- #1
Sturk200
- 168
- 17
This is out of Jackson's electrodynamics, problem 1.15.
I am trying to prove the theorem that if a number of surfaces are fixed in position and a given total charge is placed on each surface, then the electrostatic energy in the region bounded by the surfaces is an absolute minimum when the charges are placed so that every surface is an equipotential, as happens when they are conductors.
The electrostatic energy is given by the square of the field as
##W=\frac{\epsilon_0}{2}\int|E|^2d^3x.##
In terms of the potential this is
##W=\frac{\epsilon_0}{2}\int\nabla\Phi\cdot\nabla\Phi d^3x.##
We seek an extremum in the energy by varying the energy to first order
##W(\Phi + \delta\Phi)-W(\Phi) = \epsilon_0\int\nabla\delta\Phi\cdot\nabla\Phi.##
Any two scalars ##\phi## and ##\psi## satisfy
##\int_v [\nabla \phi \cdot \nabla \psi + \phi\nabla^2\psi]d^3x = \int_s \phi \frac{\partial\psi}{\partial n} da##
where the surface integral is taken over a the surface bounding the volume over which the volume integral is taken, and the derivative with respect to ##n## indicates a derivative in the direction normal to the surface. Thus the variation in the energy may be written
##W(\Phi+\delta\Phi)-W(\Phi) = \epsilon_0\bigg[ \int_s\delta\Phi\frac{\partial\Phi}{\partial n} da - \int_v \delta\Phi\nabla^2\Phi d^3x\bigg]##
We assume that the region bounded by the surface charge is charge-free, and so the potential satisfies Laplace's equation (##\nabla^2\Phi=0##) in this region and the volume integral vanishes.
##\epsilon_0 \int_s\delta\Phi\frac{\partial\Phi}{\partial n} da = 0##
This must be true for arbitrary ##\delta\Phi##, and therefore the integrand itself must be zero at all points on the surface
##\frac{\partial\Phi}{\partial n} = 0.##
This is where I run into trouble. Assuming that the steps up to this point are correct, I do not understand how to interpret this condition physically. It seems to be saying that the electric field through the bounding surface must be zero at all points on the surface. But how does that translate to every surface being an equipotential? Any hints?
Thanks!
I am trying to prove the theorem that if a number of surfaces are fixed in position and a given total charge is placed on each surface, then the electrostatic energy in the region bounded by the surfaces is an absolute minimum when the charges are placed so that every surface is an equipotential, as happens when they are conductors.
The electrostatic energy is given by the square of the field as
##W=\frac{\epsilon_0}{2}\int|E|^2d^3x.##
In terms of the potential this is
##W=\frac{\epsilon_0}{2}\int\nabla\Phi\cdot\nabla\Phi d^3x.##
We seek an extremum in the energy by varying the energy to first order
##W(\Phi + \delta\Phi)-W(\Phi) = \epsilon_0\int\nabla\delta\Phi\cdot\nabla\Phi.##
Any two scalars ##\phi## and ##\psi## satisfy
##\int_v [\nabla \phi \cdot \nabla \psi + \phi\nabla^2\psi]d^3x = \int_s \phi \frac{\partial\psi}{\partial n} da##
where the surface integral is taken over a the surface bounding the volume over which the volume integral is taken, and the derivative with respect to ##n## indicates a derivative in the direction normal to the surface. Thus the variation in the energy may be written
##W(\Phi+\delta\Phi)-W(\Phi) = \epsilon_0\bigg[ \int_s\delta\Phi\frac{\partial\Phi}{\partial n} da - \int_v \delta\Phi\nabla^2\Phi d^3x\bigg]##
We assume that the region bounded by the surface charge is charge-free, and so the potential satisfies Laplace's equation (##\nabla^2\Phi=0##) in this region and the volume integral vanishes.
##\epsilon_0 \int_s\delta\Phi\frac{\partial\Phi}{\partial n} da = 0##
This must be true for arbitrary ##\delta\Phi##, and therefore the integrand itself must be zero at all points on the surface
##\frac{\partial\Phi}{\partial n} = 0.##
This is where I run into trouble. Assuming that the steps up to this point are correct, I do not understand how to interpret this condition physically. It seems to be saying that the electric field through the bounding surface must be zero at all points on the surface. But how does that translate to every surface being an equipotential? Any hints?
Thanks!