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Chapter 3: Q3.5P (page 124)

Prove that the field is uniquely determined when the charge density 蚁

is given and either V or the normal derivative a 鈭�V/鈭�n is specified on each boundary surface. Do not assume the boundaries are conductors, or that V is constant over any given surface.

Short Answer

Expert verified

Answer:

The electric field is uniquely determined when the charge density 蚁 is given and either V or the normal derivative a 鈭�V/鈭�n is specified on each boundary surface.

Step by step solution

01

Given data

Either the potential V or the normal derivative of potential 鈭�V/鈭�nis specified on each boundary surface.

02

Second uniqueness theorem

From the proof of the second uniqueness theorem

鈭�SV3E鈫�3路da鈫�=-鈭�V(E3)2d蟿

Here, V3is the difference in potential of two points on a surface S. E3 is the difference in electric field of two points in a volume V that S encloses.

03

Uniqueness of field in a volume

If the potential is specified on a boundary surface

V3=0

If the normal derivative of potential (that is 鈭�V鈭�n=-E鈯�) is specified on a boundary surface

E3,鈯�=0

Any of these conditions in equation (1) gives

E3=0

Thus, the electric field is uniquely determined.

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Most popular questions from this chapter

A charge is distributed uniformly along the z axis from z=-atoz=+a. Show that the electric potential at a point r is given by

Vr,胃=Q4蟺蔚01r1+13ar2P2cos胃+15ar4P4cos胃+...

for r>a.

The potential at the surface of a sphere (radius R) is given by

V0=kcos3胃,

Where k is a constant. Find the potential inside and outside the sphere, as well as the surface charge density 蟽(胃)on the sphere. (Assume there's no charge inside or outside the sphere.)

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

Eave=-14蟺蔚0蚁R3

Where 蚁is 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

蚁=q/(43蟺R3), namely

14蟺蔚0(43蟺R3)鈭�qr2rd味'

Where r is the vector from r to d味

(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.

(a) Show that the quadrupole term in the multipole expansion can be written as

Vquad(r鈫�)=14蟺蔚01r3鈭�i,j-13ri^rj^Qij ............(1)

(in the notation of Eq. 1.31) where

Qij=12鈭�[3r'jr'j-(r')2未ij]蚁(r'鈬赌)d蟿' ..........(2)

Here

未ij={10ifi=jifi鈮�j ..........(3)

is the Kronecker Delta and Qijis the quadrupole moment of the charge distribution. Notice the hierarchy

Vmon=14蟺蔚0Qr;Vdip=14蟺蔚0鈭�rjpj^r2;Vquad(r^)=14蟺蔚01r3鈭�ij-13rirj^^Qij;......

The monopole moment (Q) is a scalar, the dipole moment p鈬赌 is a vector, the quadrupole moment Qij is a second rank tensor, and so on.

(b) Find all nine componentsQij of for the configuration given in Fig. 3.30 (assume the square has side and lies in the x-y plane, centered at the origin).

(c) Show that the quadrupole moment is independent of origin if the monopole and

dipole moments both vanish. (This works all the way up the hierarchy-the

lowest nonzero multipole moment is always independent of origin.)

(d) How would you define the octopole moment? Express the octopole term in the multipole expansion in terms of the octopole moment.

Find the potential in the infinite slot of Ex. 3.3 if the boundary at x = 0 consists of two metal strips: one, from y = 0 to y = a/2, is held at a constant Potential V0, and the other, from y = a/2 to y = a , is at potential V0.
See all solutions

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