Solution

Given that L ≫ 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.

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

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

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

$For \ r < \rho < R :$

$Q_{\text{enc}} = q$

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

$For\ \rho > R :$

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

If the shield were isolated and uncharged, the enclosed charge would be q and the field would remain as

$E = \frac{q}{2\pi \varepsilon_0 L \rho} .$

$E = 0 \ for \ \rho > R .$