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Chapter 3: Q3.5P (page 124)

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

is given and either V or the normal derivative a ∂V/∂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 ρ is given and either V or the normal derivative a ∂V/∂n is specified on each boundary surface.

Step by step solution

01

Given data

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

02

Second uniqueness theorem

From the proof of the second uniqueness theorem

∮SV3E→3·da→=-∫V(E3)2dτ

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 ∂V∂n=-E⊥) is specified on a boundary surface

E3,⊥=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 more elegant proof of the second uniqueness theorem uses Green's

identity (Prob. 1.61c), with T=U=V3. Supply the details.

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.

A circular ring in thexy plane (radius R , centered at the origin) carries a uniform line charge λ. Find the first three terms(n=0,1,2) in the multi pole expansion for V(r,θ).

Buckminsterfullerine is a molecule of 60 carbon atoms arranged

like the stitching on a soccer-ball. It may be approximated as a conducting spherical shell of radius R=3.5A°. A nearby electron would be attracted, according to Prob. 3.9, so it is not surprising that the ion C60-exists. (Imagine that the electron on average-smears itself out uniformly over the surface.) But how about a second electron? At large distances it would be repelled by the ion, obviously, but at a certain distance r (from the center), the net force is zero, and closer than this it would be attracted. So an electron with enough energy to get in that close should bind.

(a) Find r, in A°. [You'll have to do it numerically.]

(b) How much energy (in electron volts) would it take to push an electron in (from

infinity) to the point r? [Incidentally, the C60-ion has been observed.]

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.]
See all solutions

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