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Solução de Problema de Valor Médio da Potencial Elétrica de Jackson 1.10, Provas de Física

A solução do problema 1.10 do livro 'classical electrodynamics' de jackson, que consiste em provar o teorema do valor médio da potencial elétrica no espaço sem carga livre. A solução é baseada na equação da função verde de dirichlet e envolve a determinação da função verde para o caso específico sem carga livre.

Tipologia: Provas

2016

Compartilhado em 28/04/2016

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Jackson 1.10 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any
point is equal to the average of the potential over the surface of any sphere centered on that point.
SOLUTION:
The potential is known on the surface, so this problem can be formulated using a Dirichlet Green's
function equation:
x= 1
40xGDd3x'1
4
d GD
d n '
da '
where the Dirichlet Green's function must satisfy:
GD(x,x')= 1
xx'
+F(x,x')
where
'2Fx,x'=0
and GD = 0 on the surface
In this particular case, there is no free charge,
 x=0
, so that the equation simplifies to
Φ( x)=− 1
4π
(
Φd GD
d n '
)
da '
Because we are only measuring the potential at the center of the sphere which is centered on the origin,
x = 0 and therefore
1
xx'
=1
x'
, leading to:
GD(x,x')= 1
x'+F(x,x')
In order for the green function to disappear on the surface, GD(x' = R) = 0, we must have F = -1/R. The
Green function is now:
GD(x,x')= 1
x'1
R
Insert this into the equation:
pf2

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Jackson 1.10 Homework Problem Solution

Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: Prove the mean value theorem : For charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point. SOLUTION: The potential is known on the surface, so this problem can be formulated using a Dirichlet Green's function equation:   x =

∫  x^  GD d^ 3 x '−

∮^ d GD d n '  da ' where the Dirichlet Green's function must satisfy: GD ( x , x ')=

xx '∣

  • F ( x , x ') (^) where (^) ∇ '^2 Fx , x '= 0 and GD = 0 on the surface In this particular case, there is no free charge, ^ x =^0 , so that the equation simplifies to Φ( x )=−

4 π ∮(Φ^ d GD d n ' ) da ' Because we are only measuring the potential at the center of the sphere which is centered on the origin, x = 0 and therefore

xx '∣

x ' , leading to: GD ( x , x ')=

x '

  • F ( x , x ') In order for the green function to disappear on the surface, GD ( x ' = R ) = 0, we must have F = -1/ R. The Green function is now: GD ( x , x ')=

x '

R

Insert this into the equation: Φ( x )=−

4 π ∮(Φ^ d d x ' [^

x '

R ]) x '= R da '

Φ( x )=

4 π

[

x '

2 ])

x '= R da ' Φ( x )=

4 π

[

R

2 ] da^ '

  x =

∮ ^ da^ '

4  R

2 The divisor is the surface area of the sphere so that:   x =

∮ ^ da^ '

∮ da^ '

The right side is by definition the average value of the function over the surface and thus equals the value at its center.