Transcript Laplace`s equation

PHY 042: Electricity and Magnetism
Laplace’s equation
Prof. Hugo Beauchemin
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Introduction I
 Typical problem in E&M:
A set of conductors are maintained at a given potential to
shape a field outside the conductors in order to perform some
task. We need to know what is the field created.
 Example: drift chambers

Knowing the electric field allows to compute drift time, and therefore
to determine particle trajectories in the detector (tracks)
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Introduction II
 In the previous example: we don’t know the charge distribution nor
the total charge, but we know the voltage on the conductors
because they are maintained by hand
 We can find the E-field everywhere using Laplace’s equation
 A potential that doesn’t fully satisfy Laplace’s equation together
with a certain set of boundary conditions in a region of space
where there is no charge is not an electrostatics potential
 To fully appreciate the gain in generality provided by the Gauss’
equation (+ Helmholtz’ theorem) in its differential form, the
development of more in-depth mathematical tools to solve the
Laplace’s equation is needed
 The techniques apply equally well to Quantum and Newtonian
mechanics, astronomy, fluid mechanics, heat theory
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A 1D intuition
 To get an intuition about solving the Laplace’s equation, let study
the 1D case


Two undetermined constants
Using the boundary conditions specified in a concrete experiment
(problem), we can solve for m and b and get a unique solution
 E.g.: {V(0), V(L)}, {V(0), V’(0)}, {V(0), V’(L)}, etc.
 The solution has two simple but important properties:
1.
 Solution satisfied the superposition principle
 Similar to stationary wave solutions …
2. The solution has no max nor min, except at the end points
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Higher dimensions
 Laplace’s equation now becomes a partial differential equation

The general solution is not expressible with just 2 arbitrary constants
 It generally needs an finite number of free constants

There are no simple general solution: case by case
 The two properties in 1D are generalizable to higher dimensions

The value of V at r is the average of V over a spherical surface of
radius R centered on r
 This is used to get numerical solutions to Laplace’s equation

V cannot have any local min nor max, except at boundaries
 Will use these two properties to show that:
A set of boundary conditions suffices to
uniquely find V from Poisson/Laplace
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How to solve Laplace?
 Once again, there is no general solution to the equation and the
number of constants to be fixed completely depends on the
boundary conditions and thus on the problem to be solved
 But there are many different methods that can be used to solve
different classes of problems, classes that each account for many
different experimental setups and contexts.

There are problems that are formally and mathematically related to
each others such that the same techniques of solving Laplace can be
used for all of them.
 Problems (situations, setups) where V or s is specified on the
boundary and V must be determined inside the interior:
 Solve by a method called: separation of variables
 Still very general: this corresponds to many concrete devices such as
drift chamber detectors where an E-field is produced
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Types of boundary conditions
 The separation of variables method applies when either V or s is
specified on the boundary.
 Means two very different things
1.
2.
Empirically: Two very different experimental preparations:

Fixing V means grounding a conductor or attaching it to a battery

Fixing s means charging a conductor, and maintaining the charge
distribution constant
Mathematically: two different formalisms for fixing V or s:

V=V0  solution is known at some x0  Dirichlet conditions

s=s0  derivative of solution is known at x0  Neumann conditions
 For conductors, just need to know Qtot (theorem)
With a complete set of Dirichlet and/or Neumann
conditions the solution to Laplace’s equation is unique
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A generic example
Problem:
y
Two infinite grounded metal
plates parallel to the XZ plane
at y=0 and y=a, and an infinite
strip insulated from the two
plates situated at x=0 and
maintained at a fixed potential
V0(y)
V=0
V=V0(y)
V=0
x
 Find V between plates
z
Solution
2.
Apply separation of variables
1.
Find boundary conditions
3.
Get general solutions to each 4. Apply boundary conditions to fix
separated diff. eqn.
free constants
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A generic solution
 Solving the Laplace’s equation with the method of separation of
variables for the above 2D problem yields the general solution:
 Use boundary conditions specific to this problem to fix A,B,C,D, k
 After having applied 3 of the boundary conditions, we are left with
a family of solutions to Laplace’s equation with these boundaries
 The sum of solutions is a solution to Laplace’s equation, with
different boundary conditions

Remember V3 in our proof of uniqueness
 Fourier series provide THE solution to our generic problem
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Another generic example:
spherical symmetry I
 Laplace’s equation takes a different form:
 Assume again separation of variables (with azimuthal symmetry):
 An eigenvalue problem
 Remember spherical harmonics
 l is an integer number
 Remember quantization of angular momentum
 Solution for Q(q): Legendre’s polynomial
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Spherical symmetry II
 Solution to R(r):
 guesses from limiting cases:
and
 If this is a solution, then this is the solution…
 From this intuitive approach, we have as a general
solution:
 We then need specific boundary conditions to fix Al, Bl
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Example: hollow sphere
 An hollow sphere of radius R is maintained
at a potential V0(q) on its external surface
 Boundary conditions
Inside
Outside
V(0,q)<∞
V(∞,q)<∞
V(R,q)=V0(q)
V(R,q)=V0(q)
 Generally, the potential V0 applied to the
V0=ksin2q/2
sphere is not a Legendre’s polynomial, but
any functions decided by the experimenter
⇒ Need to use Fourier again

Assume V0(q) = ksin2(q/2)
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Another example
 Grounded hollow sphere in an
constant external electric field:
 The solution is:
E=E0 z

First term: external field
contribution

Second term: contribution due
to the induced charge
 The induced charge is:

Positive at the top, negative at
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the bottom