Mastering Physics: Exercise 22.21

Mastering Physics: Exercise 22.21

A hollow, conducting sphere with an outer radius of 0.251 m and an inner radius of 0.201 m has a uniform surface charge density of +6.55×10^(−6) C/m^2 . A charge of -0.640 μC is now introduced into the cavity inside the sphere.

Part A

Question: What is the new charge density on the outside of the sphere?

Answer: 
σ = Q / A
6.55 × 10^(−6) = Q / (0.251)^2 × 4π

=> Q = 5.1856 ×10^(−6)

New Q = 5.1856 ×10^(−6)  − 0.64 ×10^(−6)
= 4.5456 ×10^(−6)

=> σ = Q / A
σ = 4.5456 ×10^(−6) / (0.251)^2 × 4π
σ =  5.74×10^(−6)  C/m^2  

Part B

Question: Calculate the strength of the electric field just outside the sphere.

Answer: 

E⋅A = Q_enclosed / (ϵ_0)
E⋅ 4π(r^2) = 4.5456 ×10^(−6) / (8.85 ×10^(−12)
E =  6.49×105  N/C   

Part C

Question: What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Answer: 
Φ = q_enclosed / (ϵ_0)
Φ = −0.64 ×10^(−6) / (8.85 ×10^(−12)
Φ =  −7.23×10^4  N⋅m^2/C