Question

Start from Eq. 33-11 and 33-17 and show that $\mathbf{E}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{t}\mathbf{)}$,and $\mathit{B}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{t}\mathbf{)}$,the electric and magnetic field components of a plane traveling electromagnetic wave, must satisfy the wave equations

$\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{E}}{\mathbf{\partial }{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}{\mathit{c}}^{\mathbf{2}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{E}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}$and$\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{B}}{\mathbf{\partial }{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}{\mathit{c}}^{\mathbf{2}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{B}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}$

Short answer

It is proved that the electric and magnetic field components of a plane traveling electromagnetic wave must satisfy the wave equations

$\frac{{\partial }^{2}E}{\partial t^2}=c^2\frac{{\partial }^{2}E}{\partial x^2}$ and $\frac{{\partial }^{2}B}{\partial t^2}=c^2\frac{{\partial }^{2}B}{\partial x^2}$

Step-by-step solution

Listing the given quantities

$E\left(x,t\right)$ and $B\left(x,t\right)$ are the electric and magnetic field components.

Understanding the concepts electric and magnetic field components

We have to take the derivative of equation 33-11 with respect to $\mathit{x}$ and the derivative of equation 33-17 with respect to $\mathit{t}$. By comparing these two equations, we can prove that the electric field satisfies the wave equation. We can use the same concept to prove that the magnetic field satisfies the wave equation.

Formula:

$\frac{\partial E}{\partial x}=-\frac{\partial B}{\partial t}$

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

$-\frac{\partial B}{\partial x}={\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}$

Show that  ∂2E∂t2=c2∂2E∂x2 and   ∂2B∂t2=c2∂2B∂x2

From equation 33-11 , we have

$\frac{\partial E}{\partial x}=-\frac{\partial B}{\partial t}$

Differentiating this with respect to $x$ we get

$\frac{\partial }{\partial x}\left(\frac{\partial E}{\partial x}\right)=\frac{\partial }{\partial x}\left(-\frac{\partial B}{\partial t}\right)$

$\frac{{\partial }^{2}E}{\partial x^2}=-\frac{{\partial }^{2}B}{\partial x\partial t}······\left(1\right)$

Now, from equation , we have

$-\frac{\partial B}{\partial x}={\mu }_0{\epsilon }_0\frac{\partial E}{\partial t}$

Differentiating this equation with respect to $t$, we get

$$\begin{gathered} \frac{\partial }{\partial t}\left(-\frac{\partial B}{\partial x}\right)={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}\left(\frac{\partial E}{\partial t}\right) \\ \left(-\frac{{\partial }^{2}B}{\partial t\partial x}\right)={\mu }_0{\epsilon }_0\left(\frac{{\partial }^{2}E}{\partial t^2}\right)······\left(2\right) \end{gathered}$$

Comparing (1) and (2) , we get

$$\begin{gathered} {\mu }_0{\epsilon }_0\frac{{\partial }^{2}E}{\partial t^2}=\frac{{\partial }^{2}E}{\partial x^2} \\ \frac{{\partial }^{2}E}{\partial t^2}=\frac{1}{{\mu }_0{\epsilon }_0}\frac{{\partial }^{2}E}{\partial x^2} \end{gathered}$$

But, $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$, so

$\frac{{\partial }^{2}E}{\partial t^2}=c^2\frac{{\partial }^{2}E}{\partial x^2}$

Thus, the electric field satisfies the wave equation.

Now, differentiating equation 33-11 with respect to $t$ we get

$$\begin{gathered} \frac{\partial }{\partial t}\left(\frac{\partial E}{\partial x}\right)=\frac{\partial }{\partial t}\left(-\frac{\partial B}{\partial t}\right) \\ \left(\frac{{\partial }^{2}E}{\partial t\partial x}\right)=\left(-\frac{{\partial }^{2}B}{\partial t^2}\right)······\left(3\right) \end{gathered}$$

Now, differentiating equation 33-17 with respect to $x$,

$$\begin{gathered} -\frac{\partial B}{\partial x}={\mu }_0{\epsilon }_0\frac{\partial E}{\partial t} \\ \frac{\partial }{\partial x}\left(-\frac{\partial B}{\partial x}\right)={\mu }_0{\epsilon }_0\frac{\partial }{\partial x}\left(\frac{\partial E}{\partial t}\right) \\ \left(-\frac{{\partial }^{2}B}{\partial x^2}\right)={\mu }_0{\epsilon }_0\left(\frac{{\partial }^{2}E}{\partial x\partial t}\right)·····\left(4\right) \end{gathered}$$

By comparing equations (3) and (4) , we get

$$\begin{gathered} \left(-\frac{{\partial }^{2}B}{\partial x^2}\right)={\mu }_0{\epsilon }_0\left(-\frac{{\partial }^{2}B}{\partial t^2}\right) \\ \frac{{\partial }^{2}B}{\partial x^2}={\mu }_0{\epsilon }_0\frac{{\partial }^{2}B}{\partial t^2} \\ \frac{{\partial }^{2}B}{\partial t^2}=\frac{1}{{\mu }_0{\epsilon }_0}\frac{{\partial }^{2}B}{\partial x^2} \end{gathered}$$

But, $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$, so

$\frac{{\partial }^{2}B}{\partial t^2}=c^2\frac{{\partial }^{2}B}{\partial x^2}$

Thus, the magnetic field satisfies the wave equation.