Answer to Question #93518 in Electricity and Magnetism for EE

Question

Using Maxwell’s equations in free space, derive the wave equation for the z-component of
the electric field vector.

Answer 1

The Maxwell's equations in free space have the following form (Gauss units are used):

$$\nabla \cdot \vec{E} = \rho,\\ \nabla \cdot \vec{H} = 0, \\ \nabla \times \vec{E} = - \frac{1}{c} \frac{\partial \vec{H}}{\partial t},\\ \nabla \times \vec{H} = \frac{1}{c} \frac{\partial \vec{E}}{\partial t} + \frac{4 \pi}{c} \vec{j}.$$

Calculating the curl from the left side of the third equation above, one can derive:

$$\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \Delta \vec{E} = \nabla \rho - \Delta \vec{E}.$$

Taking into account that operators $\nabla$ and $\frac{\partial}{\partial t}$ commute, one can derive the corresponding expression after taking the curl from the right side of the same equation:

$$\nabla \times \left( - \frac{1}{c} \frac{\partial \vec{H}}{\partial t} \right) = - \frac{1}{c} \frac{\partial}{\partial t} \left( \frac{1}{c} \frac{\partial \vec{E}}{\partial t} + \frac{4 \pi}{c} \vec{j} \right) = - \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} - \frac{4 \pi}{c^2} \frac{\partial \vec{j}}{\partial t}$$

Combining the expressions together, we obtain:

$$\Delta \vec{E} - \frac{1}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2} = \nabla \rho + \frac{4 \pi}{c^2} \frac{\partial \vec{j}}{\partial t}.$$

Finally, projecting this equation on the z-axis, one can derive:

$$\Delta E_z - \frac{1}{c^2} \frac{\partial^2 E_z}{\partial t^2} = \frac{\partial \rho}{\partial z} + \frac{4 \pi}{c^2} \frac{\partial j_z}{\partial t}.$$