Show that the electric field of a (perfect) dipole (Eq. 3.103) can be written in the coordinate-free form

Edip(r)=14πε014πε01r3[3p·^rr-p]

Short Answer

Expert verified

Answer

The given relation is proved.

Step by step solution

01

Define functions

Write the expression for electric field.

Edipole(r,θ)=14πε01r3ρ[2cosθ^r+sinθ^θ] …… (1)

Here, ρis the dipole moment, θis the orientation of dipole electric field and ε0is the permittivity for the free space.

02

Determine electric field

Write the expression for the electric field.

Edipoler,θ=14πε01r32pcosθ^r+psinθ^θ=14πε01r32pcosθ^r-pcosθ^r+psinθ^θ=14πε01r33pcosθ^r-pcosθ^r+psinθ^θ …… (2)

Write the dipole moment vector.

p=pcosθ^θ …… (3)

p·^r=pcosθ^r-psinθ^θ·^r=pcosθ …… (4)

Substitute pcosθ^r-psinθ^θfor ρand pcosθfor p·^rin equation (2).

Edipoler,θ=14πε01r33pcosθ^r-pcosθ^r+psinθ^θEdipoler,θ=14πε01r33p·^rr-p

Thus, the given relation is proved.

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Most popular questions from this chapter

In Section 3.1.4, I proved that the electrostatic potential at any point

in a charge-free region is equal to its average value over any spherical surface

(radius R )centered at .Here's an alternative argument that does not rely on Coulomb's law, only on Laplace's equation. We might as well set the origin at P .Let Vave(R)be the average; first show that

dVavedR=14πR2V.da

(note that the R2in da cancels the 1/R2out front, so the only dependence on R

is in itself). Now use the divergence theorem, and conclude that if Vsatisfies

Laplace's equation, then,Vave(0)=V(P),forallR18.

Two long, straight copper pipes, each of radius R, are held a distance

2d apart. One is at potential V0, the other at -V0(Fig. 3.16). Find the potential

everywhere. [Hint: Exploit the result of Prob. 2.52.]

Show that the average field inside a sphere of radius R, due to all the charge within the sphere, is

Eave=-14πε0ρR3

Where ρis the total dipole moment. There are several ways to prove this delightfully simple result. Here's one method:

(a) Show that the average field due to a single chargeqat point r inside thesphere is the same as the field at r due to a uniformly charged sphere with

ρ=q/(43πR3), namely

14πε0(43πR3)qr2rdζ'

Where r is the vector from r to dζ

(b) The latter can be found from Gauss's law (see Prob. 2.12). Express the answerin terms of the dipole moment of q.

(c) Use the superposition principle to generalize to an arbitrary charge distribution.

(d) While you're at it, show that the average field over the volume of a sphere, dueto all the charges outside, is the same as the field they produce at the center.

Two point charges, 3q and -q, are separated by a distance a. For each of the arrangements in Fig. 3.35, find (i) the monopole moment, (ii) the dipole moment, and (iii) the approximate potential (in spherical coordinates) at large r (include both the monopole and dipole contributions).

The potential at the surface of a sphere (radius R) is given by
V0=kcos3θ,

Where kis a constant. Find the potential inside and outside the sphere, as well as the surface charge densityσ(θ) on the sphere. (Assume there's no charge inside or outside the sphere.)

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