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 $$1 \, \mu$$m. The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If $$\sigma_0 = 9 \, \mu$$C/m$$^2$$, then which of the following statements is/are correct:

(Take permittivity of free space $$\epsilon_0 = 9 \times 10^{-12}$$ F/m)

Using field contribution per sheet: $$E_i = \frac{\sigma_i}{2\varepsilon_0}$$

In Configuration I, summing fields from left to right:

$$E_{\text{region 4}} = \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} - \left(\frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0}\right) = 0$$

Finding fields in all regions of Configuration I:

$$E_1 = \frac{\sigma_0}{\varepsilon_0},\ E_2 = 0,\ E_3 = \frac{\sigma_0}{\varepsilon_0},\ E_4 = 0,\ E_5 = \frac{\sigma_0}{\varepsilon_0}$$

Evaluating potential difference for Configuration I with $$d = 1\ \mu\text{m}$$:

$$\Delta V_{\text{I}} = E_1 d + E_2 d + E_3 d + E_4 d + E_5 d = \frac{\sigma_0}{\varepsilon_0}d + 0 + \frac{\sigma_0}{\varepsilon_0}d + 0 + \frac{\sigma_0}{\varepsilon_0}d = \frac{3\sigma_0 d}{\varepsilon_0}$$

$$\Delta V_{\text{I}} = \frac{3 \times (9 \times 10^{-6}) \times 10^{-6}}{9 \times 10^{-12}} = 3\text{ V}$$

In Configuration II, summing fields from left to right in region 3:

$$E_{\text{region 3}} = \frac{\sigma_0/2}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \left(-\frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0/2}{2\varepsilon_0}\right) = \frac{\sigma_0}{2\varepsilon_0}$$

Finding fields in all regions of Configuration II:

$$E_1 = \frac{\sigma_0}{2\varepsilon_0},\ E_2 = -\frac{\sigma_0}{2\varepsilon_0},\ E_3 = \frac{\sigma_0}{2\varepsilon_0},\ E_4 = -\frac{\sigma_0}{2\varepsilon_0},\ E_5 = \frac{\sigma_0}{2\varepsilon_0}$$

Evaluating potential difference for Configuration II:

$$\Delta V_{\text{II}} = \left(\frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0}\right)d = \frac{\sigma_0 d}{2\varepsilon_0} \neq 0$$