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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).

Short Answer

Expert verified

Answer

  1. The total potential at a distance r including monopole term is 4θ4πε0[2qr+3qacosr2].

  2. The total potential at a distance r including monopole term is 4θ4πε0[2qr+qacosr2].

  3. The total potential at a distance r including monopole term is 4sinθ4πε0[2qr+3qasinϕr2].

Step by step solution

01

Given data

a)

From the above figure,

Monopole moment

Q=3q=2q

Dipole moment

p=3qaz+-q0p=3qaz

02

Determine monopole moment

Write the expression for total potential at a distanceincluding monopole term.

Vr=VmQnQ+Vdipole=14πε0Qr+14πε0p·^rr2=14πε02qr+3qacosθr2

Here, p·^r=3qacosθ

Therefore, the total potential at a distance r including monopole term is 14πε02qr+3qacosθr2.

03

Determine dipole moment

b)

From the above figure,

Monopole moment

Q=2q

Dipole moment

p=-qa-^z=qaz

Write the expression for total potential at a distance r including monopole term.

Vr=Vmono+Vdipole=14πε0Qr+14πε0ÒÏ·^rr2

Substitute 2qfor Q and 3qacosθfor p·^r.

Vr=14πε02qr+3qacosθr2

Therefore, the total potential at a distance rincluding monopole term is

14πε02qr+3qacosθr2.

04

Determine potential

c)

From the above figure,

Monopole moment

Q=2q

Dipole moment

p=3qay^

Write the expression for total potential at a distance r including monopole term.

Vr=Vmono+Vdipole=14πε0Qr+14πε0p·r^r2=14πε02qr+3qasinθsinϕr2

Therefore, the total potential at a distance r including monopole term is

14πε02qr+3qasinθsinϕr2

As,

r^=sinθcosϕx^+sinθsinϕy^+cosθz^

Then,

y^·r^=sinθsinϕ.

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Most popular questions from this chapter

A "pure" dipoleÒÏis situated at the origin, pointing in thezdirection.

(a) What is the force on a point charge q at (a,0,0)(Cartesian coordinates)?

(b) What is the force on q at (0,0,a)?

(c) How much work does it take to move q from(a,0,0)to (0,0,a)?

In Ex. 3.9, we derived the exact potential for a spherical shell of radius R , which carries a surface charge σ=kcosθ.

(a) Calculate the dipole moment of this charge distribution.

(b) Find the approximate potential, at points far from the sphere, and compare the exact answer (Eq. 3.87). What can you conclude about the higher multipoles?

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

Eave=-14πε0ÒÏR3

Where ÒÏis 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

ÒÏ=q/(43Ï€R3), 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.

In Prob. 2.25, you found the potential on the axis of a uniformly charged disk:

V(r,0)=σ2ε0(r2+R2-r)

(a) Use this, together with the fact that PI(1)=1, to evaluate the first three terms

in the expansion (Eq. 3.72) for the potential of the disk at points off the axis, assuming r>R.

(b) Find the potential for r<Rby the same method, using Eq. 3.66. [Note: You

must break the interior region up into two hemispheres, above and below the

disk. Do not assume the coefficientsAIare the same in both hemispheres.]

You can use the superposition principle to combine solutions obtained by separation of variables. For example, in Prob. 3.16 you found the potential inside a cubical box, if five faces are grounded and the sixth is at a constant potential V0; by a six-fold superposition of the result, you could obtain the potential inside a cube with the faces maintained at specified constant voltages . V1,V2,......V6In this way, using Ex. 3.4 and Prob. 3.15, find the potential inside a rectangular pipe with two facing sides (x=±b)at potential V0, a third (y=a)at V1. and the last at(y=a) grounded.

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