Concentric charged spheres (PH 310, Homework 3)

Two concentric conducting spheres of inner and outer radii $a$ and $b$, respectively carry charges $+/-Q$. The empty space between the spheres is  filled by a spherical shell of dielectric (of dielectric constant $\epsilon$). determine the potential and electric field in the three regions.

We know that inside a conducting sphere the electric field is zero and the potential is constant. So that for $r<a$ (region 1) we have $E(r)=0$  and
$V(r)=C (=constant)$.

Take a Gaussian surface of radius $a<r<b$ (radius between the two spheres). From the symmetric geometry of the problem one has only radial electric
field (directed along the radius). Gauss law is

$E(r)*S(r)=\frac{Q_{inside}}{\epsilon}$   but  $S(r)=4\pi r^2$

so that
$E(r)=\frac{+Q}{4\pi\epsilon r^2}$

The potential in this region ($a<r<b$) is
$V(r) =- \int_{r}^{a}E(r)d r= \frac{Q}{4\pi\epsilon}\int_{a}^{r}\frac{d r}{r^2}=\frac{Q}{4\pi\epsilon}\left(\frac{1}{r}-\frac{1}{a} \right )$

In the 3rd region where $r>b$ (outside both spheres)  the Gauss law is
$E(r)*S(r) =0$ since the net charge inside is $Q_{inside} =+Q-Q=0$

Therefore in the 3rd region
$E(r) =0$  and $V(r) =V(b) =\frac{Q}{4\pi\epsilon}\left(\frac{1}{b}-\frac{1}{a} \right )$

For region 1, where $r<a$ from the continuity conditions at $r=a$ one gets
$C =V(a)= \frac {+Q}{4\pi\epsilon}\left(\frac{1}{a}-\frac{1}{a} \right )=0$

valentin68