Three, large thin and charged sheets are placed parallel to each other. Surface I has a charge density of $6.5nC/m^2$, surface II has a charge density $-2.0nC/m^2$ and surface III has a charge density of $5.0nC/m^2$. Then, 

Magnitude of the electric field due to each sheet is given by, $E=\frac{\sigma }{2{ϵ}_0}$.

Due to positive charge it is away from the sheet and due to negative charge, it is towards the sheet.

$$\begin{gathered} {\left(\frac{F}{A}\right)}_{\mid }={\sigma }_1\left(E_{\Vert |}-E_{\Vert }\right)=\frac{{\sigma }_1}{2{\epsilon }_0}\left(\left|{\sigma }_{\Vert |}\right|-\left|{\sigma }_{\Vert }\right|\right) \\ =\frac{6.5\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2}{2\left(8.55\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)}\times \left[(5.0-2.0)\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2\right] \\ =1.1\times {10}^{-6} \mathrm{N}/{\mathrm{m}}^2\left(\mathrm{up}\right) \end{gathered}$$

$$\begin{gathered} {\left(\frac{F}{A}\right)}_{\Vert }={\sigma }_{\Vert }\left(E_{\Vert |}-E_1\right)=\frac{{\sigma }_{\Vert }}{2{\epsilon }_0}\left(\left|{\sigma }_{\Vert |}\right|-\left|{\sigma }_1\right|\right) \\ =\frac{-2.0\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2}{2\left(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)}\times \left[(5.0-6.5)\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2\right] \\ =1.7\times {10}^{-7} \mathrm{N}/{\mathrm{m}}^2\left(\mathrm{up}\right) \end{gathered}$$

${\left(\frac{F}{A}\right)}_{\text{III }}={\sigma }_{\text{III }}\left(E_{\text{III}}-E_1\right)=\frac{{\sigma }_{\text{III }}}{2{\epsilon }_0}\left(\left|{\sigma }_{\Vert }\right|-\left|{\sigma }_1\right|\right)$ 

           $=\frac{5.0\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2}{2\left(8.55\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)}\times \left[\right(2.0-6.5)\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2]$

            $=-1.3\times {10}^{-6} \mathrm{N}/{\mathrm{m}}^2\text{ (down) }$