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Question
Differential Form of Maxwell's Equations: [...]
Answer
\(\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)

Question
Differential Form of Maxwell's Equations: [...]
Answer
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Question
Differential Form of Maxwell's Equations: [...]
Answer
\(\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)
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Differential Form of Maxwell's Equations
Differential Form of Maxwell's Equations: \(\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)

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