Question

A hollow, conducting sphere with an outer radius of 0.240 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37 × 10−6 C/m2. A charge of -0.400 μC is now introduced into the cavity inside the sphere.

What is the new charge density on the outside of the sphere? Express your answer with the appropriate units.

Calculate the strength of the electric field just outside the sphere. Express your answer with the appropriate units.

What is the electric flux through a spherical surface just inside the inner surface of the sphere? Express your answer with the appropriate units.

Answer 1

Outer radius of the hollow sphere, R1 = 0.240 m

inner radius of the hollow sphere, R2 = 0.200 m

uniform surface charge density of the sphere, σ = 6.37 x 10^-6 C/m²

charge placed into a cavity inside the sphere, q = 0.400 μC = 0.400 x 10^-6 C

* A charge of q = 0.400 μC inside the sphere induces an equal quantity of positive charge on the outer surface and an equal quantity of negative charge on the outer surface due to electrostatic induction.

The surface charge density due to the induced charge on the outer surface, σ1 = q/4πR1²

σ1 = 9 x 10⁹ x 0.400 x 10^-6 /0.240² = 62.5 x 10³ C/m²

the total surface charge density of the sphere,

σ_net = σ + σ1 = 6.37 x 10^-6 + 62.5 x 10³ = 62.5 x 10³ C/m²

the total surface charge density of the sphere, σ_net = 62.5 x 10³ C/m²

electric field just outside the sphere, E = σ_net /ε₀ = 62.5 x 10³ /8.854 x 10^-12 = 7.059 x 10¹⁵ N/C

electric field just outside the sphere, E = 7.059 x 10¹⁵ N/C

the electric flux through the a spherical surface just inside the inner surface is ɸ = q /ε₀ (Gauss’s law)

ɸ = 0.400 x 10^-6 /8.854 x 10^-12 = 4.517 x 10⁴ Vm

the electric flux through the a spherical surface just inside the inner surface is ɸ = 4.517 x 10⁴ Vm