# Determining electric displacement using Gauss's law

This is an example problem from my E&M textbook that I don't quite understand:

A metal sphere of radius a carries a charge $Q$. It is surrounded, out to radius b, by linear dielectric material of permittivity $\epsilon$. Find the potential at the centre (relative to infinity).

Using $$\iint \mathbf{D}\cdot\ d\mathbf{A}=Q_{free}$$ I get $\mathbf{D}=\dfrac{Q}{4\pi r^2}\hat{\mathbf{r}}$. What I don't understand is that the author claims that this is only valid for $r>a$. Why is this the case? Couldn't I simply draw a gaussian surface for the inner surface to compute $\mathbf{D}$ in the same way? Why does this only work for the outer surface but not on the inside? I guess I'm having trouble visualising what $\mathbf{D}$ actually is since for the electric field, I can picture the 'flux' of the electric field lines when using Gauss's law but I'm lost on what I'm supposed to picture for electric displacement.

## 1 Answer

Inside the metal sphere, where $r<a$, there is no free charge so your Gaussian sphere will enclose $Q_{free}=0$. As a result, both $\vec E$ and $\vec D$ are $0$.

In addition but separately from this, your expression is problematic as it would produce $D\to\infty$ as $r\to 0$.

• Why is there no free charge? I thought it was spread all over the sphere? – Ayumu Kasugano Sep 9 '17 at 0:25
• There is no free inside a conductor in static equilibrium. All the charge is on the surface of the conductor. See math.union.edu/~barbanej/IMP13S02/PhysicsNotes/… – ZeroTheHero Sep 9 '17 at 0:30
• @Yuvraj sorry I don’t follow how something like $\log(\cos(x))/\sec(x)$ c0mes about. $Q$ here is fixed so $Q/r$ does not evaluate to $0/0$ so l’Hopital rule does not apply. – ZeroTheHero Aug 12 '20 at 12:14