Solution - homework 4

(PHQ350)

October 13, 2009

Note that problems 1 and 2 were marked by A. Takayama and problems 3, 4 and 5 by J.-F. Paquet

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2

3

4

5

Problem 3 (modified 5.15)

There is no z dependence:

V (r) = V (s, φ)(s2 = x2 + y2, tan(φ) =

y

x

)In cylindrical coordinates, the Laplacian can be written as:

∇2V (r) =1s

∂

∂s

(s∂V (r)∂s

)+

1s2∂2V (r)∂φ2

If V (r) = F (s)G(φ),

∇2V (r) =G(φ)s

d

ds

(sdF (s)ds

)+F (s)s2

d2G(φ)dφ2

s

F (s)d

ds

(sdF (s)ds

)+

1G(s)

d2G(φ)dφ2

⇒ s

F (s)d

ds

(sdF (s)ds

)= λ (1)

⇒ 1G(φ)

d2G(φ)dφ2

= −λ; (2)

with λ ∈ RThe general solution for G(φ) is of course:

G(φ) = A sin(√λφ) +B cos(

√λφ) + Cφ

If we want G(φ) to have a period of 2π, the values of λ and C are restricted to:

C = 0;√λ = n, n ∈ N

So there is actually an infinite number of solutions for (2):

Gn(φ) = An sin(nφ) +Bn cos(nφ)

Note that fixing λ affects (1) too, so for each Gn(φ), there will be an associated Fn(r) that solves (1).The solution to (1) is:

Fn(r) = Dnsn + Ens

−n +Hn ln(s) (3)

Thus:

Vn(s, φ) = (An sin(nφ) +Bn cos(nφ))(Dns

n + Ens−n +Hn ln(s)

)or:

V (s, φ) =∞∑n=0

[(An sin(nφ) +Bn cos(nφ))

(Dns

n + Ens−n +Hn ln(s)

)](4)

Inside the cylinder, we want the solution to be finite at r = 0, thus:

Vint(s, φ) =∞∑n=0

[(An sin(nφ) +Bn cos(nφ)) (Dnsn)] =

∞∑n=0

[sn (A′n sin(nφ) +B′n cos(nφ))]

Outside the cylinder, we want the potential to go to 0 at infinity, thus:

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Vext(s, φ) =∞∑n=0

[s−n (A′′n sin(nφ) +B′′n cos(nφ))

]We also need Vint(s = R,φ) = Vext(s = R,φ)

Vint/ext(s = R,φ) =∞∑n=0

[(A′′′n sin(nφ) +B′′′n cos(nφ))]

The boundary conditions at s = R are:

V (s = R,φ) ={V0 if 0 ≤ φ < π/2 or π ≤ φ < 3π/2−V0 if π/2 ≤ φ < π or 3π/2 ≤ φ < 2π

A′′′n and B′′′n can be calculated the usual way:

B′′′n = 1π

∫ π0dφV (R,φ) cos(nφ)

= 1π

[∫ π/20

dφ (V0) cos(nφ) +∫ ππ/2

dφ (−V0) cos(nφ)+∫ 3π/2

πdφ (V0) cos(nφ) +

∫ 2π

3π/2dφ (−V0) cos(nφ)

]= V0

nπ

[sin(nφ)|π/20 − sin(nφ)|ππ/2 + sin(nφ)|3π/2π − sin(nφ)|2π3π/2

]= V0

nπ [2 sin(nπ/2)− sin(0)− 2 sin(nπ) + 2 sin(n3π/2)− sin(n2π)]= 2V0

nπ [sin(nπ/2) + (−1)n sin(nπ/2)]= 2V0

nπ sin(nπ/2)︸ ︷︷ ︸0 if n is even

[1 + (−1)n]︸ ︷︷ ︸0 if n is odd

= 0

since

sin(3nπ/2) = sin(nπ/2 + nπ) = sin(nπ/2) cos(nπ) + cos(nπ/2) sin(nπ)= sin(nπ/2)(−1)n + cos(nπ/2)(0)= (−1)n sin(nπ/2)

Also:

A′′′n = 1π

∫ π0dφV (R,φ) sin(nφ)

= 1π

[∫ π/20

dφ (V0) sin(nφ) +∫ ππ/2

dφ (−V0) sin(nφ)+∫ 3π/2

πdφ (V0) sin(nφ) +

∫ 2π

3π/2dφ (−V0) sin(nφ)

]= −V0

nπ

[cos(nφ)|π/20 − cos(nφ)|ππ/2 + cos(nφ)|3π/2π − cos(nφ)|2π3π/2

]= −V0

nπ [2 cos(nπ/2)− cos(0)− 2 cos(nπ) + 2 cos(n3π/2)− cos(n2π)]= −2V0

nπ [cos(nπ/2)− (−1)n + (−1)n cos(nπ/2)− 1]= −2V0

nπ [(1 + (−1)n)(cos(nπ/2)− 1)]= −2V0

nπ (1 + (−1)n)︸ ︷︷ ︸0 if n is odd

(cos(nπ/2)− 1)︸ ︷︷ ︸0 if cos(nπ/2) = 1

=

0 if n is odd0 if n is even and a multiple of 48V0nπ if n is even and not a multiple of 4

We have used:

cos(3nπ/2) = cos(nπ/2 + nπ) = cos(nπ/2) cos(nπ)− sin(nπ/2) sin(nπ)= cos(nπ/2)(−1)n − sin(nπ/2)(0)= (−1)n cos(nπ/2)

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n even and not a multiple of 4 can be expressed as n = 2(2k + 1), k = 0, 1, 2, . . .Since

A′′′n =A′′nRn

= A′nRn (5)

we get:

Vint(s, φ) =∞∑k=0

[( sR

)2(2k+1)(

4V0

(2k + 1)πsin(2(2k + 1)φ)

)](6)

Vext(s, φ) =∞∑k=0

[(R

s

)2(2k+1)( 4V0

(2k + 1)πsin(2(2k + 1)φ)

)](7)

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Problem 5

The complex function F (z) is defined as:

F (z) = Azp + iΦ0 = A(x+ iy)p + iΦ0 (8)

It’s easy to see that this is a solution to Laplace’s equation:

∇2F (z) = d2

dx2F (z) + d2

dy2F (z)= Ap(p− 1)(x+ iy)(p−2) +Ai2p(p− 1)(x+ iy)(p−2)

= 0

Note that if we express F (z) as F (z) = f(x, y)+ ig(x, y) with f(x, y) = Re[F (z)] and g(x, y) = Im[F (z)],it means that:

∇2F (z) = ∇2f(x, y) + i∇2g(x, y) = 0⇒ ∇2f(x, y) = 0⇒ ∇2g(x, y) = 0

Thus, f(x, y) and g(x, y) are both solutions to Laplace’s equation. We are advised to use g(x, y) (theimaginary part of F (z)). It’s because it would not be possible to satisfy the boundary conditions usingf(x, y).

g(x, y) = Im[A(x+ iy)p + iΦ0] = AIm[(√

x2 + y2ei arctan(y/x))p]

+ Φ0

= A√x2 + y2

psin(p arctan(y/x)) + Φ0

In polar coordinates:

g(s, θ) = Asp sin(pθ) + Φ0; s =√x2 + y2, θ = arctan(y/x)

The boundary conditions are g(s, θ = 0) = Φ0 and g(s, θ = α) = Φ0 where Φ0 is the potential of theconductor.

g(s, θ = 0) = Asp(0) + Φ0 = Φ0

g(s, θ = α) = Asp sin(pα) + Φ0 = Φ0

⇒ Asp sin(pα) = 0 ∀s⇒ sin(pα) = 0⇒ pα = nπ

Thus, p = nπ/α satisfy the boundary conditions. Choosing n = 1 seems to be a natural choice, sincen > 1 would imply that the potential will be Φ0 for all s at some angle(s) θ (0 < θ < α), which would be arather odd behavior for the potential. Note however that a superposition of solutions with different n couldcircumvent this problem.

The problem here is that we don’t have enough information about the boundary conditions. It’s alsodifficult to find physical arguments (like the value of the potential at infinity) since the problem is itself notquite physical (infinite plates). The solution found is actually the approximate solution far from the edgesof very large plates, as explained at the end of example 5.7 in the book.

At any rate, n = 1 is actually the right solution, although it’s not obvious. Thus:

V (x, y) = g(x, y) = A√x2 + y2

π/αsin(π arctan(y/x)/α) + Φ0

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