Question

电荷以相同的面密度分布在半径为R1=10cm和R2=20cm的两个同心球面上,设无限远处电荷为零,球心处的电势为U=30

Answer 1

⑴求电荷面密度.

4π R1^2*σ / (4πε0 R1) + 4π R2^2*σ / (4πε0 R2) = 300

σ = 300 ε0 / (R1+R2) = 8.85*10^(-9) C

⑵若要使球心处的电势也为零,外球面上应放掉多少电荷?

4π R1^2*σ / (4πε0 R1) + 4π R2^2*σ2 / (4πε0 R2) = 0

σ2 = - 4.425*10^(-9) C

外球面上应放掉的电荷

(σ - σ2)4π R2^2 = 6.67*10^-9 C