Given $I=I_{m} \sin \omega t .$ We see that
$$
j=\frac{I_{m}}{S} \sin \omega t=-j_{d}=-\frac{\partial D}{\partial t}
$$
or, $D=\frac{I_{m}}{\omega S} \cos \omega t$, so, $E_{m}=\frac{I_{m}}{\varepsilon_{0} \omega S}$ is the amplitude of the electric field and is $7 \mathrm{~V} / \mathrm{cm}$