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/ee/ - Electrical Engineering


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25 Dec 2021Mathchan is launched into public

2 / 2 / 2 / ?

File: james maxwell.jpg ( 15.97 KB , 300x300 , 1641415993367.jpg )

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Gauss's law

ΩEdS=1ε0ΩρdV\qquad \oiint_{\partial\Omega}\textbf{E}\cdot\textrm{d}\textbf{S} = \frac{1}{\varepsilon_0}\iiint_\Omega\rho\,\mathrm{d}V


Gauss's law for magnetism
ΩBdS=0\qquad \oiint_{\partial\Omega}\textbf{B}\cdot\textrm{d}\textbf{S} = 0


Faraday's law

ΣEd=ddtΣBdS\qquad \oint_{\partial\Sigma}\textbf{E}\cdot\textrm{d}\ell = -\frac{\textrm{d}}{\textrm{d}t}\iint_\Sigma\textbf{B}\cdot\textrm{d}\textbf{S}


Ampere's law

ΣBd=μ0(ΣJdS+ε0ddtΣEdS)\qquad \oint_{\partial\Sigma}\textbf{B}\cdot\textrm{d}\ell = \mu_0\left(\iint_\Sigma\textbf{J}\cdot\textrm{d}\textbf{S} + \varepsilon_0\frac{\textrm{d}}{\textrm{d}t}\iint_\Sigma\textbf{E}\cdot\textrm{d}\textbf{S}\right)
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In differential form:

E=ρε0B=0×E=Bt×B=μ0(J+ε0Et) \begin{aligned} \nabla\cdot\textbf{E} &= {\rho \over \varepsilon_0}\\ \nabla\cdot\textbf{B} &= 0\\ \nabla\times\textbf{E} &= -\frac{\partial\textbf{B}}{\partial t}\\ \nabla\times\textbf{B}&=\mu_0\left(\textbf{J} + \varepsilon_0\frac{\partial\textbf{E}}{\partial t}\right) \end{aligned}
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File: Maxwell.jpg ( 2.6 MB , 2252x2195 , 1722108563626.jpg )

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These are the gibbs-heaviside truncations. This is maxwell's