$\text{Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure,}$

$\text{the sheets carry uniform surface charge densities which are indicated in terms of}{\sigma }_0.\text{The separation between any}$

$\text{two consecutive sheets is}\mathrm{ 1}\mu \text{m.}\text{The various regions between the sheets are denoted as}\mathrm{ 1},2,3,4\text{and}5.$

$\text{If}{\mathrm{ \sigma }}_0=9\mu \frac{C}{m^2},\text{then which of the following statements is/are correct:}(\text{Take permittivity of free space}{\epsilon }_0=9\times {10}^{-12}\frac{F}{m})$

The correct option is:
In region 4 of the configuration I, the magnitude of the electric field is zero.

1. Understanding the Electric Field due to an Infinite Sheet
An infinitely large, thin non-conducting sheet carrying a uniform surface charge density $\sigma$ produces a uniform electric field on both sides. The magnitude of this electric field is given by:

$E=\frac{\sigma }{2{\epsilon }_0}$

The direction of the electric field is directed away from the sheet if the charge is positive ($\sigma >0$), and towards the sheet if the charge is negative ($\sigma <0$).

Let us define the unit vector pointing to the right as $\hat{i}$.
For a system of parallel sheets, the net electric field in any region is the vector sum of the electric fields produced by each individual sheet. For a region located relative to the sheets, the net electric field is:

$\overrightarrow{E}=\frac{1}{2{\epsilon }_0}\left[\sum {\sigma }_{\text{left}}-\sum {\sigma }_{\text{right}}\right]\hat{i}$

where $\sum {\sigma }_{\text{left}}$ is the sum of the charge densities of all sheets situated to the left of the region, and $\sum {\sigma }_{\text{right}}$ is the sum of the charge densities of all sheets situated to the right of the region.

2. Analysis of Configuration I
In Configuration I, the six sheets have the following surface charge densities from left to right:
- Sheet 1: ${\sigma }_1=+{\sigma }_0$
- Sheet 2: ${\sigma }_2=-{\sigma }_0$
- Sheet 3: ${\sigma }_3=+{\sigma }_0$
- Sheet 4: ${\sigma }_4=-{\sigma }_0$
- Sheet 5: ${\sigma }_5=+{\sigma }_0$
- Sheet 6: ${\sigma }_6=-{\sigma }_0$

Let us determine the electric field in Region 4 (which lies between Sheet 4 and Sheet 5):
- The sheets to the left of Region 4 are Sheets 1, 2, 3, and 4. The sum of their charge densities is:

$\sum {\sigma }_{\text{left}}={\sigma }_1+{\sigma }_2+{\sigma }_3+{\sigma }_4={\sigma }_0-{\sigma }_0+{\sigma }_0-{\sigma }_0=0$

- The sheets to the right of Region 4 are Sheets 5 and 6. The sum of their charge densities is:

$\sum {\sigma }_{\text{right}}={\sigma }_5+{\sigma }_6={\sigma }_0-{\sigma }_0=0$

Substituting these values into the net electric field formula for Region 4:

$E_4=\frac{1}{2{\epsilon }_0}\left[0-0\right]=0$

Therefore, the magnitude of the electric field in Region 4 of Configuration I is indeed zero.