Week 4 Flashcards
2 goal equations of electrostatics
Harmonic functions
Solutions to laplaces equation
(Not only examples)
form integral form of Gauss’s law for a region of zero net charge in terms of φ
where we use E = grad*phi
Greene’s second identity
Apply greens second identity and laplaces equation to a region V which is a ball radius R and for f = 1/r g = φ
Key insight of harmonic functions
They are averaging functions and have no local extrema across V. Any extrema must occur at boundary of V
If φ satisfies Laplace’s equation in a region V and in addition is constant on boundary of V then φ must be equal to the same value across the entire region as the boundary
Dirichlet boundary conditions
Values of a function itself on a boundary (for PDE)
Neumann boundary conditions
Values of normal derivative of function along a boundary (for PDE)
Showing uniqueness of solution to Laplace’s equation in some volume V
1) let there be 2 solutions
2) construct a difference solution
3) on boundary all solutions must be equal
4) therefore both solutions equal
If Φ is specified on the boundary of V, S
Then below:
Integration by parts
Taking inner product for sin(nx)
Requirements for stable equilibrium
- force in a positive test charge Q is zero at the equilibrium point r0 (E(r0) = 0)
- when displaced from the equilibrium point forces act so as to return the test charge Q to r0
(Field lines all point towards r0)
For a charge q located at the origin ρ(r) = ?
q * δ3(r) which of course spikes at origin
In spherical coordinates, how to ‘remove’ laplacians from equation
Also grad(1/r) = 1/r * er
Relate Laplace’s equation to Poisson’s
Laplace’s equation is a special cause of Poisson’s used in regions where ρ = 0
How to go from E to Φ
You may need to split the problem
Take line integral
Uniformly charged sphere has charge inside? Outside ?
Inside: zero it’s a closed surface therefore there is no net flux and the charges are distributed symmetrically around the surface
Outside: as the sphere act like a point charge at its centre use the expression for a point charge
What is grounding
Potential is zero on the surface
Method of images: why? How?
Used in problems where we’re looking for field or potential but conductors make it difficult to do so
Additional Charges are chosen to mimic conductors zero potential
Separation of variables
Generally:
1) boundary conditions of potential
2) define variable separated potential
3) sub into Laplace’s eq and divide by variable separated Φ
4) solve individual equations with boundaries
5) determine which functions it must be
Green’s function
Using green’s function to solve Poisson’s equation
Use the Heaviside to construct a finite length between -d/2 and d/2
