Physics 217, Fall 2002 4 October 2002dmw/phy217/Lectures/Lect_14p.pdf · Physics 217, Fall 2002 4...

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Physics 217, Fall 2002 4 October 2002 (c) University of Rochester 1 4 October 2002 Physics 217, Fall 2002 1 Today in Physics 217: Laplace’s equation Solutions to the Laplace equation Simple examples Uniqueness of solutions 2 2 2 2 2 2 2 2 1 1 sin sin 1 0 sin V V r r r r V r V r θ θ θ θ θ φ = + + = 2 2 2 2 2 2 1 1 0 V V s s s s V V s z φ = + + = 4 October 2002 Physics 217, Fall 2002 2 Solving the Laplace equation In electrostatics, we’re usually looking for E. If there isn’t symmetry appropriate for the use of Gauss’ Law, we use the generalized Coulomb’s Law: Or, if that’s too hard, use Even that’s too hard sometimes, so we wind up trying 2 ˆ d ρτ = E r r , d V V ρτ = =− E r 2 4 V πρ =− 4 October 2002 Physics 217, Fall 2002 3 Solving the Laplace equation (continued) If the charge density is zero where we want to compute V and E, this becomes the Laplace equation: Example solution in 1-D: 2 2 2 2 2 2 2 0 0 V V V V x y z = + + = ( ) 2 2 0 Integrate: Integrate again: dV dx dV a dx Vx ax b = = = +

Transcript of Physics 217, Fall 2002 4 October 2002dmw/phy217/Lectures/Lect_14p.pdf · Physics 217, Fall 2002 4...

Physics 217, Fall 2002 4 October 2002

(c) University of Rochester 1

4 October 2002 Physics 217, Fall 2002 1

Today in Physics 217: Laplace’s equation

Solutions to the Laplace equationSimple examplesUniqueness of solutions

2 22

2

2

2 2 2

1

1 sinsin1 0

sin

VV rr rr

Vr

Vr

θθ θθ

θ φ

∂ ∂ ∇ = ∂ ∂ ∂ ∂ + ∂ ∂

∂+ =

2

2 2

2 2 2

1

1 0

VV ss s s

V Vs zφ

∂ ∂ ∇ = ∂ ∂

∂ ∂+ + =

∂ ∂

4 October 2002 Physics 217, Fall 2002 2

Solving the Laplace equation

In electrostatics, we’re usually looking for E. If there isn’t symmetry appropriate for the use of Gauss’ Law, we use the generalized Coulomb’s Law:

Or, if that’s too hard, use

Even that’s too hard sometimes, so we wind up trying

dρ τ= ∫E rr

,dV Vρ τ= = −∫ E

r—

2 4V πρ∇ = −

4 October 2002 Physics 217, Fall 2002 3

Solving the Laplace equation (continued)

If the charge density is zero where we want to compute Vand E, this becomes the Laplace equation:

Example solution in 1-D:

2

2 2 2

2 2 2

0

0

V

V V Vx y z

∇ =

∂ ∂ ∂+ + =

∂ ∂ ∂

( )

2

2 0 Integrate:

Integrate again:

d VdxdV adx

V x ax b

=

=

= +

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Solving the Laplace equation (continued)

This is the general solution, complete with the two integration constants a and b. To find out what they are, consult boundary conditions. Suppose we were told that V = 4 at x = 1, and 0 at x = 5. Then

( )4

1, 5 50 5

a ba b V x x

a b= +

⇒ = − = ⇒ = − += +

4 October 2002 Physics 217, Fall 2002 5

Solving the Laplace equation (continued)

Example solution in 3-D: suppose we have a spherically-symmetric situation, in which things only depend upon r, and not on the other two spherical coordinates:

22 2

2 2 2 2 21 1 1sin

sin sinr

r rr r rθ

θ θθ θ φ

∂ ∂ ∂ ∂ ∂ ∇ = + + ∂ ∂ ∂ ∂ ∂

2 22

22

2

1 0 Integrate:

Integrate again:

General solution.

VV rr rr

V V ar ar r r

dr aV a brr

∂ ∂ ∇ = = ∂ ∂ ∂ ∂

= ⇒ =∂ ∂

= = − +∫

4 October 2002 Physics 217, Fall 2002 6

Solving the Laplace equation (continued)

Boundary conditions: suppose that

The first of these implies b = 0, the second implies that a = V0R. Yet another example: cylindrical coordinates, but independent of φand z.

00 at , at :V r V V r R= →∞ = =

2 22

2 2 21 1ss s s s zφ

∂ ∂ ∂ ∂ ∇ = + + ∂ ∂ ∂ ∂

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4 October 2002 Physics 217, Fall 2002 7

Solving the Laplace equation (continued)

As before,

and boundary conditions would determine what a and b are.

2 1 0

ln ln

VV ss s s

Vs as

ds sV a a s b as b

∂ ∂ ∇ = = ∂ ∂ ∂

=∂

′= = + = ∫

4 October 2002 Physics 217, Fall 2002 8

Uniqueness of solutions

The Laplace equation, supplied with two boundary conditions, has one and only one solution. Why?

If V is known on all boundaries, then only one V in the space between the boundaries will match. Proof:Assume that there are two solutions, V1 and V2. Consider the sum of these solutions, Then, because they have the same boundary conditions,

So V3 must be zero everywhere at the boundary, and be “flat” everywhere else because its Laplacian is zero. That is, V3 is zero everywhere, so

3 2 1.V V V= −

( ) ( ) ( )3 2 12 2 2

3 2 1

boundary boundary boundary 0

0 0 0

V V V

V V V

= − =

∇ = ∇ −∇ = − =

2 1 everywhere.V V=

4 October 2002 Physics 217, Fall 2002 9

Uniqueness of solutions (continued)

For charge specified on an arbitrary arrangement of conductors, there is only one solution for V. “Proof:”• E = 0 inside conductors. Any distributed charge would

move until that were true. • There would be only one final resting place for each of

the charges.• Thus

integrates to one and only one answer at each point in space.

dV ρ τ= ∫ r

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4 October 2002 Physics 217, Fall 2002 10

Importance of uniqueness of solutions to the Laplace equation: your hunting license

One can try any means at one’s disposal to derive or guess a solution to the Laplace equation. If one can show that it fits the boundary conditions, or gives the right charge on each conductor, then one has found the only correct answer.

This fact will enable us to use several tricks that simplify theobtaining of solutions to the Laplace equations. Next time we’ll talk about the cutest one, the method of images.