Home exercise 3.3.9: Spherical capacitor

Problem 3.3.3 with $V_a=V_0$ and $V_b=0$ gives

and the charge on the inner conductor follows from ${\rho }_S=D_r(a)={\epsilon }_0{E}_r(a)$ times its area $4\pi a^2$:

By Gauss's law the field between the spheres is that of the charge $Q$ of the inner conductor,

Since $V_b=0$, the outer sphere is the reference point, and by Eq. (3.3.12)

At $r=a$ we have $V(a)=V_0$, which gives

the same as before. Substituting the values,

Both routes give the same $Q$. The values $V(0.1)=1800\times 5=9000\mathrm{V}$ and $V(0.2)=0$ reproduce the boundary values, and $4\pi {\epsilon }_{0}ab{V}_0/(b-a)$ has the unit $(F/m)\mathrm{m}\mathrm{V}=\mathrm{C}$.