Assignment 10, solutions

**Problem 1:**

A dielectric sphere radius "a" and relative permittivity
ε_{r} 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, E_{ext} = E_{0}k.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 E _{0}k. 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 = ε_{r}E 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) = (4πε_{0})^{-1}(3(p∙r)r/r^{5}
- p/r^{3}),with p = pk = dipole moment of the sphere.Then at r = ak (on the
z-axis) we have E _{out}(sphere)= pk/(2πε_{0}a^{3})
= 2Pk/(3ε_{0}),
with P = p/(4πa^{3}/3).E_{out}(sphere) - E_{in}(sphere) = (σ_{p}/ε_{0})k = (P/ε_{0})k.
Therefore E_{in}(sphere) = -Pk/(3ε_{0}),
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
E_{in} = (E_{0} - P/(3ε_{0}))k, D_{in} =
ε_{r}(E_{0} - P/(3ε_{0}))k.The total field at r = ak (on the z-axis) outside the sphere
is E _{out} = (E_{0} + 2P/(3ε_{
0}))k, D_{out} = ε_{0}(E_{0} + 2P/(3ε_{0}))k.The normal component of D is continuous.ε _{r}(E_{0} - P/(3ε_{0})) = ε_{0}(E_{0}
+ 2P/(3ε_{0})), (2ε_{0 }+ ε_{r})P/(3ε_{0})
= (ε_{r} -
ε_{0})E_{0}.P = 3ε _{0}(ε_{r} - ε_{0})E_{0}/(2ε_{0 }+ ε_{r}), p = P4πa^{3}/3 = 4πε_{0}a^{3}(ε_{r}
- ε_{0})E_{0}/(2ε_{0 }+ ε_{r}).E_{in} = E_{0}3ε_{0}/(2ε_{0 }+ ε_{r}), D_{in} = ε_{r}E_{in},
p = 4πε_{0}a^{3}(ε_{r}
- ε_{0})E_{0}/(2ε_{0 }+ ε_{r}).(b) When the sphere is conducting the total field at r = ak
(on the z-axis) inside the sphere is E _{in} = (E_{0} - P/(3ε_{0}))k
= 0.P = 3ε_{0}E_{0}. p
= 4πε_{0}a^{3}E_{0}. |

**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 ρ_{true},
a polarization or bound charge density ρ_{bound}, and a
total charge density ρ_{total}, such that ρ_{true} + ρ_{bound }= ρ_{total}, 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 = ρ_{total}/ε_{0},
∇×E = 0, (SI units).∇∙E = ρ_{total}/ε_{0}
is the source equation for E. Let ρ _{true }+ ρ_{bound
}= ρ_{total}, then
∇∙E = (ρ_{true }+ ρ_{bound})/ε_{0}
= ρ_{true}/ε_{0} -
∇∙P/ε_{0}.∇∙P = -ρ_{bound
}is the source equation for P.∇∙(ε_{0}E + P)
= ρ_{true}. D = ε_{0}E + P,
∇∙D = ρ_{true} 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 }ρ_{true}
dV = Q_{free inside}.For lih materials: P = ε_{0}χ_{e}E, D = ε_{0}(1
+ χ_{e})E = ε_{0}κ_{e}E = εE.Equations of macroscopic electrostatics: ∇∙D = ρ_{true, }∇×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/ε _{0}
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 j_{p} denote the polarization current, σ_{f}
the magnitude free surface charge density on the capacitor plates, and σ_{b}
the magnitude of the bound surface charge density on the dielectric.At any time we have V = (σ _{f}/ε_{0})d
- (σ_{b}/ε_{0})c, σ _{b} = |P∙n|, dσ_{b}/dt = dP/dt
= j_{p}, σ_{f} = D.At t = 0, σ _{b} = 0 and P = 0. E_{0} = V/d inside
the dielectric, and σ_{f} = ε_{0}V/d.As t --> ∞, σ _{b} = P = D - ε_{0}E = D(1 - ε_{0}/ε)
= σ_{f}(1 - ε_{0}/ε).Therefore σ _{f} = (ε_{0}V + σ_{b}c)/d =
ε_{0}V/(d + c(ε_{0}/ε - 1)).Inside the dielectric E = σ _{f}/ε = ε_{0}V/(ε d + c(ε_{0} -
ε )).As we connect the battery, we can assume that the circuit behaves like a RC circuit, and that σ _{b} increases from 0 to σ_{b}(∞).σ _{b}(t) = σ_{b}(∞)(1 - e^{-t/τ}), where τ
is the time constant of the circuit. For very small times dt, σ_{b}(t)
= σ_{b}(∞)dt/τ increases linearly.σ _{f}(t) = (ε_{0}V + σ_{b}(t)c)/d
also increases with time.The electric field in the vacuum region E _{out}(t) = V/d + σ_{b}(t)c/(dε_{0}) increases with time, and the electric field in the dielectric
E_{in}(t) = (V - E_{out}(t)(d - c))/c decreases with time.(c) In the conductor, as t --> ∞, E = 0. But for very small times E = j/σ _{c}, where σ_{c} 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 = dσ _{b}/dt
= j_{p} = σ_{c}E_{in}.We therefore have P = D - ε _{0}E_{in} = σ_{f} -
ε_{0}j/σ_{c}.σ _{b} = σ_{f} - ε_{0}E_{in} = σ_{f}
- (ε_{0}/σ_{c})dσ_{b}/dt = ε_{0}V/d
+ σ_{b}c/d - (ε_{0}/σ_{c})dσ_{b}/dt.dσ _{b}/dt = σ_{c}V/d - (1 - c/d)(σ_{c}/ε_{0})σ_{b.
}σ_{b}(t) = (ε_{0}V/(d - c))((1 - e^{-t/τ})
= σ_{f}(∞)(1 - e^{-t/τ}), with 1/τ = (1 - c/d)(σ_{c}/ε_{0}).For a conductor with a given conductivity, we can find the time constant τ. For very small times dt, σ _{b}(t) increases linearly. σ_{f}(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 C_{1 }and
C_{2 }establishes an electric field in the ground. One measures the voltage
V between a pair of electrodes, with P_{1 }and P_{2 }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(x^{2} - a^{2})^{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 C_{1 }plus
another radial distribution converging on C_{2}.

Solution:

Concepts: Conductivity, E = j/σ_{c} | |

Reasoning: The field in the ground is a superposition of a field E_{1}
which points radially away from point C_{1} and a field E_{2}
which points radially towards point C_{2}. Therefore the
current in the ground is the sum of a radial distribution eminating from
point C_{1} and another radial distribution converging on C_{2}.
At some point a distance r_{1} from C_{1} and a
distance r_{2} from C_{2} we haveE = j_{1}/σ_{c} + j_{2}/σ_{c}
= (r_{1}/r_{1})I/(2πσ_{c}r_{1}^{2})
- (r_{2}/r_{2})I/(2πσ_{c}r_{2}^{2}). | |

Details of the calculation: On the axis at position x we have E = [I/(2πσ_{c})][1/(x + a)^{2} - 1/(x - a)^{2}] i.E = [I/(2πσ_{c})](-4xa)/(x^{2} - a^{2})^{2}
i
= -2axI/[πσ(x^{2} - a^{2})^{2}] i.E = V/b = 2axI/[πσ(x ^{2} - a^{2})^{2}]. |

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 V_{A}^{'} is applied between the two wires at point A,
the voltage between the two wires at point B is V_{B}^{'}.
However if a voltage V_{B}" is applied between the two wires at point B,
the voltage at A is V_{A}". 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, V_{A}^{'} , V_{B}^{'}, V_{A}"
and V_{B}". Check your answer by showing that x = 19 miles if L = 50
miles, V_{A}^{'} = 200 volt, V_{A}" = 40 volt, V_{B}^{'}=
40 volt, V_{B}" = 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. V _{B'} = RV_{A'}/(2ρx + R), V_{A''} = RV_{B''}/(2ρ(L – x) + R)Eliminate R: V _{B'}(2ρx + R)
= RV_{A}, V_{B'}2ρx = R(V_{A'} – V_{B'}), R = V_{B'}2ρx/(V_{A'} – V_{B'}).
V _{A''}(2ρ(L – x)
+ R) = RV_{B''}, V_{A''}2ρ(L – x) = R(V_{B''} – V_{A''}), R = V_{A''}2ρ(L – x)/(V_{B''
}– V_{A''}).
V _{B'}x/(V_{A'} - V_{B'}) = V_{A''}L/(V_{B''} – VA_{''}) - V_{A''}x/(V_{B''} – V_{A''}). x*(V _{B'}/(V_{A' }– V_{B'}) + V_{A''}/(V_{B''} – V_{A''})) = V_{A''}L/(V_{B''} – V_{A''}).x*(V _{B'}(V_{B''} – V_{A''}) + V_{A''}(V_{A'
}– V_{B'})) = V_{A''}L(V_{A' }– V_{B'}).x = V _{A''}L(V_{A' }– V_{B'})/(V_{B'} (V_{B''} – V_{A''}) + V_{A''}(V_{A'
}– V_{B'})).Inserting the given values
L = 50 miles, V_{A}^{'} = 200 volt, V_{A}" = 40
volt, V_{B}^{'}= 40 volt, V_{B}" = 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. | |

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.
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