Solution

Given that L \gg R , we neglect edge effects in the middle region. The system behaves as infinitely long. Thus

$E= E(x)\hat{x}

The "cylindrical shield" of radius R acts as an outer conductor that shields the field.

The charge +q in the cavity induces a charge –q on the inner surface (ρ = r).

Since the conductor is neutral, a charge +q appears on its outer surface.

"We choose a cylindrical Gaussian surface of radius \rho and length L, coaxial with the system.

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}

E \cdot 2\pi \rho L = \frac{Q_{\text{enc}}}{\varepsilon_0}

E(\rho) = \frac{Q_{\text{enc}}}{2\pi \varepsilon_0 L \rho}
\]"

For r < \rho < R :

Q_{\text{enc}} = q.

E(\rho) = \frac{q}{2\pi \varepsilon_0 L \rho}

For \rho > R :

Q_{\text{enc}} = 0 \quad \rightarrow \quad E = 0.

If the shield were isolated and uncharged, the enclosed charge would be q and the field would remain as E = \frac{q}{2\pi \varepsilon_0 L \rho} .

E = 0 for \rho > R .