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Electrostatic Field Between Two Intersecting Grounded Planes

Consider two infinite grounded metal planes with an angle α between them, as shown in the following figure. A point charge Q is located at (r0,ϕ0,0) in cylindrical coordinates, where 0<ϕ0<α. Determine the electric potential in angle domain.

problem problem Figure 1: Diagram of the problem

As a classical problem in electrodynamics, I believe most people have encountered it. This problem is extraordinary easy when α=π/n,nZ+. But what about απ/n? It’s rare to see textbooks discussing this problem. Next, I’ll solve it with method of separation of variables and expand upon it.

In fact, this post was completed when I was a sophomore in college (2021). I am just reposting it this time, please see here. In addition, I also answered one of the key integrals on Zhihu, please see here.

1. Problem with Dirichlet boundary condition

1.1. Special case: α=π/n

If α=π/n, the system exhibits a high degree of symmetry, we can solve it using the method of images. Please see the following examples:

problem problem Figure 2: The positions of the charge Q and image charges for n=2,3,4 with Dirichlet boundary condition.

Obviously, the number of image charges is 2n1. The electric potential in angle domain can be obtained immediately,

(1)Φ(r,ϕ,z)=Q4πϵ0i=12n(1)n+1|rri|,

where ri=r0er+ϕieϕ and

(2)ϕi=[i2]2πn+(1)i+1ϕ0,(i=1,2,,2n).

Or we can rewrite (1) as

(3)Φ(r,ϕ,z)=Q4πϵ0i=12n(1)i+1r2+r02+z22rr0cos(ϕϕi).

1.2. General case: α can be any value

When α is arbitrary, the system no longer possesses a definite high degree symmetry, rendering the method of images ineffective. In such cases, solving the Poisson’s equation directly becomes the most straightforward approach,

(4)2Φ=ρϵ0=Qϵ0δ(rr0).

In cylindrical coordinates, this equation with Dirichlet boundary conditions (because the metal planes are grounded) can be expressed as:

(5){1rr(rΦr)+1r22Φϕ2+2Φz2=Qϵ0r0 δ(rr0)δ(ϕϕ0)δ(z),Φ|ϕ=0=Φ|ϕ=α=0.

Solving this problem is not easy, we can use the method of separation of variables. But we still need to approach it step by step.

1.2.1. Find the eigenfunctions

Suppose the solution Φ can be written as Φ=R(r)ψ(ϕ)Z(z), if rr0, the Poisson’s equation becomes Laplace’s equation:

ψZ d2Rdr2+ZψrdRdr+RZr2d2ψdϕ2+Rψ d2Zdz2=0.

Multiplying each term by r2/RψZ and rearranging appropriately, the equation becomes:

r2Rd2Rdr2+rRdRdr+r2 1Zd2Zdz2=1ψd2ψdϕ2.

The right hand side is a function of ϕ, while the left hand side is a function of r and z. So both sides must equal the same constant, denoted as m2. Therefore,

1ψd2ψdϕ2=m2ψ(ϕ)=Acosmϕ+Bsinmϕ.

With boundary condition ψ|ϕ=0=ψ|ϕ=α=0, we have

(6)ψ(ϕ)=sin(kπαϕ),m=kπα(k=1,2,).

Similarly, we have

1Rd2Rdr2+1rRdRdrm2r2=1Zd2Zdz2.

The right hand side is a function of z, while the left hand side is a function of r. Obviously, both sides must equal the same constant too. Note the boundary condition Φ|z±=0, so the constant must be μ2. Therefore,

1Zd2Zdz2=μ2Z=Asin(μz)+Bcos(μz).

Note that Φ(z=0)0 and cos(μz) is an even function, so we can assume μ>0, then

(7)Z(z)=cos(μz),(μ is continuous, μ>0)

Finally, R(r) satisfies the following equation:

d2Rdr2+1r dRdr(μ2+m2r2)R=0.

Let x=μr, it becomes

d2Rdx2+1x dRdx(1+m2x2)R=0.

This is the standard m-order modified Bessel equation, and its solution is

R(r)=AIm(x)+BKm(x)=AIm(μr)+BKm(μr),

where Im(x) and Km(x) are the modified Bessel functions of the first and second kind respectively. They have the following properties:

Im(x)x01Γ(m+1)(x2)m,Im(x)x12πxex[1+O(1x)],Km(x)m0x0Γ(m)2(2x)m,Km(x)xπ2xex[1+O(1x)].

Since r(0,), to make the electric potential not divergent and continuous except for the original point, R(r) needs to be divided into two parts by r:

(8)R(r)={Im(μr),r(0,r0)Km(μr),r(r0,)

Combine the results (6),(7),(8), we find the eigenfunctions of the original problem,

Φkμ(r,ϕ,z)={Ikπ/α(μr)cos(μz)sin(kπαϕ),r(0,r0)Kkπ/α(μr)cos(μz)sin(kπαϕ),r(r0,)},(kZ+;μ is continuous , μ>0)

So the general solution can be expressed as linear superposition of these eigenfunctions,

(9)Φ(r,ϕ,z)=k=10dμ akμΦkμ(r,ϕ,z)={k=10dμAkμIkπ/α(μr)cos(μz)sin(kπαϕ),r<r0k=10dμBkμKkπ/α(μr)cos(μz)sin(kπαϕ),r>r0

1.2.2. Calculate the coefficients

Even though we get (9), it’s not the final result we want. Because it still has undetermined coefficients Akμ,Bkμ. Note that we have two conditions we haven’t used yet,

  1. The electric potential should be continuous at r=r0;
  2. The electric potential satisfies the Poisson’s equation, and the right side of the Poisson equation is the Dirac function.

First let us consider continuity, we can get

k=10dμAkμIkπ/α(μr0)cos(μz)sin(kπαϕ)=k=10dμBkμKkπ/α(μr0)cos(μz)sin(kπαϕ).

Integrate z on both sides and use

(10)δ(μ)=12πdzeiμz=12πdzcos(μz).

It becomes

k=1AkμIkπ/α(μr0)sin(kπαϕ)=k=1BkμKkπ/α(μr0)sin(kπαϕ).

Due to the orthogonality of sin(kπϕ/α) in ϕ(0,α), the corresponding coefficients must be equal, so we obtain the first relation:

(11)AkμIkπ/α(μr0)=BkμKkπ/α(μr0).

Next, using the second condition and substituting the general solution into Poisson’s equation, we can get

k=10dμakμ[sin(kπϕα)cos(μz)d2Rdr2+sin(kπϕ/α)cos(μz)rdRdr+Rcos(μz)r2d2sin(kπϕ)dϕ2+Rsin(kπϕα)d2cos(μz)dz2]=Qϵ0r0δ(rr0)δ(ϕϕ0)δ(z).

Note that dR/dr is discontinuous at r=r0, we can integrate the neighborhood of r0 to cancel the δ(rr0),

limδ0k=10dμr0δr0+δdrakμ[sin(kπϕα)cos(μz)d2Rdr2]=Qϵ0r0δ(ϕϕ0)δ(z).

Similarly, integrate z on both sides,

limδ0k=10dμr0δr0+δdrakμ{sin(kπϕα)[dzcos(μz)]d2Rdr2}=limδ0k=10dμr0δr0+δdr2πakμδ(μ)[sin(kπϕα)d2Rdr2]=limδ0k=1r0δr0+δdrπakμ[sin(kπϕα)d2Rdr2]=Qϵ0r0δ(ϕϕ0).

Then, with the orthogonality of sin(kπϕ/α), multiply both sides by sin(kπϕ/α) and integrate ϕ,

limδ0r0δr0+δαπakμ2d2Rdr2dr=Qϵ0r0sin(kπϕ0α)=απ2(BkμdKkπ/α(μr0)drAkμdIkπ/α(μr0)dr).

It turns out to be the second relation

(12)AkμIkπ/α(μr0)BkμKkπ/α(μr0)=2Qμαϵ0r0πsin(kπϕ0α).

Combining (11) and (12), we can get the coefficient Akμ,Bkμ

(13){Akμ=2Qμαϵ0r0πsin(kπϕ0α)Kkπ/α(μr0)Kkπ/α(μr0)Ikπ/α(μr0)Kkπ/α(μr0)Ikπ/α(μr0),Bkμ=2Qμαϵ0r0πsin(kπϕ0α)Ikπ/α(μr0)Kkπ/α(μr0)Ikπ/α(μr0)Kkπ/α(μr0)Ikπ/α(μr0).

But that is not enough, we can simplify these expressions further with the Wrońskian of the modified Bessel functions:

(14)W[Km(x),Im(x)]=Km(x)Im(x)Km(x)Im(x)=1x.

So

Kkπ/α(μr0)Ikπ/α(μr0)Kkπ/α(μr0)Ikπ/α(μr0)=1μr0,

where the prime means the derivative with respect to μr0. The finally results of Akμ,Bkμ are

(15){Akμ=2Qαϵ0πsin(kπϕ0α)Kkπ/α(μr0),Bkμ=2Qαϵ0πsin(kπϕ0α)Ikπ/α(μr0).

1.2.3. Final result

After obtaining the coefficients Akμ,Bkμ, integrating (9) and (15), we can write the final solution of the original problem:

Φ(r,ϕ,z)={k=10dμ 2Qαϵ0πsin(kπϕ0α)Kkπ/α(μr0)Ikπ/α(μr)cos(μz)sin(kπαϕ),r<r0k=10dμ 2Qαϵ0πsin(kπϕ0α)Kkπ/α(μr)Ikπ/α(μr0)cos(μz)sin(kπαϕ),r>r0

Or in a more compact form

Φ(r,ϕ,z)=k=10dμ 2Qαϵ0πKkπ/α(μr>)Ikπ/α(μr<)cos(μz)sin(kπϕ0α)sin(kπϕα),

where r<=min(r0,r),r>=max(r0,r). But it can be further simplified, which requires us to use the formula1 (I will give a proof at the end)

(16)0Kν(ax)Iν(bx)cos(cx)dx=12abQν12(a2+b2+c22ab).

where Re a>|Re b|,Re ν>12, and Qν(x) is the second kind of the Legendre function. So the final result can be written in the following form

(17)Φ(r,ϕ,z)=Qαϵ0πr0rk=1Qkπα12(r2+r02+z22r0r)sin(kπϕ0α)sin(kπϕα).

This is exactly the electric potential in angle domain.

1.3. Prove they are equivalent if α=π/n

What we can do is not only get the results (3) and (17), we can also mathematically prove that the two solutions are equivalent when α=π/n, this requires us to prove

(18)k=12n(1)k+1ξcos(ϕϕk)=42nπk=1Qnk12(ξ)sin(nkϕ0)sin(nkϕ),

where

(19)ϕk=[k2]2πn+(1)k+1ϕ0,ξ=r2+r02+z22r0r.

Before our proof, we need to know

(20)1|rrk|=1r2+r02+z22r0rcos(ϕϕk),(21)2(1|rrk|)=4πr0 δ(rr0)δ(ϕϕk)δ(z).

So 1/|rrk| is also a solution of the original Poisson’s equation, but it does not satisfy the two boundary conditions. Following our previous expansion method, we can get a similar result

(22)1|rrk|=2πm=0dμeim(ϕϕk)cosμ(zzk)Im(μr<)Km(μr>)=4π0dμcosμ(zzk){12I0(μr<)K0(μr>)+m=1cosm(ϕϕk)Im(μr<)Km(μr>)}.

Let rk=r0er+ϕkek, we have

1|rrk|=1πrr0Q12(r2+r02+z22r0r)+2πrr0m=1cosm(ϕϕk)Qm12(r2+r02+z22rr0).

Then

2r0rk=12n(1)k+1r2+r02+z22r0rcos(ϕϕk)=2πk=12n(1)k+1{Q12(r2+r02+z22r0r)+2m=1cosm(ϕϕk)Qm12(r2+r02+z22r0r)}=2πQ12(r2+r02+z22r0r)k=12n(1)k+1+22πk=12n(1)k+1m=1cosm(ϕϕk)Qm12(r2+r02+z22r0r)=22πm=1Qm12(r2+r02+z22r0r)k=12n(1)k+1cos[m(ϕ[k2]2πn+(1)kϕ0)].

Calculate the blue part and simplify the rounding function. Obviously, the sum of this formula needs to be split into two parts where k is odd and k is even:

k=12n(1)k+1cos[m(ϕ[k2]2πn+(1)kϕ0)]=k oddcos[m(ϕ(k1)πnϕ0)]k evencos[m(ϕkπn+ϕ0)],

where the first part is (pZ+)

l=1ncos[m(ϕ(2l2)πnϕ0)]=Re[eim(ϕϕ0)ei2mπnl=1n(ei2mπn)l]={0,(mnp)ncosm(ϕϕ0),(mn=p)

and the second part is

l=1ncos[m(ϕ2lπn+ϕ0)]=Re[eim(ϕ+ϕ0)l=1n(ei2mπn)l]={0,(mnp)ncosm(ϕ+ϕ0),(mn=p)

So the blue part is

k=12n(1)k+1cos[m(ϕ[k2]2πn+(1)kϕ0)]={0,(mnp)2nsin(mϕ)sin(mϕ0),(mn=p)

Substituting this result into the right hand side of (18), we can get

(23)2r0rk=12n(1)k+1r2+r02+z22r0rcos(ϕϕk)=42nπp=1Qpn12(r2+r02+z22r0r)sin(pnϕ0)sin(pnϕ).

Obviously, the result of the above formula is exactly (18). Thus, for any point where the point charge Q is placed within the area sandwiched by the metal planes, we verify (18) is indeed true.

2. Problem with Neumann boundary condition

We can do much more than that, we can go further and consider Neumann boundary conditions. Similarly, we will discuss in two cases.

2.1. Special case: α=π/n

This situation can also be solved using the method of images, for example

Neumann Neumann Figure 3: The positions of the charge Q and image charges for n=4 with Neumann boundary condition.

It’s just that all the image charges and source charges have the same sign at this time. So

(24)ϕi=[i2]2πn+(1)i+1ϕ0,(i=1,2,,2n).

and

(25)Φ(r,ϕ,z)=Q4πϵ0i=12n1r2+r02+z22rr0cos(ϕϕi).

2.2. General case: α can be any value

In order to satisfy the Neumann boundary condition nΦ=0 , it is only necessary to modify the eigenfunction in the ϕ direction . One thing that needs to be noted is the situation of k=0 cannot be thrown away, then

ψ(ϕ)=cos(kπαϕ),m=kπα(k=0,1,2,)

So the general solution must also be modified,

Φ(r,ϕ,z)=k=00dμ akμΦkμ(r,ϕ,z)={k=00dμAkμIkπ/α(μr)cos(μz)cos(kπαϕ),r<r0k=00dμBkμKkπ/α(μr)cos(μz)cos(kπαϕ),r>r0

The coefficients can be solved using the same method as before:

{A0μ=Qαϵ0πK0(μr0)B0μ=Qαϵ0πI0(μr0),{Akμ=2Qαϵ0πcos(kπϕ0α)Kkπ/α(μr0)Akμ=2Qαϵ0πcos(kπϕ0α)Ikπ/α(μr0) (k=1,2,)

Thus, the final result of the electric potential is

(26)Φ(r,ϕ,z)=Qαϵ0πr0r{12Q12(r2+r02+z22r0r)+k=1Qkπα12(r2+r02+z22r0r)cos(kπϕ0α)cos(kπϕα)}.

2.3. Prove they are equivalent if α=π/n

Also use formula (22),

k=12n1ξcos(ϕϕk)=2r0rk=12n1r2+r02+z22r0rcos(ϕϕk)=2r0r{1πrr0k=12nQ12(r2+r02+z22r0r)+2πrr0m=1cosm(ϕϕk)Qm12(r2+r02+z22rr0)}=42nπ{12Q12(ξ)+m=1Qm12(ξ)k=1cos[ϕ[k2]2πn+(1)kϕ0]2n}.

For the summation of k on the right side, it can be calculated directly

12nk=12ncos[m(ϕ[k2]2πn+(1)kϕ0)]={0,(mnp)cos(mϕ)cos(mϕ0),(mn=p)

So

(27)k=12n1ξcos(ϕϕk)=42nπ{12Q12(ξ)+k=1Qnk12(ξ)cos(nkϕ0)cos(nkϕ)}.

This is the result given after the two solutions (25) and (26) are equivalent.

3. Problem with mixed boundary condition

The problem with mixed boundary condition can also be solved. See the following results.

3.1. Special case: α=π/2n

Note that this is different from the previous two cases. When α=π/(2n1), the method of images is no longer valid. So we just discuss the case when α=π/2n, see following example

mixed mixed Figure 4: The positions of the charge Q and image charges for n=4 with mixed boundary condition.

So

(28)ϕi=[i2]πn+(1)i+1ϕ0.(i=1,2,,4n),

and the electric potential is

Φ(r,ϕ,z)=Q4πϵ0k=12n{(1)k+1r2+r02+z22rr0cos(ϕϕ2k1)+(1)k+1r2+r02+z22rr0cos(ϕϕ2k)}.

3.2. General case: α can be any value

Also modify the eigenfunction in the ϕ direction by using

ψ(ϕ)=sin((2k1)π2αϕ),m=(2k1)π2α(k=1,2,).

The general solution is

Φ(r,ϕ,z)=k=10dμ akμΦkμ(r,ϕ,z)={k=10dμAkμI(2k1)π2α(μr)cos(μz)sin((2k1)π2αϕ),r<r0k=10dμBkμK(2k1)π2α(μr)cos(μz)sin((2k1)π2αϕ),r>r0

And the coefficients are

(29){Akμ=2Qαϵ0πsin((2k1)π2αϕ0)K(2k1)π2α(μr0),Bkμ=2Qαϵ0πsin((2k1)π2αϕ0)I(2k1)π2α(μr0).

And the final result is

(30)Φ(r,ϕ,z)=Qαϵ0πr0rk=1Q(2k1)π2α12(r2+r02+z22r0r)sin((2k1)πϕ02α)sin((2k1)πϕ2α).

3.3. Prove they are equivalent if α=π/2n

Also use formula (22),

k=12n{(1)k+1ξcos(ϕϕ2k1)+(1)k+1ξcos(ϕϕ2k)}=2r0r{k=12n(1)k+1r2+r02+z22r0rcos(ϕϕ2k1)+k=12n(1)k+1r2+r02+z22r0rcos(ϕϕ2k)}=22πm=1Qm12(ξ){k=12n(1)k+1cosm(ϕϕ2k1)+k=n2n(1)k+1cosm(ϕϕ2k)}=42πm=1Qm12(ξ)cos(mϕ0mπ2n)k=12n(1)k+1cos[mϕ(2k1)mπ2n].

Calculate the second sum of the above equation,

k=12n(1)k+1cos[mϕ(2k1)mπ2n]={2ncos(mϕ+mπ2n),(mn=2p1)0,(mn=2p1)

then

k=12n{(1)k+1ξcos(ϕϕ2k1)+(1)k+1ξcos(ϕϕ2k)}=82nπp=1Q(2p1)n12(ξ)cos((2p1)nϕ0(2p1)π2)cos((2p1)nϕ+(2p1)π2).

That is,

(31)k=12n{(1)k+1ξcos(ϕϕ2k1)+(1)k+1ξcos(ϕϕ2k)}=82nπk=1Q(2p1)n12(ξ)sin(2p1)nϕ0sin(2p1)nϕ,

which is what we want to verify.

4. Appendix: Prove the integral formula

We used (16) in our previous calculation without proof, and now we prove it. With Sonine-Gegenbauer integral

0πtνJν(ta2+b22abcosθ)(a2+b22abcosθ)ν/2sin2νθdθ=πΓ(2ν)2ν1Γ(ν)Jν(at)aνJν(bt)bν,

we calculate the integration of Jν(ax)Jν(bx)ecx,

πΓ(2ν)2ν1Γ(ν)1aνbν0Jν(at)Jν(bt)ectdt=0(0πtνJν(ta2+b22abcosθ)(a2+b22abcosθ)ν/2sin2νθdθ)ectdt=0πsin2νθ(a2+b22abcosθ)ν/2(0tνJν(ta2+b22abcosθ)ectdt)dθ

Then use the following two integral formulas,

0eaxJν(bx)xνdx=bνΓ(2ν+1)2νΓ(ν+1)1(a2+b2)ν+1/2,

and

0tνJν(ta2+b22abcosθ)ectdt=(a2+b22abcosθ)νΓ(2ν+1)2νΓ(ν+1)1(c2+a2+b22abcosθ)ν+1/2.

Substituting into the above formula, we can get

πΓ(2ν)2ν1Γ(ν)1aνbν0Jν(at)Jν(bt)ectdt=Γ(2ν+1)2νΓ(ν+1)0πsin2νθ(c2+a2+b22abcosθ)ν+1/2dθ=Γ(2ν+1)2νΓ(ν+1)2ν+1/2(2ab)ν+1/2Qν12(c2+a2+b22ab).

So

0Jν(at)Jν(bt)ectdt=1πabQν12(a2+b2+c22ab).

Following the derivation in this post, we can give the cylindrical coordinate expansion of Green’s function in another form (like (22)):

1|xx|=2πm=0dkeim(ϕϕ)cosk(zz)Im(kr<)Km(kr>).

By choosing different constant forms when separating variables, we can get another expansion form of Green’s function in cylindrical coordinates:

1|xx|=m=0dkeim(ϕϕ)Jm(kr)Jm(kr)ek(z>z<).

Since they are the same Green’s function, the two expansion forms are equivalent. The only difference lies in the selection of constants when separating variables. Note that eim(ϕϕ) is pairwise orthogonal when m is different. By comparing their coefficients, we can easily get:

0dk2πKm(kr>)Im(kr<)cosk(zz)=0dkJm(kr)Jm(kr)ek(z>z<).

Thus,

0Kν(ax)Iν(bx)cos(cx)dx=12abQν12(a2+b2+c22ab).

This is exactly (16).

  1. Zwillinger, D. (Ed.). Table of integrals, series, and products (8th ed.). Academic Press. p. 726. (2014) ↩︎

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