Assignment 10, solutions

Problem 1:

A dielectric sphere radius "a" and relative permittivity εr is placed in a uniform electric field.  The field region is large so the presence of the sphere does not perturb the field sources.
(a)  Assuming the field inside the sphere to be uniform and the field outside to be described by the uniform field plus a dipole field centered on the sphere, find the fields and the dipole moment induced in the sphere by applying boundary conditions at one point.
(b)  What happens if the sphere is conducting?

Solution:

 Concepts: The dipole field, boundary conditions Reasoning: Place the sphere at the origin.  Assume external field points in the z-direction, Eext = E0k. The external electric field will polarize the sphere.  The electric field of a uniformly polarized sphere is uniform inside the sphere and a pure dipole field outside the sphere.  Assume that the total field inside the sphere is the superposition of the field of a uniformly polarized sphere and the constant external field E0k.  The superposition of these two fields fulfills all the boundary conditions for D on the surface of the sphere.  We have a unique solution as long as inside D = εrE inside the sphere. Details of the calculation: (a)  Let us first concentrate on the field produced by a polarized sphere with dipole moment p. The field produced by the polarized sphere outside the sphere is E(r) = (4πε0)-1(3(p∙r)r/r5 - p/r3), with p = pk = dipole moment of the sphere. Then at r = ak (on the z-axis) we have Eout(sphere)= pk/(2πε0a3) = 2Pk/(3ε0), with P = p/(4πa3/3). Eout(sphere) - Ein(sphere) = (σp/ε0)k = (P/ε0)k. Therefore Ein(sphere) = -Pk/(3ε0), for the field produced by the polarized sphere. Now let us look at the total fieldinside and outside of the sphere. The total field at r = ak (on the z-axis) inside the sphere is Ein = (E0 - P/(3ε0))k,  Din = εr(E0 - P/(3ε0))k. The total field at r = ak (on the z-axis) outside the sphere is Eout = (E0 + 2P/(3ε 0))k,  Dout = ε0(E0 + 2P/(3ε0))k. The normal component of D is continuous. εr(E0 - P/(3ε0)) = ε0(E0 + 2P/(3ε0)),  (2ε0 + εr)P/(3ε0) = (εr - ε0)E0. P = 3ε0(εr - ε0)E0/(2ε0 + εr), p = P4πa3/3 = 4πε0a3(εr - ε0)E0/(2ε0 + εr). Ein = E03ε0/(2ε0 + εr), Din = εrEin, p = 4πε0a3(εr - ε0)E0/(2ε0 + εr). (b)  When the sphere is conducting the total field at r = ak (on the z-axis) inside the sphere is Ein = (E0 - P/(3ε0))k = 0.P = 3ε0E0. p = 4πε0a3E0.

Problem 2:

This question probes your understanding of dielectrics and associated fields and sources.  For this question, dielectric means linear isotropic homogeneous (lih) dielectric.
(a)  If one presumes that there exists a true charge density ρtrue, a polarization or bound charge density ρbound, and a total charge density ρtotal, such that ρtrue + ρbound = ρtotal, 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.

(b)  Draw a diagram of a lih dielectric of thickness c between the plates of a parallel plate capacitor of separation d with a gap of thickness e between each plate and the dielectric.  Presume that there is a constant voltage of V volts applied to the capacitor at time t = 0 (by, for example, connecting a V volt battery at time t = 0.)  Discuss and draw diagrams of the fields at t = dt for very small dt and the evolution of the fields to t = ∞.

(c)  Repeat part (b) with the lih dielectric replaced by a conductor.  Make sure that you consider all of the fields, creatively defining P in the conductor, masquerading (for purposes of this question) as a dielectric.  Comment on the similarities and differences between the lih dielectric and the conductor as a dielectric.  (It is strongly suggested that you do not consider part (c) as meaningless.)

Solution:

 Concepts: Maxwell's equations for electrostatics Reasoning: Maxwell's equations connect the source equations for E and P and yield the source equations for D. Details of the calculation: (a)  Maxwell's equations for electrostatics are ∇∙E = ρtotal/ε0,  ∇×E = 0, (SI units). ∇∙E = ρtotal/ε0 is the source equation for E. Let ρtrue + ρbound = ρtotal, then ∇∙E = (ρtrue + ρbound)/ε0 = ρtrue/ε0 - ∇∙P/ε0. ∇∙P = -ρbound is the source equation for P. ∇∙(ε0E + P) = ρtrue.  D = ε0E + P, ∇∙D = ρtrue is the source equation for D. Space derivatives of E, P, and D must exist, otherwise use the integral form. For example:   ∫V ∇∙D dV = ∮A D∙n dA = ∫V ρtrue dV = Qfree inside. For lih materials:P = ε0χeE, D = ε0(1 + χe)E =  ε0κeE = εE.Equations of macroscopic electrostatics:  ∇∙D = ρtrue,  ∇×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 = εE.  Unfavorable external conditions, however, can destroy those special properties. (b)  This is a quasi-static problem.  The voltage across the plates is constant.  For a very short time a polarization current is flowing, until the free and bound surface charge densities have reached their equilibrium values. The magnitude of the electric field produced by two large plates with surface charge densities ±σ, separated by a small distance d, is E = V/d = σ/ε0 in the region between the plates.  For the plates in this problem, at any time t, D is uniform throughout the capacitor.  D, E and P point in the x-direction.  E abruptly changes at the vacuum-dielectric boundary.  P is only non-zero in the dielectric. Let jp denote the polarization current, σf the magnitude free surface charge density on the capacitor plates, and σb the magnitude of the bound surface charge density on the dielectric.At any time we have V = (σf/ε0)d - (σb/ε0)c, σb = |P∙n|,  dσb/dt = dP/dt = jp,  σf = D.At t = 0, σb = 0 and P = 0.  E0 = V/d inside the dielectric, and σf = ε0V/d.As t --> ∞, σb = P = D - ε0E = D(1 - ε0/ε) = σf(1 - ε0/ε).Therefore σf = (ε0V + σbc)/d = ε0V/(d + c(ε0/ε - 1)). Inside the dielectric E = σf/ε =  ε0V/(ε d + c(ε0 - ε )). As we connect the battery, we can assume that the circuit behaves like a RC circuit, and that σb increases from 0 to σb(∞). σb(t) = σb(∞)(1 - e-t/τ), where τ is the time constant of the circuit.  For very small times dt, σb(t) = σb(∞)dt/τ increases linearly. σf(t) = (ε0V + σb(t)c)/d also increases with time. The electric field in the vacuum region Eout(t) = V/d + σb(t)c/(dε0) increases with time, and the electric field in the dielectric Ein(t) = (V - Eout(t)(d - c))/c decreases with time. (c)  In the conductor, as t --> ∞, E = 0.  But for very small times E = j/σc, where σc is the conductivity. Under certain circumstances, when no steady currents can flow and the free charges cannot travel large distances, we can formally treat the conductor like a dielectric with a very large ε and only bound charges. For inside the conductor we the write dP/dt = dσb/dt = jp  = σcEin. We therefore have P = D - ε0Ein = σf -  ε0j/σc. σb = σf - ε0Ein = σf - (ε0/σc)dσb/dt = ε0V/d + σbc/d - (ε0/σc)dσb/dt. dσb/dt = σcV/d - (1 - c/d)(σc/ε0)σb. σb(t) = (ε0V/(d - c))((1 - e-t/τ) = σf(∞)(1 - e-t/τ), with 1/τ = (1 - c/d)(σc/ε0). For a conductor with a given conductivity, we can find the time constant τ. For very small times dt, σb(t) increases linearly.  σf(t) also increases with time. The electric field in the vacuum region increases from V/d to V/(d - c), and the electric field in the conductor decreases from V/d to zero.

Problem 3:

One can locate resistivity anomalies in the ground as shown in the figure below.

The current I flowing between electrodes C1 and C2 establishes an electric field in the ground.  One measures the voltage V between a pair of electrodes, with P1 and P2 maintained at a fixed spacing b. With b << a, V/b is equal to E at the position x.  Anomalies in ground conductivity show up in the curve of as a function of x.
Show that if the substrate conductivity is uniform and equal to σ, then
V/b = 2axI/[πσ(x2 - a2)2].
The electrodes are of finite size.  However, you can perform the calculation on the assumption that they are infinitely small, disregarding the fact that E and j would then be infinite at their surfaces.
You can use the principle of superposition as follows.  The current in the ground is the sum of a radial distribution emanating from C1 plus another radial distribution converging on C2.

Solution:

 Concepts: Conductivity, E = j/σc Reasoning: The field in the ground is a superposition of a field E1 which points radially away from point C1 and a field E2 which points radially towards point C2.  Therefore the current in the ground is the sum of a radial distribution eminating from point C1 and another radial distribution converging on C2.  At some point a distance r1 from C1 and a distance r2 from C2 we have E = j1/σc + j2/σc = (r1/r1)I/(2πσcr12) - (r2/r2)I/(2πσcr22). Details of the calculation: On the axis at position x we have E = [I/(2πσc)][1/(x + a)2 - 1/(x - a)2] i. E = [I/(2πσc)](-4xa)/(x2 - a2)2 i = -2axI/[πσ(x2 - a2)2] i. E = V/b = 2axI/[πσ(x2 - a2)2].

#### Problem 4:

The distance between points A and B along a telegraph line, consisting of a pair of conducting wires, is L.  There is a single leak between the two wires at a distance x from point A.  If a voltage VA' is applied between the two wires at point A, the voltage between the two wires at point B is VB'.  However if a voltage VB" is applied between the two wires at point B, the voltage at A is VA".  Assuming the resistance per unit length of both wires is ρ derive a relationship giving the distance x of the leak as a function of L, VA' , VB',  VA" and VB".  Check your answer by showing that x = 19 miles if  L = 50 miles, VA'  = 200 volt, VA" = 40 volt, VB'= 40 volt, VB"  = 300 volt.

Solution:

 Concepts: Resistors in series Reasoning: With the leak between the two wires a current flows and we are asked to relate the voltage across the leak to the voltage across the ends of the wires. Details of the calculation: Let R be the resistance of the leak. VB' = RVA'/(2ρx + R),  VA'' = RVB''/(2ρ(L – x) + R) Eliminate R: VB'(2ρx + R) = RVA,  VB'2ρx = R(VA' – VB'),  R = VB'2ρx/(VA' – VB'). VA''(2ρ(L – x) + R) = RVB'', VA''2ρ(L – x)  = R(VB'' – VA''),  R = VA''2ρ(L – x)/(VB'' – VA'').  VB'x/(VA' - VB') = VA''L/(VB'' – VA'') - VA''x/(VB'' – VA''). x*(VB'/(VA' – VB') + VA''/(VB'' – VA'')) = VA''L/(VB'' – VA''). x*(VB'(VB'' – VA'') + VA''(VA' – VB')) = VA''L(VA' – VB'). x = VA''L(VA' – VB')/(VB' (VB'' – VA'') + VA''(VA' – VB')). Inserting the given values L = 50 miles, VA'  = 200 volt, VA" = 40 volt, VB'= 40 volt, VB"  = 300 volt yields x = 19 miles.

Problem 5:

In the circuit shown below, all three voltmeters are ideal and identical.  Each resistor has the same given resistance R.  Voltage V is also given.  Find the reading of each voltmeter.

Solution:

 Concepts: Resistors in parallel and in series, model of an ideal voltmeter Reasoning: We assume the voltmeters are ideal voltmeters and no current passes through the meters.  Then we have the points 1 and 2 are at potential V/3 and the point 3 is at potential 2V/3. We can then model the voltmeters in one of the following ways. (a)  Each voltmeter is represented by a capacitor. (b)  Each voltmeter is represented by a resistor r with r approaching infinity. Details of the calculation: (a)  Each voltmeter is represented by a capacitor. The voltage across the capacitor network is V' = V/3.  The total capacitance is C'.  Each capacitor has capacitance C.  1/C' = 1/C + 1/(2C).  C' = 2C/3. Q = V'C' = 2VC/9 is the magnitude of the separated charge. V3 = Q/C = 2V/9,  V2 = V1 = Q/(2C) = V/9. (b)  Each voltmeter is represented by a resistor r with r approaching infinity. > The voltage across the resistor network is V' = V/3. The total resistance is r'.  Each Resistor has resistance r.  r' = r + r/2,  r' = 3r/2. I = V'/r' = V/(3r') = 2V/(9r) is the magnitude of the current through the network. V3 = Ir = 2V/9,  V2 = V1  = V/9.