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Mortidesperatslav楼主2022/8/24 14:05

∮lH⋅dι=∬SJ⋅dS+∬s∂D∂t⋅dS\oint_lH\cdot d\iota =\iint_S J\cdot dS+\iint_s \dfrac{\partial D}{\partial t}\cdot dS

∮lE⋅dι=−∬s∂B∂t⋅dS\oint_lE\cdot d\iota =- \iint_s \dfrac{\partial B}{\partial t}\cdot dS

∮sB⋅ds=0\oint_s B \cdot ds = 0

∮sD⋅dS=∭VρdV\oint_s D \cdot dS = \iiint_V \rho dV

∇×H=J+∂Dθt\nabla \times H =J + \dfrac{\partial D}{\theta t}

∇×E=−∂Bθt\nabla \times E =- \dfrac{\partial B}{\theta t}

∇⋅B=0\nabla \cdot B = 0

∇⋅D=ρ\nabla \cdot D = \rho


$\oint_lH\cdot d\iota =\iint_S J\cdot dS+\iint_s \dfrac{\partial D}{\partial t}\cdot dS$

$\oint_lE\cdot d\iota =- \iint_s \dfrac{\partial B}{\partial t}\cdot dS$

$\oint_s B \cdot ds = 0$

$\oint_s D \cdot dS = \iiint_V \rho dV$

$\nabla \times H =J + \dfrac{\partial D}{\theta t}$

$\nabla \times E =- \dfrac{\partial B}{\theta t}$

$\nabla \cdot B = 0$

$\nabla \cdot D = \rho$

KeTeX实在太复杂了,有更简单的写法吗,求教

2022/8/24 14:05
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