electrodynamics / Companion to J.D. Jackson's Classical Electrodynamics 3rd ed. - R. Magyar

Problem 3.3

A think, flat, circular, conducting disc of radius R is located in the x-y plane with its center at the origin and is maintained at a fixed potential Φ. With the information that the charge density of the disc at fixed potential is proportional to (R2 − ρ2)− 12 , where ρ is the distance out from the center of the disk. . .

Note ρ is used where I usually use r′. a. Find the potential for r > R.

For a charged ring at z = 0 on the r-φ plane, Jackson derived the following:

−

We can replace L by 2ℓ because every other term vanishes.

Since σ (R2 − ρ2)− 12 on the disk, the total charge on the disk is

And κ = 2πRQ . Now, we solve for a disk made up of infinitely many infinitesimally small rings. Each contributes to the potential

X

ℓ ρ2ℓ

δΦ(r, θ) = σ ℓ=0 r2ℓ+1 f (2ℓ)P2ℓ(cos θ)dA, r ≥ R

where f (2ℓ) = P2ℓ(0). And integrating over the disk gives the total potential.

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Consider the integral over ρ.

Let Rρ = sin θ, dρ = R cos θdθ.

but we know f (2ℓ).

Since (2ℓ)!! = 2ℓℓ!,

The potential on the disk at the origin is V.

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A similar integration can be carried out for r < R.

b. Find the potential for r < R.

I can’t figure out what I did here. I’ll get back to this. c. What is the capacitance of the disc?

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Problem 3.9

A hollow right circular cylinder of radius b has its axis coincident with the z axis and its ends at z = 0 and z = L. The potential on the end faces is zero while the potential on the cylindrical surface is given by Φ(φ, z). Using the appropriate separation of variables in cylindrical coordinates, find a series solution for the potential everywhere inside the cylinder.

V = 0 at z = 0, L. Because of cylindrical symmetry, we will try cylindrical coordinates. Then, we have

Try Φ(r, φ, z) = R(r)Z(z)Q(φ). Separating the Laplace equation in cylindrical coordinates, we find three di erential equations which must be satisfied.

∂2Z − k2Z = 0 ∂z2

has the solution

Z = A sin(kz) + B cos(kz)

The solution must satisfy boundary conditions that Φ = 0 at z = 0, L. Therefore, B must vanish.

Z = A sin(kz)

where k = nπL .

Similarly, we have for Q

∂2 Q − m2Q = 0 ∂φ2

which has the solution

Q = C sin(mφ) + D cos(mφ)

m must be an integer for Q to be single valued.

The radial part must satisfy the frightening equation. Note the signs. This is not the typical Bessel equations, but have no fear.

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where x = kr. The solutions are just modified Bessel functions.

R(x) = EIm(x) + F Km(x)

m must be an integer for R to be single valued. Im and Km are related to other Bessel and Neumann functions via

Im(kr) = i−mJm(ikr)

Km(kr) = π2 im+1Hm(1)(ikr)

The potential is finite at r = 0 so

Let A and E be absorbed into C and D.

Now, we match boundary conditions. At r = b, Φ(φ, z) = V (φ, z). So

The Im nπL b are just a set of constants so we’ll absorb them into Cmn′ and Dmn′ for the time being. The coe cients, Cmn′ and Dmn′ , can be obtained via Fourier analysis.

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κ is determined by orthonormality of the various terms.

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Problem 3.14

A line charge of length 2d with a total charge of Q has a linear charge density varying as (d2 −z2), where z is the distance from the midpoint. A grounded, conducting, spherical shell of inner radius b > d is centered at the midpoint of the line charge.

a. Find the potential everywhere inside the spherical shell as an expansion in Legendre polynomials.

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The term Pℓ(1) + Pℓ(−1) is zero for odd ℓ and 2Pℓ(1) for even ℓ. So we can rewrite our answer.

b.Calculate the surface charge density induced on the shell.

σ= −ǫ0 Φ · nˆ

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c. Discuss your answers to parts a and b in the limit d << b.

This is what we would have expected if a point charge were located at the origin and the sphere were at zero potential. When d << b, r will most likely be greater than d for the region of interest so it will su ce to take the limit of the second form for Φ. Once again, only ℓ = 0 terms will contribute.

!

Qb − r

Φ = 4πǫ0 br

This looks like the equation for a spherical capacitor’s potential as it should!

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Problem 4.6

A nucleus with quadrupole moment Q finds itself in a cylindrically symmetric electric field with a gradient(∂E∂zz )0 along the z axis at the position of the nucleus.

a. Find the energy of the quadrupole interaction.

Recall that the quadrapole tensor is

Z

Qij = 3x′ix′j − r′2δij ρ(~r′)dV ′

For the external field, Gauss’s law tells us that a vanishing charge density

When i =6 j, there is no contribution to the energy. You can understand this

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