Electric field and boundary conditions

[KFC]

Homework Statement

One half of the region between the plates of a spherical capacitor of inner and outer radii a and b is filled with a linear isotropic dielectric of permittivity $$\epsilon_1$$ and the other half has permittivity $$\epsilon_2$$, as shown in the figure. If the inner plate has total charge +Q and the outer plate has total charge -Q, find the field everywhere in the sphere

http://img78.imageshack.us/img78/4120/bndal8.gif


2. The attempt at a solution
Well, this is an old problem and I know the solution of this problem. But it is quite confusing about the boundary condition and the form of the field.

1) Since all charges are located on outer and inner spherical surface, so there is no net charged found in the contact surface, which is illustrated with the normal direction in the figure. According to the boundary condition, we have

$$D_{2n} - D_{1n} = Q_{inc} = 0$$

and the electric field along the tangential direction always be continuous, that is

$$E_{2t} \equiv E_{1t}$$

I wonder what about the tangential direction of the displacement field? Are they continuous in general?

2) The solution manual tells that according to the boundary, it takes the form of the electric field as of radial-dependent

$$\vec{E} = \frac{A\vec{r}}{r^3}$$

I am really confuse about this: how can you tell the field is of this form by boundary condition? or how can you prove the field in this form satisfies all boundary conditions?

 

[Thaakisfox]

Due to the boundary conditions, the electric field will be radially the same in the sphere, hence due to Gausses law we have:

$$\epsilon_0\epsilon_1 E 2r^2\pi + \epsilon_0\epsilon_2 E 2r^2\pi =Q$$

So from here we obtain:

$$E=\dfrac{Q}{2\pi\epsilon_0(\epsilon_1+\epsilon_2)}\dfrac{1}{r^2}$$

In vector form (since there is only a radial component of the electric field):

$$\vec{E}=\dfrac{Q}{2\pi\epsilon_0(\epsilon_1+\epsilon_2)}\dfrac{\vec{r}}{r^3}$$

I hope its clearer noe