The electric field in a coaxial cylindrical capacitor

Question

I have a capcitor which has the shape of two coaxial cylinders, I'm asked to find the electric field in every point of space, I used Gauss law to determine for different radius the electric field :

I used a cylindrical gauss surface with a variable radius $r$ , for $r \lt r_1$ we have that cince the first cylinder is a conductor at equilibrum then $E = 0$ , then for $ r_1<r<r_2$ , the charges are on the surface of the first cylinder, thus creating an electric field, then for :

$r_2<r<r_3$ my professor told us it's zero, I know it is but I don't know which of these 2 explanations works in this case :

• The fact that $A1$ is charged induce charges on the interior suface of $A_2$ , this the sum of the charges cancels out to zero so by gauss law $ ES = Q_{int}/\epsilon $ and $Q_int = Q_{A1} + Q_{A_2ins} = 0$

• Or because it's inside a second conductor which is at equilibrum .

And for $r_3<r$ I don't know how the field would be ? and why ?

We suppose that at first $A_1$ is uniformely charged on it's surface and that between both cylinders there's air .

Answer 1

The charge of $A_1$ does induce charge on $A_2$, but a priori there is no reason why that charge should be on the interior surface of the second cylinder. The reason why this happens is exactly your second explanation: if a conductor is in eletrostatic equilibrium, then $\vec E = 0$ in its interior and all of its charge is concentrated on its surface (otherwise, there would be points inside the conductor in which $\vec E \neq 0$). Conversely, creating a cylindrical Gaussian surface with radius $r$, $r_2 < r < r_3$, you can conclude that the charge on the interior surface of $A_2$ is equal to $-Q_{A_1}.$

Answer 2

firstly the E inside the conductor will be zero and it can be proven by gauss law as the outer surface of A1 will Induce equal and opposite charge on the inner surface of A2 and due to symmetry the charge distribution will be uniform. Now outside the capacitor also the E = 0 and this can also be seen by gauss law as if total charge on a capacitor is Q , charge cylindrical plates are + Q/2 and -Q/2 . Therefore Qenclosed is 0. Or you can visualise these cylinders as infinitely long planes by cutting them . Thus E outside the planes of opposite charge = 0 as each plane produces an E of its own which is Q/2epsilon not ( take care of the sign of charges)