Electrostatic Field Between Two Intersecting Grounded Planes
Consider two infinite grounded metal planes with an angle
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
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:
If
Figure 2: The positions of the charge
Obviously, the number of image charges is
where
Or we can rewrite
1.2. General case: can be any value
When
In cylindrical coordinates, this equation with Dirichlet boundary conditions (because the metal planes are grounded) can be expressed as:
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
Multiplying each term by
The right hand side is a function of
With boundary condition
Similarly, we have
The right hand side is a function of
Note that
Finally,
Let
This is the standard
where
Since
Combine the results
So the general solution can be expressed as linear superposition of these eigenfunctions,
1.2.2. Calculate the coefficients
Even though we get
- The electric potential should be continuous at
; - 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
Integrate
It becomes
Due to the orthogonality of
Next, using the second condition and substituting the general solution into Poisson’s equation, we can get
Note that
Similarly, integrate
Then, with the orthogonality of
It turns out to be the second relation
Combining
But that is not enough, we can simplify these expressions further with the Wrońskian of the modified Bessel functions:
So
where the prime means the derivative with respect to
1.2.3. Final result
After obtaining the coefficients
Or in a more compact form
where
where
This is exactly the electric potential in angle domain.
1.3. Prove they are equivalent if
What we can do is not only get the results
where
Before our proof, we need to know
So
Let
Then
Calculate the blue part and simplify the rounding function. Obviously, the sum of this formula needs to be split into two parts where
where the first part is (
and the second part is
So the blue part is
Substituting this result into the right hand side of
Obviously, the result of the above formula is exactly
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:
This situation can also be solved using the method of images, for example
Figure 3: The positions of the charge
It’s just that all the image charges and source charges have the same sign at this time. So
and
2.2. General case: can be any value
In order to satisfy the Neumann boundary condition
So the general solution must also be modified,
The coefficients can be solved using the same method as before:
Thus, the final result of the electric potential is
2.3. Prove they are equivalent if
Also use formula
For the summation of
So
This is the result given after the two solutions
3. Problem with mixed boundary condition
The problem with mixed boundary condition can also be solved. See the following results.
3.1. Special case:
Note that this is different from the previous two cases. When
Figure 4: The positions of the charge
So
and the electric potential is
3.2. General case: can be any value
Also modify the eigenfunction in the
The general solution is
And the coefficients are
And the final result is
3.3. Prove they are equivalent if
Also use formula
Calculate the second sum of the above equation,
then
That is,
which is what we want to verify.
4. Appendix: Prove the integral formula
We used
we calculate the integration of
Then use the following two integral formulas,
and
Substituting into the above formula, we can get
So
Following the derivation in this post, we can give the cylindrical coordinate expansion of Green’s function in another form (like
By choosing different constant forms when separating variables, we can get another expansion form of Green’s function in cylindrical coordinates:
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
Thus,
This is exactly
Zwillinger, D. (Ed.). Table of integrals, series, and products (8th ed.). Academic Press. p. 726. (2014) ↩︎