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A point charge qis situated a large distance rfrom a neutral atom of

polarizability α.Find the force of attraction between them.

Short Answer

Expert verified

The force on charge q situated at a distance rfrom a neutral atom of

polarizability αis 2αq216π2ε02r5.

Step by step solution

01

Given data

A point charge qis situated a large distance rfrom a neutral atom of

polarizability α.

02

Electric field due to a dipole

The magnitude of electric field due to a dipole having dipole moment ÒÏat spherical polar coordinateθ=Ï€

Ed=2p4ττε0r3......(1)

Here, ε0is the permittivity of free space and r is the distance from the centre of the dipole.

03

Derivation of force on a charge due to a dipole

The electric field due to the charge q at the position of the dipole is

Eq=q4πε0r2

The expression for the dipole moment is

p=αEq

Substitute the expression for Eqin the above equation we get

p=αq4πε0r2

Substitute this expression for dipole moment in equation (1)

Ed=24πε0r3αq4πε0r2=2αq16π2ε02r5

The expression for the force on the charge is

F=qEd

Substitute the expression for Edin the above equation and get

role="math" localid="1657529076590" F=2αq216π2ε02r5

Thus, the force on the charge due to the dipole is F=2αq216π2ε02r5.

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Most popular questions from this chapter

The space between the plates of a parallel-plate capacitor (Fig. 4.24)

is filled with two slabs of linear dielectric material. Each slab has thickness a, sothe total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, andslab 2 has a dielectric constant of 1.5. The free charge density on the top plate is aand on the bottom plate-σ.

(a) Find the electric displacement Dineach slab.

(b) Find the electric field E in each slab.

(c) Find the polarization P in each slab.

(d) Find the potential difference between the plates.

(e) Find the location and amount of all bound charge.

(f) Now that you know all the charge (free and bound), recalculate the field in eachslab, and confirm your answer to (b).

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.

Suppose the field inside a large piece of dielectric is E0, so that the electric displacement is D0=ε0E0+P.

(a) Now a small spherical cavity (Fig. 4.19a) is hollowed out of the material. Find the field at the center of the cavity in terms of E0and P. Also find the displacement at the center of the cavity in terms of D0and P. Assume the polarization is "frozen in," so it doesn't change when the cavity is excavated. (b) Do the same for a long needle-shaped cavity running parallel to P (Fig. 4.19b).

(c) Do the same for a thin wafer-shaped cavity perpendicular to P (Fig. 4.19c). Assume the cavities are small enough that P,E0, and D0are essentially uniform. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization.]

Question: A sphere of linear dielectric material has embedded in it a uniform

free charge density . Find the potential at the center of the sphere (relative to

infinity), if its radius is R and the dielectric constant is ∈r.

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