1. Vanishing Potential from Two Point Charges
2. Electrostatics
2A. Method of Images
Vanishing Potential from Two Point Charges
Consider two charges q and q' of opposite sign
Lets assume without loss of generality that |q|  |q'|
To simplify, put q at origin and q' on the z –axis at a distance d
Potential is given by:
Where is the potential zero? q' d q

2. Case 1: Planes
Suppose the two charges are equal and opposite, then the potential vanishes at
This is a plane that is equidistant between the two points
How is this useful?
Suppose we have a plane with Dirichlet boundary conditions  = 0 on the plane
If you pretend there is an “image charge” – q on the other side of the Dirichlet boundary, the two charges will produce a potential that cancels on the boundary
It will also satisfy 2 = –/0 in the allowed region –q d q

3. Sample Problem 2.1 (1)
An infinite line of charge with linear charge density  is at z = h, x = 0 above a plane conductor at z = 0. Find the electric field everywhere for z > 0, and the surface charge.
The conductor is at constant potential
Since it extends forever, assume it has  = 0
Add an opposite charge line charge on the other side of the conductor
This guarantees that  = 0 on the conductor
Electric field from the original line charge
 is a vector pointing from the line charge directly to the point of interest
So the electric field from the original line is
The image line charge causes an electric field
So the total field is:

4. Sample Problem 2.1 (2)
An infinite line of charge with linear charge density  is at z = h, x = 0 above a plane conductor at z = 0. Find the electric field everywhere for z > 0, and the surface charge.
To find the surface charge on the conductor, first find the electric field there
Note that it is perpendicular to the surface there, as it must be
Related to surface density by
Can show that, in this case, integrated surface charge cancels the original line

5. Case 2: Spheres Suppose |q| > |q'| then the potential vanishes when
Complete the square
This is a sphere of radius:
Distance from q to center:
Distance from q' to center: a r' r q' d q

6. Conducting Spheres a
Let’s put these formulas together in a useful way:
How is this useful?
Suppose we had a grounded ( = 0) conducting sphere of radius a
Plus a point charge q at distance r > a
Pretend there is an image charge q' at distance r', where
This will exactly cancel the potential from the charge q on the surface of the sphere
And it will satisfy 2 = 0 outside the sphere
So we have solved the problem! r' r q' q

7. Sample Problem 2.2a
A conducting sphere of radius a with potential  = 0 at the origin has a charge q on the z-axis at z = 2a. Find the potential (outside), and the total charge on the sphere. q
Add an image charge of magnitude
It will be at
The potential is then
To get the charge:
The hard way:
Find the electric field from the potential
Get the surface charge everywhere and integrate
The easy way:
Use Gauss’s Law on a surface surrounding the sphere
Electric field looks like point charge –q/2
So charge on sphere must be –q/2 2a
– q/2
a/2 a

8. Sample Problem 2.2b
A neutral conducting sphere of radius a at the origin has a charge q on the z-axis at z = 2a. Find the potential. q
First guess: We can do this exactly the same way as Problem 2.2a
This is wrong because the sphere had charge – q/2
So far we have  = 0 on the sphere
To cancel this charge, add additional charge + q/2 on the sphere
It will distribute itself uniformly over the surface
This creates additional potential that looks like a point charge of magnitude + q/2 at the center
Total potential (outside) is sum of these three contributions 2a
– q/2
a/2
q/2 a

9. Hollow Spheres
We can use the same formulas for spherical cavity in a conductor
Place an image charge at r' of magnitude q'
Combination will cause vanishing potential on the interior surface
You don’t have to worry about whether surrounding conductor is neutral, since excess charge flows away from interior surface
If you want to make   0 on interior surface, just add a constant to it
It is irrelevant anyway. q q'

10. Green’s Functions for Planes and Spheres
Since we know how to get the potential for a point charge near a conducting plane or sphere we can find the Green’s functions for them
Green’s functions (with Dirichlet boundary conditions) satisfy
For a plane at z = 0 for the region z > 0, one source at x, one at xR = (x, y,–z)
For a sphere at origin with radius a, one point at x of size 1, one point at xR = a2x/x2 with magnitude –a/|x|
With some work can rewrite this as
This formula also works for interior of a sphere as well

11. Using Green’s Functions for Spheres (1)
Probably easiest to use in spherical coordinates
 is angle between x and x'
We can then use this formula for Dirichlet problems with spherical boundary
The normal derivative is:
On the surface r' = a this becomes

12. Using Green’s Functions for Spheres (1)
Finally, the surface integral will take the form:
So we have

13. Sample Problem 2.3 (1)
The potential on the surface of a sphere of radius a centered at the origin is given by  = Ez. Find the potential outside the sphere. x z
We use our formula for the potential, ignoring the  term
For purposes of this integral, treat x as if it were the z-axis
This makes  = '
With this choice of z-axis, what we normally call z-direction becomes
 is angle between x and ordinary z-axis
We are on the surface of the sphere, so
So we have:
Set up the integrals

14. Sample Problem 2.3 (2)
The potential on the surface of a sphere of radius a centered at the origin is given by  = Ez. Find the potential outside the sphere. x z
Do the ' integral
Let

15. Sample Problem 2.4
A neutral conducting sphere of radius a is placed in a background electric field given by . Find the potential everywhere outside the sphere.
Recall:
This suggests
But this can’t be right because it is not constant at r = a
Let’s try to find a solution of the form:
Since there’s no charge outside, we must have
We need to cancel the potential from the first term, so
We already solved this problem:
So the answer is:

16. 2B. Orthogonal Functions and Expansions
General Theory of Orthogonal Functions
Let Un(x) be a set of functions in some region in d dimensions
d may be less than 3, even if the problem is 3-dimensional!
If we have enough functions Un(x), we can often approxi- mate any function as linear combinations of these functions
Often, finite sum is a good approximation for the infinite
It is useful to normally arrange the Un’s to be orthogonal
This can always be arranged
See quantum notes for details
Assuming the integral is finite when m = n, you can normalize them to make them orthonormal
Given f(x), it is not hard to find the coefficients an:

17. Completeness Relation, and Other Comments
Substitute the expression on the right into the left
Since this is true for any function f(x), we must have the completeness relation
I’m not sure what  means, because I’m not sure how many dimensions we’re in
Sometimes, we might not want all possible functions
We might want functions that vanish at certain locations, so we’d choose
Completeness still works in interior
We might want functions that have vanishing Laplacian, so we’d choose
This will ruin completeness

18. 1D Discrete Fourier Transforms
Consider functions on the region (0, a) with periodic boundary conditions, f(0) = f(a)
It is well known that a complete set of functions that satisfy this equation are the Fourier modes:
n =0, 1, 2, 3, 
These are not quite orthonormal
Any function can be written in terms of them:
Coefficients are
Sometimes, it is easier to work with real functions
n = 0, 1, 2, 3, 
Then we can write any function in terms of them:
Coefficients are:

19. 1D With Vanishing Boundaries
If we instead demand that f(0) = 0 = f(a), we should use some different functions
Use real functions, but want them to vanish at these points:
n = 1, 2, 3, 
Not quite orthonormal
We can write an arbitrary function in the form:
The coefficients are given by

20. 2D With Vanishing Boundaries
Suppose we have a 2D problem with f(x,y) where 0 < x < a and 0 < y < b
Let’s suppose we want f to vanish on all four boundaries
Since f vanishes at x = 0 and x = a, we must be able to write
Since f vanishes at y = 0 and y = b, the same must be true of An
Therefore, An must be writable in the form
We therefore have
Coefficients are: b a

21. Continuous Bases
It is sometimes the case that you need to use a basis where n takes on continuous values
We might write the basis functions as U(x)
Then all sums become integrals
The relations above get changed to:

22. Continuous Fourier Transforms
Theorem from quantum:
Consider the set of functions:
They are orthonormal:
Also complete
So we can write arbitrary functions in terms of these
Amplitude is given by
Can find similar formulas in 3D:

23. Decomposition in Different Coordinates
Consider the Laplacian in 3D in different coordinate systems:
Cartesian:
Cylindrical:
Spherical:
Often good to solve problem explicitly by writing function as a complete function in some of these variables, and unknown function in others

24. 2C. Solving Problems in Cartesian Coordinates
The Method
Suppose, for example, you have a box with potential zero on some surfaces
And let’s say it’s specified on others
Write the potential in terms of complete functions on x and y
If there is no charge inside, then

25. Modes That Satisfy Laplace’s Equation
Substitute this into Laplace’s equation:
Since these wave functions are independent, we must have
Define
Then we have
The general solution of this equation is

26. Matching the Boundary Equations
We haven’t matched the boundary conditions
On the top and bottom, we can also write the (known) potential as
We can get the coefficients
To get it to match at z = 0, we must have
To get it to match at z = c, we must have
Two equations in two unknowns
Solve for nm and nm and we have 

27. Sample Problem 2.5 (1)
A cube of size a has potential 0 on five faces and potential  = xy on the last face. What is the potential everywhere, and at the center?
We have
We need it to vanish on z = 0, so nm = –nm
To match on z = a, we have
Find the coefficients

28. Sample Problem 2.5 (2)
A cube of size a has potential 0 on five faces and potential  = xy on the last face. What is the potential everywhere, and at the center?
Now just substitute in the center

29. 2D. Solving Probs. in Cylindrical Coordinates
Potential in a Sharp Corner
Laplacian in cylindrical coordinates
Let’s do a 2D problem
Everything independent of z
Consider a sharp conducting angle of size  with vanishing potential
Presumably other sources/potentials somewhere else
The potential must vanish at  = 0 and  = 
We therefore can write the potential as
Substitute into
To make this vanish, we must have 

30. Solving the Radial Equation
Second order differential equation should have two linearly independent solutions
Guess solutions of the form
We therefore have
The general solution is
If we want finite potential near  = 0, must choose Bn = 0
Therefore:
Largest contribution from n = 1:

31. Electric Field near a Sharp Corner
The electric field is the derivative of the potential
This expression diverges at small  if  > 
The sharper the corner, the faster it diverges
In 3D, electric fields maximum at pointy places
Lightning rods have large fields, cause charge to flow from atmosphere
Drains away charge buildup in vicinity of the rod