| Topic: |
Science > Physics |
| User: |
"Harry" |
| Date: |
20 Oct 2005 03:45:48 PM |
| Object: |
Electrodynamics' problem--Earnshaw's theorem. |
Hi there,
The problem is below, (from Introduction to electrodymanics, David
J=2EGriffiths)
Problem 3.2 In one sentence, justify Earnshaw=E2=80=99s Theorem=EF=BC=9AA =
charged
particle cannot be held in a stable equilibrium by electrostatic forces
alone. As an example, consider the cubical arrangement of fixed charges
(on each corner). It looks off hand, as though a positive charge at the
center would be suspended in midair, since it is repelled away from
each corner. Where is the leak in this =E2=80=9Celectrostatic bottle=E2=80=
=9D=EF=BC=9F
And I got the solution as below,
A stable equilibrium is a point of local minimun in the potential
energy. Here the potential energy is qV. But we know that Laplace's eq
allows no local minima for V. In this case, a minimum must be a saddle
point and the box "LEAKS" through the center of each face.
But I don't know how the potential energy is qV. What does this mean?
I tried to set a Laplace's eq, but I don't know how to set it.
Obviously, the "leak" must be in the centre of each face. Since I tried
to draw a 3-D field diagram, I'm still confused how to deal with this.
Can it be proven in mathematically work?
.
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|
| User: "Timo Nieminen" |
|
| Title: Re: Electrodynamics' problem--Earnshaw's theorem. |
20 Oct 2005 05:16:24 PM |
|
|
On Fri, 20 Oct 2005, Harry wrote:
Hi there,
=20
The problem is below, (from Introduction to electrodymanics, David
J.Griffiths)
Problem 3.2 In one sentence, justify Earnshaw=E2=80=99s Theorem=EF=BC=9A=
A charged
particle cannot be held in a stable equilibrium by electrostatic forces
alone. As an example, consider the cubical arrangement of fixed charges
(on each corner). It looks off hand, as though a positive charge at the
center would be suspended in midair, since it is repelled away from
each corner. Where is the leak in this =E2=80=9Celectrostatic bottle=E2=
=80=9D=EF=BC=9F
=20
And I got the solution as below,
A stable equilibrium is a point of local minimun in the potential
energy. Here the potential energy is qV. But we know that Laplace's eq
allows no local minima for V. In this case, a minimum must be a saddle
point and the box "LEAKS" through the center of each face.
=20
But I don't know how the potential energy is qV. What does this mean?
Just the definition of the electrostatic potential: the potential energy=20
per unit charge. So to find the potential energy of a charge in a field,=20
just multiply by the potential.
I tried to set a Laplace's eq, but I don't know how to set it.
Obviously, the "leak" must be in the centre of each face. Since I tried
to draw a 3-D field diagram, I'm still confused how to deal with this.
Can it be proven in mathematically work?
Write down the Laplace eqn in Cartesian coordinates. What is the condition=
=20
for a local maximum/minimum? Can this be satisfied along the x, y, and z=20
directions simultaneously.
Now try to distill that down to one sentence!
(Actually, it's a little more complicated than that, but I doubt the=20
question expects more. What about the case when the 2nd derivatives in the=
=20
Laplace eqn are zero at a point?)
As for the cube example, why not try some calculations? You know the=20
potential due to a single point charge, so it's easy enough to find the=20
potential due to 8 of them. What is the potential at the centre of the=20
cube? Now, what is the potential a small distance from the centre (try=20
towards one of the faces, towards one of the edges, and towards one of the=
=20
corners)? Estimate the gradient of qV. Then you should see where the leak=
=20
is.
Use the Laplace eqn for a general proof, but just use the potential of a=20
point charge for calculating the cubic arrangement case.
--=20
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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