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続・ビジュアル系高専生
- 1. ∇
·B
∇
·D =
∇ 0
× =
E ρ
∇ + ∂
× B
H ∂t
+ ∂ =
D 0
∂t
=
j
- 7. ∇
·B
∇
·D =
∇ 0
× =
E ρ
∇ + ∂
× B
H ∂t
+ ∂ =
D 0
∂t
=
j
- 25. ∇ · (φA) = (∇φ) · A + φ(∇ · A)
∇ · (A × B) = (∇ × A) · B + A · (∇ × B)
∇ · F dV = F · ndS
V ∂V