- #1

jin8

- 24

- 0

Well, actually it's not a homework problem, I just got really confused about bound charges.

Originally, I thought it was just a special technique to do the integral, but somehow Griffiths suggest that bound charges are phsically exist.(chap. 4.2.2) Well, I can accept his argument, but later when we do the boundary condition problem, we have the equation

Integral D dv = Q (free)

So why it's Q free, not Q free plus the surface bound charge ? I know the math why the equation is true, but if surface bound charge physically exist, I don't think the equation make any sense...

Any help is appreciated..

Thanks

-Jin