Answer
$E_z = (1200~V/m)~sin~[(6.67\times 10^6~m^{-1})~y+(2.00\times 10^{15}~s^{-1})~t]$
Work Step by Step
We can find the magnitude of the electric field:
$E = B~c$
$E = (4.00\times 10^{-6}~T)(3.0\times 10^8~m/s)$
$E = 1200~V/m$
We can find $k$:
$k = \frac{\omega}{c}$
$k = \frac{2.00\times 10^{15}~s^{-1}}{3.0\times 10^8~m/s}$
$k = 6.67\times 10^6~m^{-1}$
We can write an expression for the electric field:
$E_z = (1200~V/m)~sin~[(6.67\times 10^6~m^{-1})~y+(2.00\times 10^{15}~s^{-1})~t]$
