# What is the net macroscopic electric field at any point inside a dielectric?

This question is based on section 4.2.3 of Introduction to Electrodynamics by David J. Griffiths (Third Edition).

Consider a point $$\vec{r}$$ inside a dielectric where we are interested in finding the macroscopic electric field. To that end, draw a sphere of radius $$R$$ with the point $$\vec{r}$$ as the centre. Griffiths claims that the macroscopic field at $$\vec{r}$$ is the sum of two parts: the average field over the sphere due to all charges outside, plus the average field due to all charges inside the sphere: $$\vec{E}=\vec{E}_{\rm out}+\vec{E}_{\rm in}$$.

However, I am not completely convinced this is the net macroscopic field at $$\vec{r}$$. For example, suppose the dielectric is polarized by putting it inside an external field $$\vec{E}_{\rm ext}$$. Then, I think, by the principle of superposition, the correct expression for the net macroscopic field at $$\vec{r}$$ will be the sum of three parts (not two): $$\vec{E}=\vec{E}_{\rm out}+\vec{E}_{\rm in}+\vec{E}_{\rm ext}.$$ Please tell me whether I am wrong and Griffiths is right.

You are both right, as @basics points out. Any external electrostatic field is produced by a charge density somewhere in space. Whether you want to divide that in to two parts ($$E_{ext} + E_{out}$$) as you do, or collect the two in one term (as Griffiths does) is only a matter of notation.

Dividing the outside field into two terms can make sense in a lot of practical situations, however fundamentally it is correct when Griffiths says that $$E_{out}$$ (using his definition) is the field generated by all charges outside the sphere

What is producing the component you're defining as $$\vec{E}_{ext}$$? Here Griffiths is dealing with electrostatics and, for what we know, namely Gauss' law for the electric field $$\nabla \cdot \mathbf{d} = \rho$$, we need charges to produce a (di)electric field.

• Let the external field be due to two oppositely charged metal plates, and let the chunk of the dielectric is kept in between the plates. Commented Jun 22 at 13:36
• If the plates are outside the sphere, It does go into the component $\vec{E}_{out}$, isn't it? Commented Jun 22 at 13:40
• I can hardly believe that an answer starting with "as @basics points out" and adding nothing got +1, while basics' answer got 0. LEL Commented Jun 30 at 8:05

You both are correct. The external field should be produced by some other charges whose influence are already included in the part $$\vec{E}_{out}$$. Because $$\vec{E}_{out}$$ is the field due to outside charges averaged over the volume of the imagined sphere.

External field may orient the tiny dipoles so now the charge distribution has been changed. Now you have to calculate the average fields for the new charge distribution (for outside and inside). Then you will be left with the macroscopic field for that point (for that region)