91Ó°ÊÓ

a) There are two parallel non-conducting sheets with identical charges.

b) The surface density,$\mathrm{σ}=1.77×{10}^{-22}\mathrm{C}/{\mathrm{m}}^2$

Using the concept of Gauss law-planar symmetry, we can get the value of the electric field between, above, and below the non-conducting sheets.

Formula:

The electric field of a non-conducting sheet, $\mathrm{E}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$ (1)

Using the superposition principle, the electric field present at points above the sheets is given using equation (1) as follows:

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0} \\ =\frac{1.77×{10}^{-22}\mathrm{C}/{\mathrm{m}}^2}{8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2} \\ =2.00×{10}^{-11}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is$2.00×{10}^{-11}\mathrm{N}/\mathrm{C}$ , which is pointing in the upward direction.

According to the concept, the charge is present on the sheets, hence, the electric field using equation (1) is 0 N/C .

Using the superposition principle, the electric field present at points above the sheets is given using equation (1) as follows:

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0} \\ =\frac{1.77×{10}^{-22}\mathrm{C}/{\mathrm{m}}^2}{8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2} \\ =2.00×{10}^{-11}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is$2.00×{10}^{-11}\mathrm{N}/\mathrm{C}$ , which is pointing in a downward direction.