# Griffiths E&M: Do bound charges physically exist?

Hello,
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

The bound charges do exist. You see them in:

$$\epsilon_0 \nabla \cdot \vec{E} = \rho = \rho_{free} + \rho_{bound}$$

Don't forget:

$$\vec{D} = \epsilon_0 \vec{E} + \vec{P}$$

So the bound charges are absorbed in the polarization vector.

tiny-tim