# Confusion about the total bound charge of a sphere placed in dielectric

There is a solved exercise in which a charged sphere of radium $r=a$ is inside a dielectric material. The surface bound charge and the volume bound charge are respectively:

$$p_{sp}=\vec{P}\cdot \hat{n}|_{r=a},\quad p_{vp}=-\nabla \cdot \vec{P}$$

where the polarization $P$ is known.

The exercise says that the total bound charge is:

$$Q_{T}=\int_{S}p_{sp}dS+\int_{V}p_{vp}dV=f(\epsilon_{r},q)\neq 0$$

But for the total bound charge it is true that:

$$Q_{T}=\int_{S}p_{sp}dS+\int_{V}p_{vp}dV=\oint_{S} \vec{P}\cdot \hat{n}dS-\int_{V}\nabla \cdot \vec{P}dV=0$$

How can this be when the total bound charge should always be equal to 0?

• What is S what is V? Commented Mar 8, 2018 at 19:23

The integral over the total surface bound charge , inner and outer, of the dielectric is zero. Equivalently, the volume integral of the bound charge is zero. It is not a good idea to mix surface and volume integrals to avoid double counting.

You are right that net charge of a neutral dielectric body that is just polarized by external electric field is zero. However, this follows from the assumption just stated, not from any such calculation as you did.

The integrals can be assigned various values depending on where the integration surface is placed. If the integration surface is inside the dielectric, there is no single meaning to "bound charge associated with the surface", because one can mean 1) charge inside the surface or 2) charge of molecules associated with the region inside the surface (which include the surface contribution). These two interpretation do not have same value, and they are not in general zero.

You should really ask the source of the exercise what they mean, or elaborate on the exercise in the question. There is nothing there suggesting the dielectric body should have net non-zero charge.

Dielectrics are electrically neutral as they have no free electrons, which would otherwise make them a conductor. So the total charge should be zero with the surface and volume bound charges cancelling. When it says:

$Q_T = f(\epsilon_r,q)$

does the $q$ perhaps stand for the charge and is a more general expression applicable for non-dielectrics?

• I know that they should be neutral that is why I am confused. No from what I see q is for the dielectric.