Answer
$
\frac{d E}{d t}=2.3 \times 10^{11} \frac{\mathrm{V}}{\mathrm{m} \cdot \mathrm{s}}
$
Work Step by Step
The rate of change of the electric field is
$i_{d}=2.0A$ From Part (1).$$$
\frac{d E}{d t}=\frac{1}{\varepsilon_0 A}\left|\mathrm{i} \varepsilon_0 \frac{d \Phi_E}{d t}\right|=\frac{i_d}{\varepsilon_0 A}\\=\frac{2.0 \mathrm{~A}}{\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m} \| 1.0 \mathrm{~m}\right]^2}=2.3 \times 10^{11} \frac{\mathrm{V}}{\mathrm{m} \cdot \mathrm{s}}
$$
