吴远清-2018300001031-电动力学第五次作业1
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电动力学-第五次作业
吴远清-2018300001031
2020 年 4 月 12 日
Problem 4.10
Answer:
(a)
σ
b
= P · ˆn = kR (1.1)
ρ
b
= −∇ · P = −
1
r
3
∂
∂r
(r
2
kr) = −3k (1.2)
(b)
For r < R:
E =
1
3ϵ
0
ρrˆr = −
k
ϵ
0
r (1.3)
For r > R, we can treat it as all charge at center:
Q
t
= (kR)(4πR
2
) + (−3k)(
4
3
πR
3
) = 0 (1.4)
Then:
E = 0 (1.5)
Problem 4.18
Answer:
(a):
For the surface, we have:
∫
D · da = Q (2.1)
Then:
DA = σA (2.2)
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