Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of \(\sigma_0\). The separation between any two consecutive sheets is \(1 \mu \mathrm{~m}\). The various regions between the sheets are denoted as \(1,2,3,4\) and 5. If \(\sigma_0=9 \mu \mathrm{C} / \mathrm{m}^2\), then which of the following statements is/are correct: (Take permittivity of free space \(\epsilon_0=9 \times 10^{-12} \mathrm{~F} / \mathrm{m}\)):

\(\begin{aligned}\left(\mathrm{E}_4\right)_{\mathrm{I}}= & \frac{\sigma_0}{2 \epsilon_0}[1-1+1-1-1+1]=0 \\ \left(\mathrm{~V}_{\text {First }}\right)_{\mathrm{I}} & =\frac{-\sigma_0}{2 \epsilon_0}[-1+2-3+4-5] \mathrm{d} \\ & =\frac{-\sigma_0}{2 \epsilon_0}[-3] \mathrm{d}=\frac{\sigma_0 3 \mathrm{~d}}{2 \epsilon_0}\end{aligned}\)
\(\left(V_{\text {Last }}\right)_I =\frac{-\sigma_0}{2 \epsilon_0}[1-2+3-4+5] \)
\( =\frac{\sigma_0}{2 \epsilon_0}[-3 \mathrm{~d}]\)
\(\left(V_{\text {First }}-V_{\text {Last }}\right)_I =\frac{3 \sigma_0 \mathrm{~d}}{\epsilon_0} \)
\( =\frac{3 \times 9 \times 10^{-6} \times 1 \times 10^{-6}}{9 \times 10^{-12}}=3 \mathrm{volt}\)

\(\left(\mathrm{E}_3\right)_{\text {II }}= \frac{\sigma_0}{2 \epsilon_0}\left[\frac{1}{2}-1+1+1-1+\frac{1}{2}\right]=\frac{\sigma_0}{2 \epsilon_0} \)
\( =\frac{-\sigma_0}{2 \epsilon_0}\left[-1+2-3+4-\frac{5}{2}\right] \mathrm{d} \)
\( =\frac{-\sigma_0}{2 \epsilon_0}[2-2.5] \mathrm{d}=\frac{\sigma_0 \mathrm{~d}}{4 \epsilon_0} \)
\( \left(\mathrm{V}_{\text {Last }}\right)_{\text {II }} =\frac{-\sigma_0}{2 \epsilon_0}\left[1-2+3-4+\frac{5}{2}\right] \mathrm{d} \)
\( =\frac{-\sigma_0}{2 \epsilon_0}[6.5-6] \mathrm{d}=\frac{-\sigma_0 \mathrm{~d}}{4 \epsilon_0} \)
\( \left(\mathrm{~V}_{\text {First }}-\mathrm{V}_{\text {Last }}\right)_{\text {II }}=\frac{\sigma_0 \mathrm{~d}}{2 \epsilon_0} \neq 0\)