Classical Electromagnetism

Chapter 6: Laplace's equation in Cylindrical Coordinates

By Mark Lawrence

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Siméon Denis Poisson, 1781 to 1840

Pierre-Simon Laplace, 1749 to 1827

Two Dimensional Boundary Value Problems in Cylindrical Coordinates

In this chapter we'll be working in cylindrical coordinates. We'll look for solutions to Laplace's equation. Later we'll apply boundary conditions to find specific solutions. Traditionally, ρ is used for the radius variable in cylindrical coordinates, but in electrodynamics we use ρ for the charge density, so we'll use s for the radius. φ will be the angular dimension, and z the third dimension.

s lies in the x-y plane, an angle φ up from the x axis. The relevant equations are:

x = s cos φ
y = s sin φ
z = z
s = √ (x2 + y2)
φ = arctan(y/x)
z = z

Laplace's equation for the potential in the absence of sources is:

Ñ2Φ = 0

Ñ2 in cylindrical coordinates is:

1 1 2 2
Ñ2  = 


s ∂s ∂s s2 ∂φ2 ∂z2

First we'll solve a pseudo two dimensional problem. We'll presume the given geometry extends infinitely in the z direction, so that the solution does not depend on z. We'll also presume that our solution is separable, that is Φ( s, φ, z ) = S(s)Q(φ). We can check the separability assumption later. Laplace's equation is a second order differential equation, so if we find two unique solutions we've done it. Generally speaking, if the boundary conditions are separable, there's a good chance the solution is separable. If the boundary conditions are not separable, most likely we're hosed. Now,

Ñ2Φ(s,φ,z) = Ñ2S(s)Q(φ).

Q is independent of s, so it can be pulled through the s partial derivatives. Similarly S is independent of φ, so it can be pulled through the φ partial derivatives. Φ is independent of z, so the z partial derivatives are 0.

Q S 2
Ñ2SQ  = 

 S + 

 Q   = 0 
s ∂s ∂s s2 ∂φ2

Now multiply the entire equation by s2 / SQ:

s 1 2
Ñ2SQ  = 

 S + 

 Q   = 0 
S ∂s ∂s Q ∂φ2

Since there's an expression that depends only on s and another that depends only on φ, and since the sum is zero for all s and φ, it must be that the individual expressions are constants. We'll start by setting them to 0.

s ∂S

 S   = 0 = 
 S   Þ
S ∂s ∂s ∂s ∂s ∂s s

1 2 2Q

 Q   = 0 = 
Q ∂φ2 ∂φ2

We can solve these immediately. S = a0 + b0 ln s.  Q = A0 + B0 φ.

Another possibility is that the expression for S is equal to some constant, and the expression for Q is equal to the negative of that same constant. Since these are second order equations, we'll anticipate and set the constant to η2. Now,


 S   = η2 Þ
 S   = η2
S ∂s ∂s ∂s ∂s


 Q   = -η2

We can solve the second equation immediately. Q = Aη cos ηφ + Bη sin ηφ.

For the equation in S, we'll try a polynomial sα. Immediately we see that the solutions to this equation are S = aηsη + b ηs.

We've arbitrarily put the plus sign on the S expression, and the minus sign on the Q expression. Let's see what happens if we reverse this.


 S   = -η2 Þ
 S   = -η2
S ∂s ∂s ∂s ∂s


 Q   = η2

Again, we can solve the Q equation immediately - this is just eηφ and e -ηφ. We'll choose to represent the solutions as Q = Aη cosh ηφ + Bη sinh ηφ.

The S equation is a bit trickier. Again, we'll try a polynomial sα. Immediately we see that the solutions to this equation are S = aηs + bηs-iη. This is not very pleasant, so we'll work on these solutions a bit. s = e iη ln s. This suggests, perhaps "obviously" in retrospect, that we should have tried cos( η ln s ). So, our solutions to the equation will be

S = aη cos( η ln s ) + bη sin( η ln s ).

These solutions oscillate - unlike the solutions for S above, which are strictly non-negative, these solutions are both positive and negative. Both the Q and the S solutions are either symmetric or anti-symmetric, so we can assign these solutions to negative values of η and any minus sign will be absorbed into the coefficients. Similarly, the other solutions we got were either symmetric or anti-symmetric, so we can assign those solutions to the positive values of η.

Now, our complete solution to the 2d + z Laplace equation is:

Φ  =   Σ (aη cos( η ln s ) + bη sin( η ln s ))  (Aη cosh ηφ + Bη sinh ηφ )    +  Σ ( aηsη + bηs( Aη cos ηφ + Bη sin ηφ )   +   (a0 + b0 ln s) (A0 + B0φ) 
-∞ 1

A Worked Problem

Suppose we have conductor filling all of space except for a wedge cut out of it. OK, this is a pretty artificial problem. Unfortunately, artificial problems are pretty much what we can actually solve. Importantly, if you had a piece of metal with a wedge cut out of it, this solution would be quite accurate as long as you were reasonable far from the edges.

Figure 5.1: A boundary value problem

Suppose the metal is at a potential V. Now, we have three boundary conditions:

  1. Φ(s,φ,z) = V    for s = 0.  This is the inside corner.
  2. Φ(s,φ,z) = V    for φ = 0. This is the lower face.
  3. Φ(s,φ,z) = V    for φ = β. This is the upper face.

Condition 1 tells us:

   aη = bη = 0 for η < 0, as the sine and cosine functions are not well behaved as their arguments go to minus infinity.

   bη = 0 for η > 0, as otherwise we have a rather embarrassing divide by zero.

   b0 = 0 or we have another embarrassing infinity

   B0 = because the constant term must be independent of φ.

   a0 A0 = V

Now our remaining solution is:

Φ  =  Σ ( aηsη( Aη cos ηφ + Bη sin ηφ )   +   V 

Condition 2 tells us:

   Aη = 0, otherwise the voltage on the lower face will be a function of s. We could have set aη to 0 instead, but this takes all the fun out of the solution.

Condition 3 tells us:

   ηβ = nπ, n integer.

Now, our complete solution is:

Φ  = V +  Σ ansnπ/β  sin( nπφ/β )

Now let's have a closer look at this solution. If s is very small, then we can ignore all the terms with n > 1. The electric field is E = -ÑΦ, which is

E = -a1 s(π/β-1) ( sin πφ/β s + cos πφ/β φ )

As φ goes to 0, E goes to -a1 s(<π/β-1) = 4πσ.

Immediately we see that if β < π, E -> 0; if β > pi;, E -> ∞. This is the well known result that the electric field and charge density go to zero at an inside corner, and they go to infinity at a point. This is why lightning rods work: the lightning is attracted to the very high field at a point on an object. We should not be too concerned by the infinity - after all, there's no such thing as an infinitely sharp point.