Statement
12.1.3. The figure illustrates the electric field of a plane sinusoidal wave at time $t = 0$ . The arrow indicates the direction of wave propagation. How does the electric field intensity depend on the coordinate $z$ and time $t$ ?
Solution
For a plane sinusoidal wave, the general equation of the electromagnetic wave is given by:
\begin{equation} E(z,t)=E_0\sin(At+Bz+\varphi_0) \end{equation}
where $A$ ,$B$ and $\varphi_0$ are constants.
From the condition of the problem we know that:
\begin{equation} E(z,0)=E_0\sin(\frac{2\pi}{\lambda}z) \end{equation}
Using eq(1) we get:
\begin{equation} E_0\sin{(Bz+\varphi_0)}=E_0\sin(\frac{2\pi}{\lambda}z) \end{equation}
By comparing the phases we obtain that:
\begin{equation} B=\frac{2\pi}{\lambda};\varphi_0=0 \end{equation}
As the wave travels in the positive direction:
\begin{equation} \frac{dz}{dt}=c \end{equation}
Setting the phase($\varphi=At+Bz+\varphi_0$ ) to a constant value allows us to derive the phase velocity($c$ ):
\begin{equation} A+B\frac{dz}{dt}=\frac{d\varphi}{dt}=0 \end{equation}
So we get the constant $A$ :
\begin{equation} A=-\frac{2\pi c}{\lambda} \end{equation}
Plugging all the constants into the eq(1) gives us:
\begin{equation} E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct)) \end{equation}
Answer
$E(z,t)=E_0\sin(\frac{2\pi}{\lambda}(z-ct))$