Poisson’s Equation and Green’s Functions

Static Potentials with Sources—Poisson’s Equation

6.3 Green’s Functions

6.3.2 Poisson’s Equation and Green’s Functions

Poisson’s equation, ∇²V = −ρ/ε₀, can be converted to an integral equation by choosing Ψ in (6–30) such that Ψ satisfies ∇²Ψ = −δ(r−r). Green’s second identity then becomes

τ

−δ(r−r)V(r) + Ψρ(r) ε₀

d³r=

S

V∂Ψ

∂n −Ψ∂V

∂n

dS (6–31) or, carrying out the integration of theδ function,

V(r) =

τ

Ψρ(r) ε₀ d³r+

S

Ψ∂V

∂n −V∂Ψ

∂n

dS (6–32)

To gain some insight into this expression, let us use a familiar solution to∇²Ψ =

−δ(r−r), namely Ψ = 1/(4π|r−r|). Equation (6–32) then becomes V(r) = 1

4πε₀

τ

ρ(r)

|r−r|d³r+ 1 4π _S

∂V /∂n

|r −r|−V ∂

∂n 1

|r−r|

dS (6–33) The first integral is well known to us as the Coulomb integral over the source charge density. The surface integrals reflect the contribution of induced charges or fields on the boundary to the potential within the volume. The numerator of the first term of the surface integral, ∂V /∂n, is just the normal component of the field at the boundary. The potential due to this term is the same as that from surface charge σ =−ε0∂V /∂n. The last term is the potential corresponding to a dipole layer D = ε₀Vˆn on the surface. The surface charge and dipole layer need not be real; we have shown only that the potential inside a volume can always be expressed in these terms. It is interesting that higher multipole fields from outside the boundary do not affect the potential inside the boundary. If the surface integral vanishes as it would whenS→ ∞, we recover the familiar Coulomb result.

The choice of Ψ(r, r) is not unique, as∇²Ψ = −δ(r−r) is satisfied by Ψ(r, r) = 1

4π|r−r|+ F(r, r) (6–34) where F satisfies∇²F(r, r) = 0. This freedom makes it possible to choose Ψ such that the potential,V(r), calculated inside the boundary depends explicitly only on either V or on the value of its normal derivative∂V /∂n(the electric field), on S. We handle the Dirichlet and Neumann problems separately.

If the boundary condition specifiesV onS (Dirichlet problem), we choose Ψ = 0 onS, in which case (6–32) reduces to

V(r) = 1 ε₀

τ

ρ(r)Ψ(r, r)d³r−

S

V(r)∂Ψ(r, r)

∂n dS (6–35)

and, assuming Ψ is known, solving Poisson’s equation is reduced to evaluating two definite integrals.

If instead the boundary condition specifies the normal derivative∂V /∂n onS (the Neumann problem), it would be tempting to choose∂Ψ/∂n= 0, but this would lead to an inconsistency with the definition of Ψ. Because the demonstration of the inconsistency holds the resolution, we produce it below. Since

∇²Ψ(r, r) =∇ · (∇Ψ) = −δ(r−r) (6–36) we find, on integrating∇²Ψ overr, that

τ

∇²Ψd³r=−1 =

S

∇Ψ ·dS =

S

∂Ψ

∂ndS (6–37)

which cannot be satisfied when∂Ψ/∂n= 0 onS. In fact, choosing

∂Ψ

∂n

ronS

=−1

S (6–38)

does satisfy theδfunction normalization, and with this replacement (6–32) becomes V(r) = 1

ε₀

τ

ρ(r)Ψ(r, r)d³r+

S

V(r)1

S +∂V(r)

∂n Ψ(r, r)

dS

= V+

τ

ρ(r)

ε₀ Ψ(r, r)d³r+

S

∂V(r)

∂n Ψ(r, r)dS (6–39) whereVis the average potential on the boundary,

V= 1

S _SV(r)dS (6–40)

Once the Green’s function is known, the solution of Poisson’s equation for all problems with that boundary reduces to several integrations. We now proceed to obtain the Green’s function for the special case of the Dirichlet problem with a spherical (inner or outer) boundary. We will henceforth drop the masquerade and denote the Green’s function Ψ by G(r, r).

6.3.3 Expansion of the Dirichlet Green’s Function in Spherical Harmonics The potential of a point chargeqlocated atr in the vicinity of a grounded sphere solves the equation

∇²V =−qδ(r−r)

ε₀ (6–41)

with the boundary condition V = 0 at the surface of the sphere. Except for the size of the source inhomogeneity, this is precisely the same equation and boundary condition that the Green’s function must satisfy. We may therefore conclude that the expression (6–13) gives us the Green’s function for the problem of a charge

distribution in the neighborhood of (but exterior to) a spherical boundary of radius a, namely

G(r, r) = 1 4π

1

|r−r| − a

r|r−(a²/r²)r|

(6–42) We have previously expressed the first term in spherical polar coordinates (Section 2.1.2), using as an intermediate step the generating function for Legendre polyno- mials (F–35)

1

|r−r| = 1 r>

∞

=0

r_<

r>

P(cosγ) (6–43)

where γ is the angle betweenrandr. A similar expansion may be found for the second term.

a

r|r−(a²/r²)r| = a

r²r²+a⁴−2a²rrcosγ (6–44) If the charge distribution and the field pointrlie outside the sphere, bothrandr are larger thanaand we factor these from the radical in order to obtain a convergent expansion

a

r|r−(a²/r²)r| = a rr

&

1 + a⁴

r²r² −2a² rr cosγ

= a rr

∞

=0

a² rr

P(cosγ) (6–45)

Combining the two terms and rewriting the productrr asr>r<, we have G(r, r) = 1

4π ∞

=0

r_<

r⁺¹_> − a²⁺¹ r⁺¹_< r⁺¹_>

P(cosγ)

= 1 4π

∞

=0

1 r⁺¹_>

r_<−a²⁺¹ r⁺¹_<

P(cosγ) (6–46)

Now using (F–47)) to rewrite P(cosγ) in terms of the polar coordinates ofrandr, we obtain the desired Green’s function for a charge distribution outside the sphere.

G(r, r) = ∞

=0

1 2+ 1

m=−

1 r⁺¹_>

r_<−a²⁺¹ r⁺¹_<

Y^m(θ, ϕ)Y^∗m(θ, ϕ) (6–47) It might be observed that G vanishes on the boundary, as of course it should.

When the source and field points are inside the sphere, the expansion above no longer converges. We distinguish the result for the source and field points interior to the sphere by renaming the radiusb and regain a convergent series by factoring b²from the modified radical in (6–44). We proceed in the same fashion to obtain

b

r|r−(b²/r²)r| = 1 b

&

1 + r²r² b⁴ −2rr

b² cosγ

= 1 b

rr b²

P(cosγ) (6–48)

As before, we combine the two terms in G and rewrite the productrr as r>r< to get

G(r, r) =

,m

1 2+ 1

1 r>

r_<

r>

−1 b

r_<r_>

b²

Y^m(θ, ϕ)Y^∗m (θ, ϕ)

=

,m

r_<

2+ 1 1

r⁺¹_> − r_>

b²⁺¹

Y^m(θ, ϕ)Y^∗m (θ, ϕ) (6–49) Again it is clear that the Green’s function vanishes at the surface of the enclosing sphere wherer =r> =b.

Example 6.5: Find the potential due to a uniformly charged thin circular disk of radiusabearing charge Q placed in the center of a grounded conducting sphere of radiusb > a.

Solution: The charge density on the disk forr < amay be written ρ(r) = Qδ(cosθ)

πa²r (Ex 6.5.1)

as is readily verified by integrating the density over the volume of a sphere containing part of the disk. In general, the potential inside the sphere may be found from

V(r) = 1 ε₀

τ

ρ(r)G(r, r)d³r−

S

V(r)∂G

∂ndS (Ex 6.5.2)

As the sphere is grounded, the surface integral vanishes, reducingV to V = 1

ε₀

τ

ρ(r)

m

Y^m (θ, ϕ)Y^∗m (θ, ϕ) 2+ 1 r_<

1

r⁺¹_> − r_>

b²⁺¹

d³r (Ex 6.5.3) SinceV cannot depend onϕ, only the Y⁰(θ, ϕ) =

(2+ 1)/4πP(cosθ) can enter into the sum. We therefore write

V = 1 4πε₀

τ

P(cosθ)P(cosθ)ρ(r)r_<

1

r_>⁺¹ − r_>

b²⁺¹

d³r

= Q

4πε₀πa²

P(cosθ)

τ

1

rδ(cosθ)P(cosθ)r_<

1

r_>⁺¹ − r_>

b²⁺¹

×r²drd(cosθ)dϕ (Ex 6.5.4) We perform the integration overθ andϕ to get

V = Q 2πε₀a²

P(0)P(cosθ) a

0

r_<

1

r_>⁺¹ − r_>

b²⁺¹

rdr (Ex 6.5.5)

We focus our attention on the integral in (Ex 6.5.5) which we abbreviate (r).

Whenr > a,r isr< throughout the entire range of integration. Hence (r) =

1

r⁺¹ − r b²⁺¹

a 0

r⁺¹dr= a⁺² + 2

1

r⁺¹ − r b²⁺¹

(Ex 6.5.6) Whenr < a, the form of the integrand changes asr passes r. To accommodate this, we split the range of integration into two segments; one whererisr_< and one wherer isr_>. When= 1,

(r) = 1

r⁺¹ − r b²⁺¹

r 0

r⁺¹dr+r _a

r

1

r − r⁺¹ b^2,+1

dr

= 1

(−1)(+ 2)

(2+ 1)r− r a⁻¹

+ 2−(−1)a²⁺¹ b²⁺¹

(Ex 6.5.7) The explicit values of P(0), P₂₊₁(0) = 0, and P₂(0) = (−1)(2−1)!!/2! may be used to eliminate all odd terms from the series. Making this substitution we get

V(r > a) = Q 2πε₀

∞

=0

(−1)(2−1)!!a² 2(2+ 2)!

1

r²⁺¹ − r² b⁴⁺¹

P₂(cosθ) (Ex 6.5.8)

V(r < a) = Q 2πε₀a²

∞

=0

(−1)(2−3)!!

2!(2+ 2) 3

(4+ 1)r

− r² a²⁻¹

(2+ 2)−(2−1)a⁴⁺¹ b⁴⁺¹

4

P₂(cosθ) (Ex 6.5.9)

Example 6.6: Find the potential due to a uniformly charged ring of radiusaand total chargeQ enclosed concentrically within a grounded conducting sphere of radius b (Figure 6.9).

Figure 6.9: A conducting sphere of radiusb has a uniformly charged con- centric ring of radiusaenclosed.

Solution: The charge density on the ring is ρ(r) = Q

2πarδ(r−a)δ(cosθ) (Ex 6.6.1) The sphere is grounded, therefore the surface integral in (6–30) vanishes, reducing the expression for the potential to

V = 1 ε₀

τ

ρ(r)

m

Y^m (θ, ϕ)Y^∗m(θ, ϕ) 2+ 1 r_<

1

r_>⁺¹ − r_>

b²⁺¹

d³r (Ex 6.6.2) and the lack of ϕ dependence again allows us to replace the spherical harmonics with Legendre polynomials to give

V = 1 4πε₀

τ

P(cosθ)P(cosθ)ρ(r)r_<

1

r_>⁺¹ − r_>

b²⁺¹

d³r

= Q

4πε₀2πa

P(cosθ)

τ

δ(r−a)δ(cosθ)P(cosθ)r_<

1

r⁺¹_> − r_>

b²⁺¹

×rdrd(cosθ)dϕ (Ex 6.6.3) The integration overθ andϕ is easily performed to give

V = Q 4πε₀

P(0)P(cosθ)

δ(r−a)r_<

1

r_>⁺¹ − r_>

b²⁺¹

dr (Ex 6.6.4) As in the previous example, the explicit values of P(0) eliminate all odd terms from the series. The δfunction ensures that r =a. Forr < a , r=r<, allowing us to write

V(r < a) = Q 4πε₀

∞

=0

(−1)(2−1)!!

2! r² 1

a²⁺¹ − a² b⁴⁺¹

P₂(cosθ) (Ex 6.6.5)

while forr > a,r =r<, giving V(r > a) = Q

4πε₀ ∞

=0

(−1)(2−1)!!

2! a² 1

r²⁺¹ − r² b⁴⁺¹

P₂(cosθ) (Ex 6.6.6) In the limit asb→ ∞, this should reduce to the expansion for the potential of a charged ring in free space as found in Example 5.8.

This result (Ex 6.6.6) could of course also have been obtained using a ring of image charge at a =b²/a bearing total chargeQ = (−b/a)Q.

6.3.4 Dirichlet Green’s function from the Differential equation

When there are both an inner and an outer spherical bounding surfaces there would be an infinite number of images as the image produced by each surface is reflected

in the other. Under these conditions it proves necessary to solve the defining differ- ential equation for G,∇²G =−δ(r−r). Expressing∇²G in spherical polars using (42) we have

1 r

∂²(rG)

∂r² + 1 r²sinθ

∂

∂θ

sinθ∂G

∂θ

+ 1

r²sin²θ

∂²G

∂ϕ² =−δ(r−r) (6–50) Considering G(r, r) as a function of r we may generally expand it in spherical polars as

G(r, r) =

,m

A,m(θ, ϕ)R(r, r)Y^m (θ, ϕ) (6–51) Substituting this expansion into (6–50) with the help of (F–39) we obtain

,m

1 r

d²(rR)

dr² −(+ 1)R

r²

A,m(θ, ϕ)Y^m(θ, ϕ) =−δ(r−r) (6–52)

We can also expand the right hand side of (6–52) first in terms of one dimensional δfunctions

δ(r−r) = 1

r²δ(r−r)δ(cosθ−cosθ)δ(ϕ−ϕ) (6–53) and then in terms of spherical harmonics with the aid of the completeness relation (F-46)

δ(r−r) = 1

r²δ(r−r)

,m

Y^m (θ, ϕ)Y^∗m(θ, ϕ) (6–54) We immediately identify the coefficientA,m(θ, ϕ) with Y^∗m (θ, ϕ) allowing us to remove these two terms from the equation. The remaining equation is

,m

1 r

d²(rR)

dr² −(+ 1)R r²

Y^m (θ, ϕ) =−1

r²δ(r−r)

,m

Y^m (θ, ϕ) (6–55) Finally we appeal to the linear independence of the Y^m to equate the coefficients on the left and the right, resulting in

1 r

d²(rR)

dr² −(+ 1)R

r² =−1

r²δ(r−r) (6–56) So long asr=r, the right hand side of (6–56) vanishes it may be solved forrR in terms ofr

rR=Ar⁺¹+B

r (6–57)

however, the δ function inhomogeneity atr =r means the expansion coefficients will differ forr < r andr > r so that

R(r < r) =Ar+ B

r⁺¹ and R(r > r) =Cr+ D

r⁺¹ (6–58)

The boundary condition on G, and hence on R, is that it vanish on the boundary.

At the inner sphere of radiusa,r < r so that R(a) =Aa+ B

a⁺¹ ⇒ B=−Aa²⁺¹ (6–59)

and

R(b) =Cb+ D

b⁺¹ ⇒ C=− D

b²⁺¹ (6–60)

so that, explicitly recognizing that R is a function of both r and r, we rewrite (6–58) as

R(r < r) =A(r)

r−a²⁺¹ r⁺¹

(6–61) and

R(r > r) =D(r) 1

r⁺¹ − r b²⁺¹

(6–62) recalling that the Green’s function is invariant under interchange of r and r, we can also express these results (6–61) and (6–62) in terms of r and r interchanged

R(r < r) =D(r) 1

r⁺¹ − r b²⁺¹

(6–63) and

R(r > r) =A(r)

r−a²⁺¹ r⁺¹

(6–64) (6–61) and (6–63) taken together imply that

R(r < r) =B 1

r⁺¹ − r b²⁺¹

r−a²⁺¹ r⁺¹

(6–65) and (6–62) and (6–64) similarly lead to

R(r > r) =C 1

r⁺¹ − r b²⁺¹

r−a²⁺¹ r⁺¹

(6–66) Finally, the continuity of G (required to make V continuous at r = r implies B=C and we can write (6–65) and (6–66) in the more economical form

R(r, r) =C 1

r⁺¹_> − r_>

b²⁺¹

r_<− a²⁺¹ r<+1

(6–67) The constant C is yet to be determined from the size of the source term, in this case theδfunction inhomogeneity. To evaluate the effect of the δfunction, we must integrate (6–56) over a vanishingly small interval aboutr.

_r+

r−

d²(rR)

dr² −(+ 1)R r

dr=−

_r+

r−

δ(r−r)

r dr (6–68)

or

d(rR) dr

^r

+

r−−

5(+ 1)R

r 6

2=−1

r (6–69)

The term

lim→0

5(+ 1)R

r 6

2= 0 (6–70)

leaving

d(rR) dr

r₊

−d(rR) dr

r₋

=−1

r (6–71)

Forr=r₊,r > r and we use (6–66) to compute the required derivative d(rR)

dr

r₊

=C

r−a²⁺¹ r⁺¹

d dr

1

r − r⁺¹ b²⁺¹

r=r

=C

r−a²⁺¹ r⁺¹

−

r⁺¹ −(+ 1)r b²⁺¹

(6–72) In the same fashion we use (6-64) (with C substituted for B) to compute the derivative ofrR atr=r₋

d(rR) dr

r₋

=C 1

r⁺¹ − r b²⁺¹

d dr

r⁺¹−a²⁺¹ r

r=r

=C 1

r⁺¹ − r b²⁺¹

(+ 1)r+a²⁺¹ r⁺¹

(6–73) Inserting (6–72) and (6–73) into (6–71) we obtain after some algebra

C = 1

(2+ 1)

1− a

b

₂₊₁ (6–74)

Collecting terms we use (6–74) in (6–67) to construct the Green’s function in (6–51) G(r, r) =

∞

=0

m=−

Y^m (θ, ϕ)Y^∗m(θ, ϕ) (2+ 1)

1−

a b

₂₊₁

r_<−a²⁺¹ r_<⁺¹

1 r_>⁺¹− r_>

b²⁺¹

(6–75)

When there is either no inner boundary or no outer bounding sphere, it suffices to let a go to zero (no inner boundary) or let b go to ∞ (no outer boundary) to recover our earlier solutions.

Exercises and Problems

Figure 6.10:The bent, infinite, grounded con- ducting sheet has a chargeq midway between the plates at distancebfrom the vertex.

Figure 6.11: A line chargeλ =Q/2b spans the enclosing sphere along thez axis.

6-1 Find the force between an isolated conducting sphere of radius R and a point chargeq in its vicinity.

6-2 It is tempting to write the poten- tial of a charge q situated a distance z above a grounded conducting plane as V =q/(4πε₀2z). Show that one obtains twice the correct electric field when com- puting−∇V. Explain the reason for the too-large result.

6-3 Show that when the potential at a charge located atrnear a grounded con- ducting sphere due to its image charge is expressed in terms of the charge position

r, −∇V gives twice the correct electric field atr.

6-4 Use image charges to obtain the force and the torque on a electric dipole in the vicinity of a grounded conducting plane.

6-5A point charge is placed at distance a from the left plate between two par- allel conducting grounded plates spaced distance D. Find a series expression for the force on the charge.

6-6 A point electric dipole is placed at the center of a conducting sphere. Find

the resulting electric field both inside and outside the sphere.

6-7Use image charges to find the poten- tial along the z axis due to a uniformly charged thin ring of radiusa concentric with a larger grounded sphere of radius b (illustrated in Figure 6.9). Generalize this result forr ∈(a,b).

6-8 A small (point) electric dipole is placed in the vicinity of a neutral con- ducting sphere. Find the energy of the dipole due to the induced dipole field of the sphere. This potential has the gen- eral form of the dipole-induced dipole en- countered in molecular physics.

6-9 Find the force and torque on the dipole in problem 6-8.

6-10 Find the image of a uniform (straight) line charge in the vicinity of a grounded conducting sphere.

6-11 Show that the problem of an un- charged conducting sphere placed in an initially uniform field can be solved by images. (Hint: Place charges Q =

±2πε0L²E₀ at sufficiently large ±L to produce a uniform electric fieldE₀in the region the sphere will be placed.)

6-12A charge is placed distancebalong the bisector of the vertex of a bent, infi- nite plane conductor as shown in Figure 6.10. Find the force on the charge.

6-13 Generalize problem 6-9 to vertex with a small angle that divides 360^◦ evenly.

6-14 Determine the images of a point charge q placed above the center of a hemispherical bump (radiusa) in an oth- erwise flat plate extending to infinity.

Assume all surfaces are grounded con- ductors.

6-15 A line charge λ is placed inside a grounded, conducting cylinder paral- lel to the axis of the cylinder. Find the potential inside the cylinder.

6-16 Calculate the capacitance of two nested cylinders whose radii are a < b and whose parallel axes are separated by D < b−a.

6-17 A long wire of radius a is placed with its axis parallel to a large conduct- ing (assume infinite) sheet at distanced from the sheet. Find the capacitance of this arrangement.

6-18 Find the capacitance of two long wires of differing radii a and b carry- ing equal charges of opposite sign dis- tributed over their surfaces.

6-19 A hollow conducting sphere of ra- dius b encloses a uniform line chargeλ

=Q/2b along thez axis (Figure 6.11).

Use Green’s functions to find the poten- tial interior to the sphere. Hint: ρ(r)

=Q/4bπr²[δ(cosθ−1) +δ(cosθ+ 1)]

and care should be taken integrating b

0

r_<

1 r

+1

> − r_>

b²⁺¹

dr Break it up into intervals 0 ≤ r ≤ r.

andr≤r≤bto obtain _b

0

· · ·

dr = 2+ 1 (+ 1)

1−

r b

This result is indeterminate for = 0;

direct integration works best in the = 0 case.

6-20 A point charge Q is placed mid- way between two concentric conducting grounded spheres of radiusa andb. For convenience assume the charge to be at θ= 0. Find the potential at other points between the two spheres.

6-21 Obtain the Green’s function for Poisson’s equation in cylindrical polar coordinates using a Bessel function ex- pansion for the Diracδfunction.

Chapter 7