Show that the energy of an ideal dipole p in an electric field E isgiven by

U=pE

Short Answer

Expert verified

Thepotential energy for the dipole moment ispE .

Step by step solution

01

Determine the formulas

Consider the formula for the torque on the dipole moment as

τ=pEsinθ

Here, pis the dipole moment and E is the electric field.

02

Determine the formula for the energy of the dipole moment 

Consider the formula for work done in the rotating dipole moment as:

dw=τdθ

Substitute the values and solve as

dw=pEsinθdθw=a1a2pEsinθdθw=pE(cosθ1cosθ2)

Consider the change in the potential for the two dipole positions and the corresponding working done as

w=U(θ2)U(θ1)=pE(cosθ2θ1)=pEcosθ=pE

Therefore, the potential energy for the dipole moment is pE.

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Most popular questions from this chapter

In a linear dielectric, the polarization is proportional to the field:

P=0χeE.If the material consists of atoms (or nonpolar molecules), the induced

dipole moment of each one is likewise proportional to the fieldp=αE . Question:

What is the relation between the atomic polarizabilityand the susceptibility χe? Since P (the dipole moment per unit volume) is P (the dipole moment per atom)times N (the number of atoms per unit volume),P=Np=NαE, one's first inclination is to say that

χe=Nα0

And in fact this is not far off, if the density is low. But closer inspection reveals

a subtle problem, for the field E in Eq. 4.30 is the total macroscopicfield in the

medium, whereas the field in Eq. 4.1 is due to everything except the particular atom under consideration (polarizability was defined for an isolated atom subject to a specified external field); call this field Eelse· Imagine that the space allotted to each atom is a sphere of radius R ,and show that

E=1-Nα30Eelse

Use this to conclude that

χe=Nα/01-Nα/30

Or

α=30Nr-1r+2

Equation 4.72 is known as the Clausius-Mossottiformula, or, in its application to

optics, the Lorentz-Lorenzequation.

In Fig. 4.6,P1andP2are (perfect) dipoles a distance rapart. What is

the torque onP1due toP2? What is the torque onP2due toP1? [In each case, I want the torque on the dipole about its own center.If it bothers you that the answers are not equal and opposite, see Prob. 4.29.]

Two long coaxial cylindrical metal tubes (inner radius a,outer radiusb)stand vertically in a tank of dielectric oil (susceptibility χe,mass density ρ).The inner one is maintained at potential V,and the outer one is grounded (Fig. 4.32). To what height (h) does the oil rise, in the space between the tubes?

Calculate W,using both Eq. 4.55 and Eq. 4.58, for a sphere of radius

Rwith frozen-in uniform polarization P (Ex. 4.2). Comment on the discrepancy.

Which (if either) is the "true" energy of the system?

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·Ecosθ

(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)e-(u/kT)

where =sinθdθdϕ, and the integration is over all orientations θ:0π;ϕ:02πUse 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.

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