Question

A sphere of radius 1=0.280 m and uniform charge density −43.5 μC/m³ lies at the center of a neutral, spherical, conducting shell of inner and outer radii 2=0.558 m and 3=0.685 m, respectively. Find the surface charge density on the inner and outer surfaces of the shell.

a) +13.4 μC/m^2 ,−13.4 μC/m^2
b) −13.4 μC/m^2 ,+13.4 μC/m^2
c) +6.7 μC/m^2 ,−6.7 μC/m^2
d) −6.7 μC/m^2,+6.7 μC/m^2

Answer 1

To find the surface charge density on both the inner and outer surfaces of the conductive shell, evaluate the total charge on the sphere, which equals the charge density times its volume, and then calculate the corresponding charge on the shell's surfaces, taking into account the shell's neutrality and the surface areas.

Step-by-step explanation:

The student's question involves finding the surface charge density on the inner and outer surfaces of a conductive shell that contains a uniformly charged sphere. By using the properties of conductors in electrostatic equilibrium, we know that the total charge on the inner surface of the shell must equal the charge of the sphere but with opposite sign to neutralize the electric field inside the conductor. The outer surface must carry the same charge to keep the shell neutral overall. Given the radius of the sphere (0.280 m) and the charge density (-43.5 μC/m³), the total charge on the sphere is the charge density times the volume of the sphere. Since the shell must neutralize this charge internally, the inner surface charge density will be the opposite of this total charge divided by the surface area of the inner surface of the shell (4πr22). The outer surface charge density will be the same as the inner surface but opposite in sign since the shell is neutral overall.