has a dielectric constant very close to that of a vacuum, ε₀. Thus

E = D/ε₀ = Q/4pe₀R² .

The polarization P is related to D and E by the formula P = D – ε₀E. Thus the polarization has the same radial symmetry as D and E. Since the electric field has three regions, so does the polarization.

1. a £ R

Both D and E are zero, therefore so is P. This makes sense as there is no dielectric to polarize.

2. a £ R £ b

   We have D and E so

   P = D – ε₀E = Q/4π R² – ε ₀(Q/4peR²) = (1 – ε₀/ε) Q/4πR² .

3. b £ R

We have assumed that there is very nearly a vacuum outside the sphere, so there will be nothing to polarize. However we do know D and E, so checking to be sure we get

P = D – ε₀E = Q/4πR² – ε₀(Q/4pe₀R²) = (1 – ε₀/ε₀) Q/4πR² = 0 .

Summing our results for D, E, and P in a graph we have



The polarization charge density in the dielectric is determined by ρ_P = -Ñ·P. We should use the particular form of the divergence in spherical coordinates as given in our text.

ρP	= -Ñ·P
	=
	=
	=
	= 0

Since is the expected result since the polarization is uniform.

The surface polarization charge density is given by Σ_P = Q_P/A = P·n, where n is the outward looking normal to the surface, and P is evaluated at the point in question. We have two surfaces here. For the inner face of the dielectric n_inner = -r. For the outer face the surface, n_outer = +r.

Σ_inner = P · n_inner|_R=a = (1 – ε/ε₀) Qr/4πR² · –r|_R=a = – (1 – ε/ε₀) Q/4πa²,

and

Σ_outer = P · n_outer|_R=b = (1 – ε/ε₀) Qr/4πR² · r|_R=b = (1 – ε/ε₀) Q/4πb² .

The electrostatic potential difference between two points when the electric field is known is given by

ΔV = V_B – V_A = –ò_A^B E·dL .

When we are asked for the potential of a point B, we must remember that potential is always relative to some arbitrary reference position A. The most common reference point is at infinity. We will use that reference here. Therefore we define the potential as

V = –ò_¥^r=B E·dL .

We have determined given the form of the electric field. The electric field is radial, that is E(r) = E(r)r. The path element is dL = rdr + θrdθ + φsin(θ)dφ where r, θ, and φ are the spherical coordinate unit vectors. Therefore the potential reduces to

V = –ò_¥^r=B E(r) dr .

Since there are discontinuities in the electric field at r = a and b, we need to consider three cases.

1. The first is when B > b. Here we know the form of E(r) and we find

   
   V = –ò_¥^r=B Q/4pe₀r² dr = Q/4pe₀r |_¥^r=B = Q/4pe₀B .

2. In the second case, a < B < b and we have to break up the integral for the two regions where E(r) has different behaviour:

   V	= –ò r=br=B E(r) dr – ò¥r=b E(r) dr
   	= –ò r=br=B Q/4per2 dr – ò¥r=b Q/4pe0r2 dr
   	= Q/4per |r=br=B + Q/4pe0r |¥r=b
   	=

3. In the third case, 0 < B < a and we have to break up the integral for the three regions where E(r) has different behaviour:

   V	= –òr=ar=B E(r) dr – òr=ar=b E(r) dr – ò¥r=b E(r) dr
   	= 0 – ò r=br=a Q/4per2 dr – ò¥r=b Q/4pe0r2 dr
   	= Q/4per |r=br=a + Q/4pe0r |¥r=b
   	=

   Note that the first integral was zero since E = 0 inside the sphere. As expected, the potential is constant everywhere inside the sphere.

   The potential looks like

   

Questions?mike.coombes@kpu.ca