Static Potentials in Vacuum – Laplace’s Equation
5.1 Laplace’s Equation
5.2.2 Plane Polar Coordinates
In plane polar coordinates,∇2V assumes the form 1
r
∂
∂r
r∂V
∂r
+ 1 r2
∂2V
∂ϕ2 = 0 (5–16)
We assume a separable solution and try V of the form V = R(r)Φ(ϕ) with R independent ofϕ and Φ independent of r. Substituting this form into (5–16), we get
1 r
∂
∂r
r∂R
∂r
R +
1 r2
∂2Φ
∂ϕ2
Φ = 0 (5–17)
or
r ∂
∂r
r∂R
∂r
R =−
∂2Φ
∂ϕ2
Φ = constant =m2 (5–18)
where we have chosen a positive separation constant, m2, in anticipation of a re- quirement by the boundary condition on Φ. The equation
∂2Φ
∂ϕ2 =−m2Φ (5–19)
is easily solved to give
Φ =Acosmϕ+Bsinmϕ (5–20)
The physical requirement that Φ be periodic with period 2πjustifies our choice of a positive separation constant and further restrictsm to integer values.
The remaining equation r ∂
∂r
r∂R
∂r
=m2R (5–21)
is easily solved. Clearly the equation suggests a monomial solution as differentiating r lowers the exponent of r by one and multiplication byr promptly restores the original exponent. We try R =kr, which, when substituted into (5–21), gives
2kr=2R =m2R ⇒ =±m (5–22)
We conclude that R takes the form R =Crm+Dr−m. The linear combination of the two linearly independent solutions is the general solution whenm = 0. When m = 0, however, only one solution is obtained. To obtain the second, we integrate (5–21) directly withm = 0.
∂
∂r
r∂R
∂r
= 0 ⇒ r∂R
∂r =C ⇒ ∂R
∂r =C
r (5–23)
Thus, whenm = 0, R =C0lnr+D0 is the solution.
Combining the two forms of the solution, we find the most general solution to
∇2V = 0 in plane polar coordinates to be V(r, ϕ) =C0lnr+D0+
∞ m=1
Cmrm+Dm
rm
(Amcosmϕ+Bmsinmϕ) (5–24) It might be noted that there are more arbitrary constants in (5–24) than can be uniquely determined. The two trigonometric terms might, for instance, be added to give a single phase-shifted cosine at which pointAmandBmbecome superfluous since the amplitude is already determined byC andD. In order to avoid a prolif- eration of constants, we will freely rename constants and product of constants as convenient.
Example 5.2: Find the potential when a neutral, long conducting circular cylinder is placed in an initially uniform electric field with its axis perpendicular to the field (Figure 5.2).
Solution: We choose thex axis along the initially uniform field so that at sufficiently large r, E = E0ˆı. To obtain this field at sufficiently large r, we must have V
→ −E0x=−E0rcosϕ.
We now apply the boundary conditions to the general solution (5–24). Reflection symmetry about the x axis requires that V(ϕ) = V(−ϕ), implying for m = 0, Bm= 0. The termC0 = 0 implies a net charge on the cylinder. The termsCmrm
Figure 5.2: The coordinate system is chosen so that the axis of the cylinder lies along thez axis whereas the electric field at large distances is purelyx directed.
all grow too rapidly as r → ∞, requiring that Cm = 0 for m = 1. The general solution then reduces to
V(r, ϕ) =D0−E0rcosϕ+ ∞ m=1
Dm
rm cosmϕ (Ex 5.2.1) where the products of arbitrary constants, Am andDm, have been replaced by single constants,Dm.
On the surface of a conductor, the potential must be a constant, sayV0; therefore V(a, ϕ) =V0=D0−E0acosϕ+
∞ m=1
Dm
am cosmϕ (Ex 5.2.2)
0 = (D0−V0) + D1
a −E0a
cosϕ+ ∞ m=2
Dm
am cosmϕ (Ex 5.2.3) As the{cosmϕ}are linearly independent functions, each coefficient in the expansion must vanish (i.e.,D0=V0,D1=E0a2,Dm= 0 form≥2). The final result is then
V(r, ϕ) =V0−E0rcosϕ+E0a2
r cosϕ (Ex 5.2.4)
It is straightforward to verify that (Ex 5.2.4) satisfies the boundary conditionsV = V0 atr=aandV → −E0rcosϕasr→ ∞.
More generally, the boundary conditions may be applied to two nested cylinders sharing a common axis. We sketch the procedure in the following example.
Example 5.3: Two coaxial, nonconducting cylinders have surface charge densities σa(ϕ) and σb(ϕ) on the inner and outer cylinders, giving rise to potentialsVa(ϕ) andVb(ϕ) on the two surfaces (Figure 5.3). Find the potential (a) for r < a, (b) forr > b, and (c) forr∈(a, b).
Solution: We assume the general solution (5–24) and tailor it to the appropriate boundary conditions in each of the three regions.
Figure 5.3:The two concentric cylinders of radiusaandbare x at potential Va(ϕ) andVb(ϕ) respectively.
(a) Forr≤a: terms in lnror 1/rmdiverge atr= 0 and must be eliminated from the sum by setting their coefficients equal to 0. Hence
V =D0+
rm(Amcosmϕ+Bmsinmϕ) (Ex 5.3.1) Atr=a, this specializes to
Va(ϕ) =A0+
(amDmcosmϕ+amBmsinmϕ) (Ex 5.3.2) We recognize the righthand side as the Fourier expansion of Va(ϕ). The ex- pansion coefficients are
A0= 1 2π
2π
0
Va(ϕ)dϕ (Ex 5.3.3)
amAm= 1 π
2π
0
Va(ϕ) cosmϕdϕ (Ex 5.3.4)
amBm= 1 π
2π
0
Va(ϕ) sinmϕdϕ (Ex 5.3.5) (b) Forr≥b: This time the terms inrmdiverge at∞(lnralso diverges, but such a term would be required by a nonzero net charge on the cylinders). Eliminating all the divergent terms except lnr, we write the expansion of the potential as
V(r≥b) =C0lnr+D0+ ∞ m=1
1
rm(Amcosmϕ+Bmsinmϕ) (Ex 5.3.6) Matching the boundary condition atb, we have
Vb(ϕ) = (C0lnb+D0) + ∞ m=1
1
bm(Amcosmϕ+Bmsinmϕ) (Ex 5.3.7)
and we again identify the coefficients of the trigonometric terms with the coef- ficients of the Fourier series ofVb(ϕ) to obtain
C0lnb+D0= 1 2π
2π
0
Vb(ϕ)dϕ (Ex 5.3.8)
Am= bm π
2π
0
Vb(ϕ) cosmϕdϕ (Ex 5.3.9)
Bm= bm π
2π
0
Vb(ϕ) sinmϕdϕ (Ex 5.3.10) It is not possible to tell from Vb alone whether there is a net charge on the cylinders, hence we cannot distinguish between C0 and B0 without further information.
(c) For a ≤ r ≤ b: There is now no reason to eliminate any of the terms from (5–24). Fortunately we have twice as many boundary conditions; one ataand one at b. Equating the potential (5–24) to the values on the boundaries, we have
Va =C0lna+D0+ ∞ m=1
Cmam+Dm
am
(Amcosmϕ+Bmsinmϕ) (Ex 5.3.11) and
Vb=C0lnb+D0+ ∞ m=1
Cmbm+Dm bm
(Amcosmϕ+Bmsinmϕ) (Ex 5.3.12)
leading form= 0 to C0lna+D0= 1
2π 2π
0
Va(ϕ)dϕ C0lnb+D0= 1 2π
2π
0
Vb(ϕ)dϕ (Ex 5.3.13) and form= 0 we get
Cmam+Dm
am Am
Bm
= 1 π
2π
0
Va(ϕ)
cosmϕ sinmϕ
dϕ (Ex 5.3.14)
Cmbm+Dm
bm Am
Bm
= 1 π
2π
0
Vb(ϕ)
cosmϕ sinmϕ
dϕ (Ex 5.3.15) These equations may now be solved (two at a time) to obtainAm, Bm, Cm, andDm for a complete solution. For example, equations (Ex 5.3.13) yield
C0lnb a = 1
2π 2π
0
[Vb(ϕ)−Va(ϕ)]dϕ (Ex 5.3.16)
Figure 5.4: A current flowing around a resistive cut cylinder establishes a potential that varies linearly withϕaround the cylinder.
and
D0lna b = 1
2π 2π
0
[Vb(ϕ) lna−Va(ϕ) lnb]dϕ (Ex 5.3.17)
Example 5.4: A resistive cylinder of radius a with a narrow longitudinal gap at ϕ=πcarries a current in the azimuthal direction, giving rise to a linearly varying potentialV(a, ϕ) =V0ϕ/2πfor−π < ϕ < π(Figure 5.4). Find the potential inside the cylinder.
Solution: Interior to the cylinder, the solution is of the form (we have redefined the constantsA andB slightly in order to writer in normalized form):
V(r < a) =
r a
(Acosϕ+βcosϕ) (Ex 5.4.1)
giving, atr=a,
V(a, ϕ) =
Acosϕ+Bsinϕ= V0
2π ϕ (Ex 5.4.2)
We obtain the coefficientsAandBin the usual fashion; multiplying both sides of the equation by cosmϕand integrating from−πto +π, we get
π
−π(Acosϕ+Bsinϕ) cosmϕ dϕ= V0 2π
π
−πϕcosmϕ dϕ (Ex 5.4.3) orAm= 0, which could have been predicted from the symmetry ofV. Multiplying by sinϕand integrating, we have
π
−πBsinϕsinmϕ dϕ= V0 2π
π
−πϕsinmϕ dϕ (Ex 5.4.4) or
πBm= V0 2π
π
−πϕsinmϕ dϕ=−V0cosmπ
m = (−1)m+1V0
m (Ex 5.4.5)
The potential interior to the cylinder is then V(r, ϕ) =V0
∞
=1
(−1)+1 π
r a
sinϕ (Ex 5.4.6)
Frequently, even when there are free charges or currents on some surface, it is feasible to solve for the potentials inside and outside the surface, and to convert the presence of charges or currents into a boundary condition that relates the difference between the inside and outside solutions to the source term.
Example 5.5: A hollow cylindrical shell bearing no net charge of radiusahas a surface chargeσ =σ0cosϕ distributed on it. Find the potential both inside and outside the shell.
Solution: The interior and exterior potentials are given by V(r < a, ϕ) = r
a
(Acosϕ+Bsinϕ) (Ex 5.5.1) and
V(r > a, ϕ) = a r
(Ccosϕ+Dsinϕ) (Ex 5.5.2) The boundary conditions can be obtained from Maxwell’s electric field equations,
∇ × E = 0 and ∇ · E =ρ/ε0, as follows.
Figure 5.5: A thin box enclosing a small segment of the cylindrical surface.
(a) We integrate ∇ · E over the thin cylindrical shell segment of Figure 5.5 that encloses an areaAof the cylinder:
∇ · Ed 3r= ρ
ε0d3r= σA¯
ε0 (Ex 5.5.3)
where ¯σ is the average charge density on the surface A. With the aid of the divergence theorem, we recast this as
E·dS=σA¯
ε0 (Ex 5.5.4)
Figure 5.6: The loop encloses a section of the cylinder on a plane perpen- dicular to the cylinder.
We decompose the integral in terms of integrals along the top, the bottom, and the thin side
E ·dS=
bottom
−ErdS+
top
ErdS+
sides
E·dS (Ex 5.5.5)
As we diminish the thickness of the box, the integral over the sides vanishes, leaving only ¯Er(a+)A−E¯r(a−)A= ¯σA/ε0. Now, letting the area tend to zero, E¯r→Er and ¯σ→σ. ThusEr(a+) =Er(a−) +σ/ε0.
(b) The second of Maxwell’s E equations, ∇ × E = 0, integrated over the area of the thin loop with long sides straddling the cylinder as shown in Figure 5.6, gives, using Stokes’ theorem (18)
Σ
(∇ × E) ·dS= E ·d
=
L1
−Eϕd+
L2
Eϕd+
short sides
E ·d (Ex 5.5.6) Shrinking the short sides to vanishingly small gives us
−E¯ϕ(a−)L+ ¯Eϕ(a+)L= 0 (Ex 5.5.7) which, whenL→0, becomesEϕ(a−) =Eϕ(a+).
To summarize, the boundary conditions areEϕis continuous ata, or
− 1 a
∂V(r, ϕ)
∂ϕ
a+
=− 1 a
∂V(r, ϕ)
∂ϕ
a−
(Ex 5.5.8) and Er is discontinuous, increasing by σ/ε0 as it crosses the surface out of the enclosed cylinder, or
− ∂V(r, ϕ)
∂r
a+
=− ∂V(r, ϕ)
∂r
a−
+ σ
ε0 (Ex 5.5.9)
Applying the first of these conditions (Ex 5.5.8) to the series forV, we obtain (Asinϕ−Bcosϕ) =
(Csinϕ−Dcosϕ) (Ex 5.5.10) Grouping terms, we have
(A−C) sinϕ+
(B−D) cosϕ= 0 (Ex 5.5.11) The linear independence of the trigonometric functions then requires that the co- efficients of each term in the sum above equal zero; therefore A =C and B = D.
The second boundary condition (Ex 5.5.9) applied to the series gives
a(Ccosϕ+Dsinϕ) =−
a(Acosϕ+Bsinϕ) +σ0 ε0cosϕ
(Ex 5.5.12) We again equate the coefficient of each linearly independent function to zero, giving for the cosϕterm
C1+A1−σ0a ε0
cosϕ= 0 (Ex 5.5.13)
while the remaining terms give B +D = 0 and A+C = 0. Together with the equations from (Ex 5.5.11) these latter imply that except for A1 and C1, the coefficients vanish. SubstitutingA1 =C1 in the equation above yieldsA1=C1 = σ0a/2ε0. Finally, then,
V(r < a) = σ0r
2ε0 cosϕ (Ex 5.5.14)
and
V(r > a) =σ0a2
2ε0rcosϕ (Ex 5.5.15)
It might be noted that Exercise 5.5 is essentially a Neumann problem in that the derivatives of the potential are specified, albeit indirectly, on the boundary. It should be obvious that an arbitrary constant may be added toV without changing the fields.