Question #93561
Using Maxwell’s equations in free space, derive the wave equation for the z-component of
the electric field vector.
1
Expert's answer
2019-09-03T12:22:56-0400

The Maxwell's equations in free space


×E=1cBt\nabla\times {\bf E}=-\frac{1}{c}\frac{\partial {\bf B}}{\partial t}×B=1cEt\nabla\times {\bf B}=\frac{1}{c}\frac{\partial {\bf E}}{\partial t}divE=0\rm div{\bf E}=0divB=0\rm div{\bf B}=0

From the second equation we obtain


×Bt=1c2Et2\nabla\times \frac{\partial {\bf B}}{\partial t}=\frac{1}{c}\frac{\partial^2 {\bf E}}{\partial t^2}

Plug first equation into last, we get


c×(×E)=1c2Et2-c\nabla\times (\nabla\times {\bf E})=\frac{1}{c}\frac{\partial^2 {\bf E}}{\partial t^2}

Since

×(×E)=grad(divE)2E=2E\nabla\times (\nabla\times {\bf E})={\rm grad(div {\bf E})}-\nabla^2 {\bf E} =-\nabla^2 {\bf E}

we finally obtain


2E1c22Et2=0\nabla^2 {\bf E}-\frac{1}{c^2}\frac{\partial^2 {\bf E}}{\partial t^2}=0

For the z-component of the electric field vector the wave equation


2Ez1c22Ezt2=0\nabla^2 {E_z}-\frac{1}{c^2}\frac{\partial^2 {E_z}}{\partial t^2}=0


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS