Differential Form of Maxwell's Equations in Vacuum: [...]
Answer
\(\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)
Question
Differential Form of Maxwell's Equations in Vacuum: [...]
Answer
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Question
Differential Form of Maxwell's Equations in Vacuum: [...]
Answer
\(\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)
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Differential Form of Maxwell's Equations in vacuum Differential Form of Maxwell's Equations in Vacuum: \(\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)
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