Question

An infinitely long cylindrical conducting shell of outer radius r₁ = 0.10 m and inner radius r₂ = 0.08 m initially carries a surface charge density σ = -0.25 μC/m². A thin wire, with linear charge density λ = 1.1 μC/m, is inserted along the shells' axis. The shell and the wire do not touch and there is no charge exchanged between them. What is the new surface charge density, in microcoulombs per square meter, on the inner surface of the cylindrical shell?

Answer 1

The new surface charge density on the inner surface of the cylindrical shell is -4.38 μC/m², which neutralizes the charge carried by the inserted wire.

Step-by-step explanation:

The question pertains to the concept of electric field and charge distribution in conductors, particularly a cylindrical conducting shell and a wire inserted along its axis. To determine the new surface charge density on the inner surface of the shell, one can apply Gauss's law. Before the wire is inserted, the conducting shell has a surface charge density σ. Inserting a thin wire with linear charge density λ does not change the total charge on the shell, but it does induce a new charge distribution.

For an infinitely long cylindrical conducting shell, the electric field inside the shell is zero. So, the charge on the inner surface must neutralize the charge carried by the wire. Since the wire carries a positive charge density, the inner surface must carry an equal and opposite charge.

Therefore, the new surface charge density on the inner surface σ' is equal in magnitude to the linear charge density λ of the wire but with the opposite sign when distributed over the inner surface area. This translates to σ' = -λ/(2πr2), where r2 is the inner radius of the shell. Plugging in values: σ' = -1.1 μC/m / (2π * 0.08 m) ≈ -4.38 μC/m2.