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When you polarize a neutral dielectric, the charge moves a bit, but the total remains zero. This fact should be reflected in the bound charges σb and ÒÏb· Prove from Eqs. 4.11 and 4.12 that the total bound charge vanishes.

Short Answer

Expert verified

The value of total charge of a piece of neutral dielectricQtotal is 0.

Step by step solution

01

Write the given data from the question

Consider thereflected in the bound charges σband ÒÏb·

Consider when you polarize a neutral dielectric, the charge moves a bit, but the total remains zero.

02

Determine the formula of total charge of a piece of neutral dielectric Qtotal.

Write the formula of total charge of a piece of neutral dielectric Qtotal.

Qtotal=∮SσbdS+∫νÒÏbdÏ„ …… (1)

Here, σb and ÒÏb are the reflected in the bound charges.

03

 Determine the value of total charge of a piece of neutral dielectric Qtotal.

Determine thetotal charge of a piece of neutral dielectric.

SubstituteP→⋅n^ for σand∇⋅P→ forÒÏ into equation (1).

Qtotal=∮SσdS+∫ν∇⋅P→dτ

However, according to divergence law, the two terms are equivalent and cancel one another out.

Qtotal=0

Therefore, the value of total charge of a piece of neutral dielectricQtotal is 0.

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Most popular questions from this chapter

A sphere of radius R carries a polarization

P(r)=kr,

Where k is a constant and r is the vector from the center.

(a) Calculate the bound charges σband ÒÏb.

(b) Find the field inside and outside the sphere.

Calculate the potential of a uniformly polarized sphere (Ex. 4.2) directly from Eq. 4.9.

E2→Find the field inside a sphere of linear dielectric material in an otherwise uniform electric field E0→(Ex. 4.7) by the following method of successive approximations: First pretend the field inside is just E0→, and use Eq. 4.30 to write down the resulting polarization P0→. This polarization generates a field of its own, E1→ (Ex. 4.2), which in turn modifies the polarization by an amount P1→. which further changes the field by an amount E2→, and so on. The resulting field is E→0+E→1+E→2+.... . Sum the series, and compare your answer with Eq. 4.49.

The Clausius-Mossotti equation (Prob. 4.41) tells you how to calculatethe susceptibility of a nonpolar substance, in terms of the atomic polariz-ability. The Langevin equation tells you how to calculate the susceptibility of apolar substance, in terms of the permanent molecular dipole moment p. Here's howit goes:

(a) The energy of a dipole in an external field E isu=-p··¡³¦´Ç²õθ

(Eq. 4.6), whereθ is the usual polar angle, if we orient the z axis along E.

Statistical mechanics says that for a material in equilibrium at absolute temperature

T, the probability of a given molecule having energy u is proportional to

the Boltzmann factor,

exp(-u/kT)

The average energy of the dipoles is therefore

<u>=∫ue-(u/kt)»åΩ∫e-(u/kT)»åΩ

where »åΩ=²õ¾±²Ôθ»åθ»åÏ•, and the integration is over all orientations θ:0→π;Ï•:0→2Ï€Use this to show that the polarization of a substance

containing N molecules per unit volume is

P=Np[cothpE/kT-kT/pE] (4.73)

That's the Langevin formula. Sketch as a function ofPE/KT .

(b) Notice that for large fields/low temperatures, virtually all the molecules arelined up, and the material is nonlinear. Ordinarily, however, kT is much greaterthan p E. Show that in this regime the material is linear, and calculate its susceptibility,in terms of N, p, T, and k. Compute the susceptibility of water at 20°C,and compare the experimental value in Table 4.2. (The dipole moment of wateris 6.1×10-30C·m) This is rather far off, because we have again neglected thedistinction between E and Eelse· The agreement is better in low-density gases,for which the difference between E and Eelse is negligible. Try it for water vapor

at 100°C and 1 atm.

The space between the plates of a parallel-plate capacitor is filled

with dielectric material whose dielectric constant varies linearly from 1 at the

bottom plate (x=0)to 2 at the top plate (x=d).The capacitor is connectedto a battery of voltage V.Find all the bound charge, and check that the totalis zero.

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