Problems

Problem 1:

If one presumes that there exists a true charge density rtrue, a polarization or bound charge density rbound, and a total charge density rtotal, such that rtrue + rbound = rtotal, write the source equations for D, E, and P.  Explain the meaning of these equations.  Briefly address the question:  Which of the fields D or E might be considered the more fundamental field?  Why?  Write the equation(s) describing the relationships between the three field quantities.

Solution:

Concepts, principles, relations that apply to the problem:
Maxwell's equations for electrostatics
Why do they apply?
Maxwell's equations connect the source equations for E and P and yield the source equations for D.
How do they apply?
Maxwell's equations for electrostatics are
Ñ×E = rtotal/e0,   Ñ´E = 0, (SI units).
Ñ×E = rtotal/e0 is the source equation for E.
Let rtrue + rbound = rtotal, then Ñ×E = (rtrue + rbound)/e0 = rtrue/e0 - Ñ×P/e0.
Ñ×P = -rbound is the source equation for P.
Ñ×(e0E + P) = rtrue.  D = e0E + P, Ñ×D = rtrue is the source equation for D.
Space derivatives of E, P, and D must exist, otherwise use the integral form.
For example: 
For lih materials:
,  .
Equations of macroscopic electrostatics:  Ñ×D = rtrue,  Ñ´E = 0.

D should not be regarded as a fundamental field of the same status as E.  It is rather a mathematical construct related to the way in which we seek a macroscopic solution for E from the basic equations.  D is a useful construct if the material has special properties, for example if it is a lih dielectric.  Then D = eE.  Unfavorable external conditions, however, can destroy those special properties.
Details of the calculation:
None

Problem 2:

A spherical capacitor with conducting surfaces of radii R1 and R2 has a material of dielectric constant 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.

Image634.gif (1707 bytes)

Solution:

Concepts, principles, relations that apply to the problem:
Gauss’ law for D.
Why do they apply?
The problem has spherical symmetry.  We can find the field between the surfaces in terms of the charge Qfree on the surfaces using Gauss’ law for D.
How do they apply?
(a)  Symmetry:
 
if the origin is located at the center of the spheres.
Gauss’ law for D:  D4pr2 = Qfree inside in SI units.

Free charges reside only on the surfaces of the capacitor.  Between the surfaces we have
D = Qfree/( 4pr2),  D(r) = e(r)E(r), 

.

C = Qfree/V = 4pe0R12/(R2-R1) is the capacitance of the capacitor.

Details of the calculation:
(b)  U = (1/2)Q2/C = Q2(R2-R1)/(8pe0R12) is the total energy stored in the capacitor.

Problem 3:

A dielectric sphere of radius a has a uniform isotropic permittivity, ke0, and is located in an electric field that is uniform at infinity.

(a)  Solve for the electric potential everywhere by boundary value methods.
(b)  Show that the electric field inside the sphere is uniform and find its value relative to the field E0 at infinity.

Solution:

Concepts, principles, relations that apply to the problem:
Most general solution to Laplace's equation, boundary conditions
Why do they apply?
Ñ2f = 0 inside and outside the sphere since rf = 0 inside and outside the sphere.  We are looking for a solution for Ñ2f = 0, subject to the boundary conditions.
How do they apply?
We have azimuthal symmetry.
.
Assume f1 = A0 + A1rcosq + B0/r + (B1/r2)cosq inside,
f2 = A0' + A1'rcosq + B0'/r + (B1'/r2)cosq outside.
We assume all higher-order coefficients are zero.  If we find a solution with this assumption, it is the only solution.
Boundary conditions:
(i)  f1 is finite at the origin, B0 = B1 = 0.  f1 = A0 + A1rcosq.
E1 = -Ñf1,  E1r = -A1cosq,  E1q = A1sinq,  E1 = -A1k.
D1 = eE1 = ke0E1.
(ii)  f is continuous at r = a.  A0 = A0' + B0'/a,  A1a = A1'a + B1'/a2.
(iii)  The radial component of D is continuous at r = a.  (There are no free charges.)
E2 = -Ñf2.  E2r = -A1'cosq + B0'/r2 + 2(B1'/r3)cosq,  E2q = A1'sinq + (B1'/r3)sinq,
D2 = e0E2.
-ke0A1cosq = -e0A1'cosq + e0B0'/a2 + 2e0(B1'/a3)cosq.
B0' = 0, kA1 = A1' - 2(B1'/a3) = A1' - 2(A1 - A1')
(iv)  E = E0 at r >> a.
-A1'cosq = E0cosq,  A1'sinq = -E0sinq,  A1' = -E0.
A1 = 3A1'/(k+2) = -3E0/(k+2).
B1' =[(k-1)/(k+2)]E0a3.
A0 = A0' is anarbitrary constant.  We choose this constant to be zero.  A0 = A0' = 0.
Details of the calculation:
finside = -(3E0/(k+2))rcosq = -3E0z/(k+2),  foutside = -E0rcosq + [(k-1)/(k+2)](a3/r2)E0cosq.

(b)  Ein = -Ñfin = 3E0/(k+2).

Problem 4:

Consider a homogeneous dielectric e, of infinite extent, in which there is a uniform field E0.  A spherical cavity of radius a is cut out of this dielectric.  Find:

(a)  f in the cavity and on its surface.
(b)  The polarization charge density sp on the walls.
(c)  The field outside the cavity.

Solution:

Concepts, principles, relations that apply to the problem:
Boundary value problems,  is the general solution to Laplace's equation for problems with azimuthal symmetry.
Why do they apply?
Let the center of the cavity be at the origin.  Then the problem has azimuthal symmetry.
is the most general solution inside and outside of the cavity, since r = 0 inside and outside of the cavity.  To find the specific solution we apply boundary conditions.
How do they apply?
(a)  Assume f1 = A0 + A1rcosq + B0/r + (B1/r2)cosq inside,
f2 = A0' + A1'rcosq + B0'/r + (B1'/r2)cosq outside.
We assume all higher-order coefficients are zero.  If we find a solution with this assumption, it is the only solution.
Boundary conditions:
(i)  E2 = E0 at r >> a,  f2 = -E0z = -E0rcosq  at r >> a,  A1' = -E0,  we choose A0' = 0.
Then, from symmetry, A0 = 0.
(ii)  f1 is finite at the origin, B0 = B1 = 0.  f1 = A1rcosq,  
E1 = -Ñf1,  E1r = -A1cosq,  E1q = A1sinq,  E1 = -A1k.
D1 = e0E1.
(iii)  f is continuous at r = a.  B0' = 0,  A1a = -E0a + B1'/a2.
(iv)  The radial component of D is continuous at r = a.  (There are no free charges.)
E2 = -Ñf2.  E2r = -A1'cosq + 2(B1'/r3)cosq,  E2q = A1'sinq + (B1'/r3)sinq,  D2 = ke0E2 = eE2.
-e0A1cosq = ke0E0cosq + 2ke0(B1'/a3)cosq.
A1 = -kE0 - 2k(B1'/a3) = -kE0 - 2k(A1 + E0)
A1 = -3E0k/(2k+1) = -3E0e/(e0+2e)
B1' = [(e0-e)/(e0+2e)]E0a3.
f1 = [-3E0e/(e0+2e)]rcosq.
On the surface f(a) = [-3E0e/(e0+2e)]acosq.
Details of the calculation:
(b)  sp = e0(E2r(a)- E1r(a) ) = e0E0cosq[1 + 2 (e0-e)/(e0+2e) + 3e/(e0+2e)]
sp = e0E0cosq[1 + (2e0+e))/(e0+2e)] = e0E0cosq[3(e0+e))/(e0+2e)].

(c) f2 = -E0rcosq + [(e0-e)/(e0+2e)]E0a3cosq/r2,
E2r = -¶f2/r = E0cosq[1 + 2 [(e0-e)/(e0+2e)]a3/r3],
E2q = -(1/r)¶f2/¶q = -E0sinq[1 - [(e0-e)/(e0+2e)]a3/r3].

Problem 5:

A "dielectric" material consists of a number of brass spheres of diameter d, spaced 3d apart, in a regular lattice.  Assuming that each sphere is influenced only by the imposed external electric field, i.e. neglecting the effect of the neighboring spheres or the redistribution of induced charges, find the dielectric constant k for this material.

Solution:

Concepts, principles, relations that apply to the problem:
Polarization: P = e0ceE  (SI units), dielectric constant: k = (1+ce), boundary value problems
Why do they apply?
We need to find the polarization to find the dielectric constant.
How do they apply?

We need to find the dipole moment p of each sphere.
We then have P = Np,  N = 1/(3d)3.  We find the dipole moment of each sphere by finding the electric field produced by each sphere.  To find this field we solve a boundary value problem.
Details of the calculation:
For a conducting sphere at the origin in an external field E = Ek we have outside the sphere and f = constant, E = 0 inside the sphere.  Choose  f1 = 0 inside the sphere.
f2 = A1'rcosq  + (B1'/r2)cosq  outside.
We assume all other coefficients are zero.  If we find a solution with this assumption, it is the only solution.
Boundary conditions:
(i)  f is continuous at r = a = d/2.  0 = A1'a + B1'/a2.
(ii)  E2 = E0 at r >> a.
-A1'cosq = E0cosq,  A1'sinq = -E0sinq,  A1' = -E0,  B1' = E0a3.
f2 = -E0rcosq  + (E0a3/r2)cosq.
E2 = -Ñf2.  E2r = E0cosq  + 2(E0a3/r3)cosq,  E2q = -E0'sinq + (E0a3/r3)sinq.
E2 = external field + field due to polarized sphere.
Esphere_r = 2(E0a3/r3)cosq,  Esphere_q = (E0a3/r3)sinq.
This is a dipole field.  For a dipole at the origin we have
.
Therefore p = [(4pe0)E0a3]k = [(4pe0)E0d3/8]k.
P = Np = [(4pe0)E0/216]k, P = e0ceE, ce = 4p/216.
k = (1+ce) =1.058, e = e0k.

Problem 6:

A capacitor is made of two concentric cylinders of radii r1 and r2 (r1 < r2) and length L >> r2.  The region between r1 and  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?

Image1261.gif (2414 bytes)

Solution:

Concepts, principles, relations that apply to the problem:
Gauss' law, relationship between E, P, and D, capacitance, U = (1/2)CV2.
Why do they apply?
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.
How do they apply?
(a)  D = (lf/2pr)(r/r) (r1 < r < r2) from Gauss' law for D.  Here r is a cylindrical coordinate and we use SI units.

C = Qf/DV = lfL/DV = [4pe0kL/(k+1)][1/ln(r2/r1)].
Details of the calculation:
(b)  D = (lf/2pr)(r/r), E = (lf/2pe0kr)(r/r),
P = e0ceE = e0(k-1)E = (lf/2pr)[(k-1)/k](r/r).
(c)  The energy stored in the the capacitor is U = (1/2)CV2.  Without the dielectric the capacitance is Cout = 2pe0L/ln(r2/r1) < C.
The energy stored in the capacitor decreases by DU = (1/2)(Cout - C)V2.
The charge on the capacitor is Q = CV.
When the dielectric is removed, DQ = (Cout - C)V is removed from the cylinders.
An amount of (negative) work W = DQ V = (Cout - C)V2 is done by the battery.
The energy stored in the capacitor decreases only by (1/2)(Cout - C)V2.  The mechanical work done must therefore equal W = (1/2)(C - Cout)V2.
W =  [2pe0L/ln(r2/r1)]V2(k-1)/(k+1).

Problem 7:

(a)  A container is made of two square metal plates of side w held a distance of d apart by insulator ends and bottom.  Assume that the end and bottom insulator pieces have a dielectric constant of 1.  The two metal sides of the container form a capacitor and are attached to a battery of potential difference V.  Find the total energy stored in the electric field when container is empty.  You may assume that d << w.
(b)  A dielectric liquid of dielectric constant k is poured into the box until it is half full.  What will then be the total energy in the capacitor?
(c)  Calculate the amount of energy given up by the source during the dielectric-filling operation in part (b).
(d)  If the dielectric has a density r and is poured from a height w above the top of the container, with what velocity will the first droplets of dielectric strike the bottom of the container?

Solution:

Concepts, principles, relations that apply to the problem:
Capacitor with dielectric, energy conservation
Why do they apply?
We are asked to find the energy stored in a capacitor partially filled with a dielectric.
How do they apply?

(a)  U = (1/2)CV2,  C = e0w2/d,  U = (1/2)(e0w2/d)V2.
(b)  U' = (1/2)C'V2,  C' = C1 + C2,  C1 = e0w2/(2d),  C2 = ke0w2/(2d),  U = (1/2)(1+k)(e0w2/(2d))V2.
(c)  Before filling:  Q = CV,  after filling:  Q' = C'V.
Work done by source:  (Q' - Q)V = (C' - C)V2 = (1-k)(e0w2/(2d))V2.
Details of the calculation:
(d)  Assume the container contains dielectric up to a height h.  The energy stored in the capacitor is
U' = (1/2)C'V2,  C' = C1 + C2 = ke0hw/d + e0(w-h)w/d.
Now add a drop of dielectric, so that the height increases from h to h + dh.
The energy stored in the capacitor now is
U'' = (1/2)C''V2,  C'' = ke0(h + dh)w/d + e0(w - h - dh)w/d.
DU = U''-U' = (1/2)(C'' - C')V2 = (1/2)V2(k-1)(e0 dh w/d).
The work done by the battery is (C'' - C')V2 = 2DU.
Therefore the kinetic energy of the drop increases by an amount DU over what it would be from just free-falling.
T = T1 + T2,  T1 = mg2w,  T2 = DU.
(1/2)mv2 = mg2w + DU.  v2 = 4gw + 2DU/m.
m = r d w dh,  v2 = 4gw + (k-1)e0V2/(rd2).

Problem 8:

A charge q is situated at the point x = h > 0, y = z = 0 outside a homogeneous dielectric which fills the region x < 0.
(a)  Write the electric fields just outside and just inside the dielectric in terms of the charge q and surface charge density sb of bound charges on the surface of the dielectric.
(b)  Express sb in terms of the electric field just inside the dielectric.  Let e be the dielectric constant of the dielectric.
(c)  By using the equations obtained in (a) and (b), show that
sb = -[1/(2p)][(e-1)/(e+1)][qh/(h2+y2+z2)3/2].
(d)  Calculate the electric field E’ due to sb at the position (h,0,0) of the charge q.  Show that it can be interpreted as the field of an image charge q’ situated at the point (-h,0,0).
(e) Show that the charge q experiences the force F = -[(e-1)/(e+1)] (q2/4h2)i.

Solution:

Concepts, principles, relations that apply to the problem:
Polarization, polarization charge densities
Why do they apply?
The field of the point charge polarizes the dielectric and induces a  surface polarization charge density, sb = P×n.
How do they apply?
(a)  The field due to a surface charge density s is 2ps n (Gaussian units) just outside the surface.
Therefore Eout(0,y,z) = 2psb(y,z) i + q/(h2 + y2 + z2)3/2 [-hi +yj +zk].
Just inside the surface Ein(0,y,z) = -2psb(y,z) i + q/(h2 + y2 + z2)3/2 [-hi +yj +zk].
 
(b) P = [(e-1)/4p]Ein,  sb = P×n = [(e-1)/4p]Ex_in(0,y,z)

(c)  sb = -[(e-1)/4p][2psb + qh/(h2 + y2 + z2)3/2].
sb = -[1/2p][(e-1)/(e+1)][qh/(h2 + y2 + z2)3/2].
Details of the calculation:
(d) 
is the field due to sb at the position of q.
.
This is the same field as that of a point charge q' = -q(e-1)/(e+1) located at (-h, 0, 0).
(e) F = qE' = -q2(e-1)/[(e+1)4h2]i.

Problem 9:

Two spherical clouds, each of 1 km diameter, are separated by a distance of 2 km (center-to-center).  Each cloud is uniformly charged with charge density r0. They are floating with their centers 1 km above a fresh-water mountain lake of non-conducting pure water with dielectric constant e = 1.7e0. The bottom surface of each cloud is at
a potential of 106 volts compared to the lake immediately below it.
(a)  What is the charge density r0?
(b)  What is the electrostatic force on each cloud?
(c)  What is the index of refraction of the lake water?

Solution:

Concepts, principles, relations that apply to the problem:
The method of images, Coulomb's law
Why do they apply?
The electrostatic force on each cloud is the same as that due to the charges on the other cloud and due to the images charges behind the dielectric-dielectric interface.
How do they apply?

(a)  Pick a coordinate system:
Center of cloud 1: r1 = (0, 0,1 )
Center of cloud 2: r2 = (2, 0, 1)
The potential in the region z > 0 can be found by postulating two image charges.
Center of image cloud 1: r1' = (0, 0, -1)
Center of image cloud 2: r2' = (2, 0, -1)
The charge density of each image cloud is
r' = -r0(e-e0)/(e+e0) = -(0.7/2.7)r0  = -0.26r0.
Find the potential at r = (0, 0, 0):
f(r) = [r0(4/3)p(0.5 km)3/(4pe0)][1/(1 km) - 0.26/(1 km) + 1/(Ö5 km) - 0.26/(Ö5 km)]
= 4.47*104(r0/e0)m2.
Find the potential at r = (0, 0, 0.5):
f(r) = [r0(4/3)p(0.5 km)3/(4pe0)][1/(0.5 km) - 0.26/(1.5 km) + 1/(Ö(4.25) km) - 0.26/(Ö(6.25) km)]
= 9.20*104(r0/e0)m2.
Df = 106 V = (9.20 -4.47)104(r0/e0)m2,  r0 = 1.87*10-10 C/m3.
Details of the calculation:
(b)  Take cloud 1.  The force on cloud 1 is the same as that due to cloud 2 and image clouds 1 and 2.
The force on cloud 1 due to each of these clouds is the same as that due to point charges with the same total charge at the center of the clouds.  I can be calculated assuming the total charge of cloud 1 is at its center.
Force on cloud 1: 
Fx = [r0V/(4pe0)][r0V/(4 km2) + (2/Ö8)r'V/(8 km2)]
Fz = [r0V/(4pe0)][r'V/(4 km2) + (2/Ö8)r0V/(8 km2)]
Here V = (4/3)p(0.5 km)3.
F = Fxi + Fzk
Force on cloud 2: 
F = -Fxi + Fzk  from symmetry.
(c)  n = c/v = (em/e0m0)1/2.  Assume m = m0, then n = (e/e0)1/2 = Ö(1.7).

Problem 10:

Prove that the interaction energy of two dipoles P1 and P2 separated by a relative position R is given by
P1×P2/R3 - 3(P1×R)(P2×R)/R5.

Solution:

Concepts, principles, relations that apply to the problem:
The dipole field, a dipole in an external field
Why do they apply?
We have one dipole in the external field of another dipole.  The energy of a dipole in an external field is U = -P×E.
How do they apply?
Consider dipole P2 in the external field of dipole P1.  Choose your coordinate system such that P1 is at the origin and P2 is at R.
Field of P1 at R:  E(R) = 3(P1×R)R/R5 - P1/R3,  (Gaussian units).
Energy of  in this field: U = -P2×E(R) = P1×P2/R3 - 3(P1×R)(P2×R)/R5.
Details of the calculation:
None