Question

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

Answer 1

Answer on Question #85321- Physics / Electromagnetism

Question: Using Maxwell's equation in free space, derive the wave equations for the $z$-component of electric field vector

Answer:

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 in (1), 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 $\partial/\partial t$ 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:

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

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