On Wed, 13 Aug 2003, Borun Dev Chowdhury wrote:

What are the boundary conditions in presence of dielectrics ? For
conductors we take the discontinuity of electric field equal to the
surface charge density and the potential equal when approached from either
side of the boundary.
In the case of dielectric Jackson asks us to take the discontinuity of
dispacement vector at the boundary to be the charge density and equate the
tangential componenets of the electric field.
He solves the problem of the sphere in uniform field with that.
Griffiths solves the same problem by taking the potential continuous at
the boundary apart from condition first. They both get the same result but
that is due to the fact that an angular derivative on both sides of
equation to get the electric field from potential doesnt alter the
coffecients. But who has the right physics ?

Why are boundary conditions needed in the first place? Because a
discontinuity in \epsilon or \mu causes a certain mathematical problem at
the discontinuities if you are using the usual differential form of
Maxwell's equations.

Now, you don't have to use the differential form of Maxwell's equations -
you can use the integral form instead. There's no difficulty in
integrating across a boundary. Therefore, one can find the boundary
conditions for the differential equations by using the integral equations.
Choose suitable infinitesimal contours, and integrate, and one gets the
usual boundary conditions for the fields, n x (E1 - E2) = 0 etc. So the
boundary conditions for the fields directly follow from Maxwell's
equations. Therefore Jackson certainly has the right physics.

However, you don't need to write the Maxwell equations in terms of fields;
you can write them in terms of potentials instead. So one can either write
Maxwell's equations in terms of potentials, and use the above procedure to
obtain the boundary conditions in terms of potentials. Or just convert the
fields in the boundary conditions for fields into potentials. For the
scalar potential in electrostatics, one obtains the boundary condition
used by Griffiths. So clearly Griffiths also has the correct physics.

Both methods give the same result since both methods are equivalent, and
both are correct.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

Thanks but I still cant see how n x (E2-E1) = 0 ensures the continuity of
the potential across the boundary and vice versa. But I havent yet been
able to come up with a situation where this doesnt work. Unfortunately the
text book problems are often symmetric making atleast one the boundary
conditions trivial. Do you mean to say, if suppose I sollve the problem in
terms of expansion of potential in orthogonal functions, I can just equate
potentials on both sides instead of finding out tangential derivatives and
equating them ?

Borun Dev Chowdhury

1011 F Smith Lab
174 West 18th Avenue
Columbus, OH, USA

Phone 1-614-292-0067

On Thu, 14 Aug 2003, Timo Nieminen wrote:

On Wed, 13 Aug 2003, Borun Dev Chowdhury wrote:

What are the boundary conditions in presence of dielectrics ? For
conductors we take the discontinuity of electric field equal to the
surface charge density and the potential equal when approached from either
side of the boundary.
In the case of dielectric Jackson asks us to take the discontinuity of
dispacement vector at the boundary to be the charge density and equate the
tangential componenets of the electric field.
He solves the problem of the sphere in uniform field with that.
Griffiths solves the same problem by taking the potential continuous at
the boundary apart from condition first. They both get the same result but
that is due to the fact that an angular derivative on both sides of
equation to get the electric field from potential doesnt alter the
coffecients. But who has the right physics ?

Why are boundary conditions needed in the first place? Because a
discontinuity in \epsilon or \mu causes a certain mathematical problem at
the discontinuities if you are using the usual differential form of
Maxwell's equations.

Now, you don't have to use the differential form of Maxwell's equations -
you can use the integral form instead. There's no difficulty in
integrating across a boundary. Therefore, one can find the boundary
conditions for the differential equations by using the integral equations.
Choose suitable infinitesimal contours, and integrate, and one gets the
usual boundary conditions for the fields, n x (E1 - E2) = 0 etc. So the
boundary conditions for the fields directly follow from Maxwell's
equations. Therefore Jackson certainly has the right physics.

However, you don't need to write the Maxwell equations in terms of fields;
you can write them in terms of potentials instead. So one can either write
Maxwell's equations in terms of potentials, and use the above procedure to
obtain the boundary conditions in terms of potentials. Or just convert the
fields in the boundary conditions for fields into potentials. For the
scalar potential in electrostatics, one obtains the boundary condition
used by Griffiths. So clearly Griffiths also has the correct physics.

Both methods give the same result since both methods are equivalent, and
both are correct.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html