# Average field inside a sphere

Griffiths, problem 3.47: show that the average field inside a sphere of radius R, due to all the charge within the sphere, is $$\langle {\bf E} \rangle = - {1 \over 4 \pi \epsilon_0} {\bf p_{\tt tot}\over \rm R^3}$$

Here his solution (or at least what I’ve understood):

First of all, let us show that the average field due to a single charge q at a generic point $$\bf r$$ inside the sphere ($$\bf r$$ is the position vector of the charge from the centre of the sphere) is the same as the field evaluated at $$\bf r$$ due to a uniformly charged sphere with charge density $$\rho = \rm - {q \over {{4 \over 3} \pi R^3}}$$

The average field is by definition: $$\bf E_{\tt ave} \rm := { 1 \over {4 \over 3 }\pi R^3} \int_{\tau} \bf E_ \rm \space d\tau'$$ where $$\bf E \rm = {1 \over 4 \pi \epsilon_0} {q \over r'^2} \bf {\hat r}'$$ is the field perceived by the infinitesimal volume $$\rm d \tau'$$ at a distance $$\vec {\bf r }'$$ from charge q (actually, I suppose, the field is perceived as electric force per unit charge by any positive test charge I'd put in $$\rm d \tau'$$).

Now, the field at the same point $$\bf r$$ due to a uniform spherical distribution of charge $$\rho$$ is $$\bf E_{\rho} \rm = {1 \over 4 \pi \epsilon_0}\int_{\tau} {\rho \over r'^2}( -\hat {\bf {r}}' )\rm \space d\tau'$$ the minus sign is justified because this time the vector $$\vec {\bf r}'$$ is pointing in the opposite direction, from the infinitesimal amount of charge $$\rho \rm d \tau'$$ to the point at position $$\bf r$$. So if we have $$\rho = {- \rm q \over {4 \over 3} \pi R^3}$$ the two expressions agree: $$\bf E_{\tt ave} \rm = \bf E_{\rho}$$

Lastly, let us point out that $$\bf E_{\rho}$$ can be easily calculated using Gauss' law: $$\bf E_{\rho} \rm = {\rho \bf r \over \rm 3 \epsilon_0} = - { \rm q \over 4 \pi \epsilon_0} {\bf r \over \rm R^3} = - { \bf p \over \rm 4 \pi \epsilon_0 R^3}$$

If there are many charges inside the sphere, the previous equation still holds using superposition principle, with $$\langle \bf E \rangle$$ and $$\bf p_{\tt tot}$$ as sum of individual average fields and dipole moments respectively. QED

Here what I do not understand (for sure):

• Griffiths didn't specify if the sphere in question is made of a conductor or a dielectric. I assume the second one, otherwise the charge q would have been evenly spread over the spherical surface (as any charge excess inside a conductor would do) in such a way that (the only way actually, for the existence and uniqueness of the solution to Poisson's equations for any given boundary conditions) inside the conductor $$\rho = 0$$ precisely. If then I have a (say) positive charge q inside a dielectric I would expect a polarization; on the other hand, the dielectric itself has to stay neutral. So why are we considering JUST THE NEGATIVE CHARGE DENSITY? Where are the positive "components" of all the microscopic dipoles of the dielectric? Have they all disappeared just because I've put a positive charge inside it?

• Another explanation I gave myself is that this is a purely theoretical result, I should not see it in terms of polarization. Indeed, these formulas are used afterwards in the book to investigate polarization more deeply, so I really shouldn't explain a cause with its consequence. OK, but then why is "$$\rm q \space \bf r \rm = \bf p$$"? In this proof, we're ultimately saying that I can measure the same thing ($$\bf E_{\tt ave}$$ / $$\bf E_{\rho}$$) in two DISTINCT and DIFFERENT situations (right?), one in which I have a positive charge q at $$\bf r$$ and the other in which there is a uniform negative charge density $$\rho$$. But $$\bf p$$ makes sense only if I have SIMULTANEOUSLY two opposite charges +q and -q at a distance $$\bf r$$ each other.

I hope I've been clear...

Given that I solved this problem about a week ago in this other answer, I still have it more or less fresh in my mind, so I'll try to clarify some misconceptions.

1. The dipole moment $$\mathbf p$$ does not only make sense for the case of two charges $$+q$$ and $$-q$$ a distance $$\mathbf r$$ apart. The dipole moment is a more general quantity defined as

$$\mathbf p=\int\mathbf{r'}\rho(\mathbf r')d\tau'$$

and it depends on the origin we choose. In particular, the dipole moment for the case of two opposite charges a distance $$\mathbf r$$ apart (say, one at $$\mathbf r/2$$ and the other at $$-\mathbf r/2$$) with respect to their middle point is $$\mathbf p=q\mathbf r$$. Similarly, the $$\mathbf p$$ for the case of only one charge $$+q$$ at position $$\mathbf r$$ with respect to some origin is also $$\mathbf p=q\mathbf r$$.

2. The fields $$\mathbf E_{\rm ave}$$ and $$\mathbf E_{\rm \rho}$$ are conceptually NOT the same thing. They are operationally the same, but $$\mathbf E_{\rm \rho}$$ is something you can define pointwise if that volume density $$\rho$$ were really there; on the other hand, $$\mathbf E_{\rm ave}$$ is just an average of the field over a volume. It only makes sense when referring to the volume as a whole.

3. Regarding your first point, I don't see the problem with having a perfect conductor or a dielectric. Griffiths is going to apply this result later on to calculate the $$\mathbf E$$ field in a dielectric, but in the conductor case you propose where all the charge is at the surface, then $$\rho=0$$ inside the sphere and therefore $$\mathbf p_{\rm tot}$$ is also $$0$$. We get then that the average field vanishes, which is no contradiction.

4. Regarding the last part of your first bullet point, I think you are missing the point of the exercise by overthinking it. The problem makes no reference to a polarization vector, neither has Griffiths defined whatever that is at the moment he proposes the exercise. You only need to know the definition of average field and the definition of dipole moment, which is Griffiths' equation $$(3.98)$$.

5. The confusion about the minus sign is because the notation is misleading. In your question, you seem to be using $$r'$$ and $$\hat{\mathbf r}'$$ the way Griffiths uses the script $$r$$, which is short for $$|\mathbf r- \mathbf r'|$$ and $$(\mathbf r- \mathbf r')/|\mathbf r- \mathbf r'|$$ respectively. For example, the expression

$$\mathbf E(\mathbf r)=\frac{q}{4\pi\varepsilon_0}\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}$$

is the electric field at $$\mathbf r$$ due to a point charge $$q$$ at $$\mathbf r'$$. So when you calculate the average of this field over a sphere, $$\mathbf r'$$ is fixed, and the integration is over the unprimed coordinates, i.e. over $$d\tau$$, not $$d\tau'$$

\begin{align}\bf E_{\tt ave} \rm :&= { 1 \over {4 \over 3 }\pi R^3} \int_{\tau} \bf E(\bf r) \rm \space d\tau\\ &=\frac{1}{4\pi\varepsilon_0}\int_\tau \frac{q}{\frac{4}{3}\pi R^3}\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}d\tau\tag{1} \end{align}

Now, compare this last expression with the general form of Coulombs's law for a continuous charge distribution

$$\mathbf{E (r)}=\frac{1}{4\pi\varepsilon_0}\int_{\tau'}\rho(\mathbf r')\frac{\mathbf r-\mathbf r'}{|\mathbf r-\mathbf r'|^3}d\tau'\tag{*}$$

or equivalently 1

$$\mathbf{E (r')}=\frac{1}{4\pi\varepsilon_0}\int_{\tau}\rho(\mathbf r)\frac{\mathbf r'-\mathbf r}{|\mathbf r'-\mathbf r|^3}d\tau\tag{**}$$

It's easy to see then, that $$(1)$$ is a special case of $$(**)$$, obtained by setting

$$\rho(\mathbf r)=\frac{-q}{\frac{4}{3}\pi R^3},$$

so the minus sign comes naturally when we write carefully all the $$\mathbf{r}$$ and $$\mathbf{r'}$$ dependences. I'm sorry to say that I lack the kind of intuition to claim whether this is because we moved the origin to the point where we placed the charge $$q$$, or something else.

1 Make sure you see why $$(*)$$ and $$(**)$$ are the same. It's just renaming variables, interchange $$\mathbf{r\leftrightarrow r'}$$.

• Thank you very much for your answer. So, in this case, the minus sign does not refer to a negative distribution density, but only because I moved my origin, from the centre of the sphere to the point where there was the charge q, right? Nov 8 '20 at 10:45