Maxwell's Equations: Difference between revisions

Revision as of 18:20, 8 March 2020

Joe Schmoe (b. xxxx)

Title xxxx

$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}$$

$$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$$

$$\nabla \cdot \mathbf{B} = 0$$

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

In the example of an ideal vacuum with no charge or current, (i.e., $$\rho=0$$ and $$J=0$$), these equations reduce to:

$$\nabla \times \mathbf{B} = +\frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}$$

$$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$$

$$\nabla \cdot \mathbf{B} = 0$$

$$\nabla \cdot \mathbf{E} = 0$$

@@ Line 10: / Line 10: @@
 : $$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$
-This formulation assumes no charge $$\rho=0$$ and $$J=0$$. One common example of these conditions is a vacuum.
+In the example of an ideal vacuum with no charge or current, (i.e., $$\rho=0$$ and $$J=0$$), these equations reduce to:
 : $$\nabla \times \mathbf{B} = +\frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}$$