A cylindrical capacitor is made up from two concentric cylinders ($R$ and $2R$). Inside the 2R cylinder there is a dielectric from 2R to 1.5R of constant $k=3$. The charges are Q on the R cylinder and -Q on the 2R cylinder.

a) Please find the electric field between r=0 and r=3R.

b) Please find the capacitance.

The condition for the electric field at interface between two mediums 1 and 2 is

$E1n*\epsilon1 =E2n*\epsilon2$ , for the normal component

(the normal polarization vector is constant)

$E1t = E2t =0$ , for the tangential component

(the tangential component of field is zero, otherwise the field will move existent virtual electric charges on the interface surface)

Consider the entire capacitor empty. Consider a cylindrical gaussian surface of radius r and length (cylinder height) L. The Gauss law is

$E*S =Q(inside)/\epsilon0$ (1)

In all cases $S = 2*\pi*r*L$

a)

For $0<r<R$

$Q(inside) =0$

From (1) we get

$E(r) =0$

b)

For $R <r<2R$

$Q(inside) = +Q$

From (1) we get

$E(r) = Q/(2\pi*r*L\epsilon0)$ (2)

The relation (2) in real case is true for $R<r<1.5R$

Outside the dielectric when $r=1.5 R$

$E(1.5*R) =Q/(3\pi*R*L*\epsilon0)$

For $1.5R<r<2R$ we have

$E(r) = 3*E'(r)$

$Q/(2*\pi*r*L*\epsilon0) = 3*E'(r)$

$E'(r) = Q/(6\pi*r L\epsilon0)$ value of the field inside the dielectric for $1.5R<r<2R$

Inside the dielectric when $r=1.5*R$

$E'(1.5*R) =Q/(9*\pi*R*L*\epsilon0)$

c) For $r> 2R$

$Q(inside) =0$

From (1) we get

$E(r) =0$

Now for the capacitance. For a cylindrical capacitance of inner radius $a$ and outer radius $b$ the capacitance is

$C =(2\pi\epsilon0*L)/ln(b/a)$

Because of the dielectric there are two cylindrical capacitor in series. One with radius R and 1.5R (and pemittivity $\epsilon0$) and the second with radius $1.5R$ and $2R$ (and permittivity $3*\epsilon0$)

$C1/L = (2\pi\epsilon0)/ln(1.5) =4.932*\pi*\epsilon0$

$C2/L =(2\pi*3\epsilon0) / ln(2/1.5) =20.856*\pi*\epsilon0$

$Ctot =C1*C2/(C1+C2)$

$Ctot/L=3.989*\pi\epsilon0 = 111 pF/m$