About the electric field inside a capacitor If I have a metal sphere (a conductor) of radius $R$, carrying charge $Q_1$ with potential $V_1$, surrounded by a
thick concentric metal shell (a conductor with inner radius $a$, outer radius $b$, as in Figure), which carries, a charge $Q_2$ on is outer surface, and $Q_{2i}$ on is inner surface. The
shell has potential $V_2$. I would like to understand why the field inside depends only on $Q_1$ and not on $Q_2$ ? In other terms, why = 
$$V_1-V_2=\int_{1}^{2}\vec{E}\cdot \vec{dl}$$
where $E=\frac{Q_1}{4\pi\epsilon_0 r^2}$ where $E$ is the magnitude of $\vec{E}$. 
Any help would be appreciated,
 A: If you try to apply Gauss's law to a sphere concenteric with the shells and having a radius $r$ such that $c<r<a$(where $c$ is the radius of the charged sphere and $a$ is the inner radius of the charged shell), then you'd get,
$$\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q_1}{\epsilon_0} \tag{1} $$
You can also see that the field is radially symmetric due to the symmetry of the charge distribution and also it always points in the radial direction. So,
$$\mathbf{E} \cdot d\mathbf{S} = |\mathbf{E}| |d\mathbf{S}|cos(0)= |\mathbf{E}| |d\mathbf{S}|$$
Also since $|\mathbf{E}|$ is constant, so,
$$ \oint_S |\mathbf{E}||d\mathbf{S}|= |\mathbf{E}|\oint_S|d\mathbf{S}|= |\mathbf{E}|×4\pi r^2 $$
Therefore,
$$|\mathbf{E}|=\frac{Q_1}{4\pi\epsilon_0 r^2} \tag{2}$$
As you can clearly see that the magnitude and the direction of the electric field ($\mathbf{E}$) only depends on $Q_1$ and $r$. The direction of the electric field can reverse if the sign of the charge $Q_1$ is reversed. To be precise,
$$\mathbf{E}=\frac{Q_1}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}$$
where $\hat{\mathbf{r}}$ is the unit vector pointing radially outwards.
Caution :- The reason why I did not conclude that the electric field($\mathbf{E}$) is independent of other charges except $Q_1$ directly from $(1)$ is because it only suggests that $\int_S \mathbf{E} \cdot d\mathbf{S}$ is independent of $Q_1$. You cannot conclude anything more about the electric field only by using $(1)$. You need to invoke symmetry arguments to prove the independence of the electric field with respect to $Q_2$ and $Q_{2i}$.
A: Imagine that we first charge the metal shell up to a positive charge $Q_2'$ by drawing electrons from it. The remaining electrons (against the background of nuclei) will redistribute themselves by Coulomb forces such that a minimum magnitude of charge density (minimum energy) is achieved. This means that all the positive charges will be distributed on the exterior surface of the shell. The charge distribution inside will be zero everywhere. Next we charge up the inner sphere up to positive charge $Q_1$. Here in a similar fashion, the charges will redistribute themselves on the surface of the sphere to create the minimum magnitude of charge density (which happens when all charges are distributed on the outer surface of the sphere). The positive charges $Q_1$, by Coulomb forces will attract negative charges from the shell. As a result, a negative surface charge, equal in magnitude to $Q_1$ but opposite in sign will form on the inner surface of the shell. Thus $Q_{2i}=-Q_1$ and the charge on the outer surface of the shell will be $Q'_2+Q_1$. The outer surface of the inner sphere and the inner surface of the shell form a spherical capacitor whose field can be easily calculate. Thus, this field is only determined by $Q_1$.
We arrive at the same conclusion if we look at any point inside  at some radius $r$ greater than the radius but smaller than the inner radius of the shell we note that the total charge inside the sphere defined by $r$ is $Q_1$ by Gauss' Theorem this charge determines the field up to the inner radius of the shell. Inside the shell the field is of course $0$ since the totals charge enclosed in a similar Gaussian sphere is zero. 
A: Since electromagnetism obeys the superposition principle the problem can be decomposed. It is clear that the inner sphere contributes  $E_1=\frac{Q_1}{4\pi\epsilon_0 r^2}$ to the field in between the spheres. Because of superposition all you need to establish is the field of the empty sphere 2, without sphere 1 inside. By Gauss's law you know that the static field of a charged surface enclosing an empty volume is zero inside that volume. The end result is that only sphere 1 determines the field.
The physical reason is that electrostatic potential on a conductor is constant. Therefore no E field lines can connect any two points on the conductor as on the inside of the surface any field line must land back on the surface. Loops are impossible as the electrostatic potential is conservative. On the outside of the sphere a field line can extend to infinity. 
