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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

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=14πR2∫∇V.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.

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)=1to 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 coefficients A1are the same in both hemispheres.]

Find the potential outside a charged metal sphere (charge Q, radius R) placed in an otherwise uniform electric field E0 . Explain clearly where you are setting the zero of potential.

A rectangular pipe, running parallel to the z-axis (from -∞to +∞), has three grounded metal sides, at y=0,y=aand x=0The fourth side, at x=b, is maintained at a specified potential V0(y).

(a) Develop a general formula for the potential inside the pipe.

(b) Find the potential explicitly, for the case V0(y)=V0(a constant).

For the infinite rectangular pipe in Ex. 3.4, suppose the potential on

the bottom (y= 0) and the two sides (x= ±b) is zero, but the potential on the top

(y=a) is a nonzero constant V0•Find the potential inside the pipe. [Note:This is a

rotated version of Prob. 3.15(b), but set it up as in Ex. 3.4, using sinusoidal functions in yand hyperbolics in x.It is an unusual case in which k= 0 must be included. Begin by finding the general solution to Eq. 3.26 when k= 0.]
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