Question

Figure 23-47 shows cross-sections through two large, parallel, non-conducting sheets with identical distributions of positive charge with surface charge density$\mathbf{\sigma }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{77}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. In unit-vector notation, what is the electric field at points (a) above the sheets, (b) between them, and (c) below them?

Short answer

a) The electric field at points above the sheets is $2.00\times {10}^{-11}\mathrm{N}/\mathrm{C}$.

b) The electric field at points between the sheets is 0 N/C .

c) The electric field at points below the sheets is $2.00\times {10}^{-11}\mathrm{N}/\mathrm{C}$.

Step-by-step solution

The given data

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

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

Understanding the concept of Gauss law

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, $\mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}$ (1)

a) Calculation of the electric field above the sheets

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

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ =\frac{1.77\times {10}^{-22}\mathrm{C}/{\mathrm{m}}^2}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2} \\ =2.00\times {10}^{-11}\mathrm{N}/\mathrm{C} \end{gathered}$$

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

b) Calculation of the electric field between the sheets

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

c) Calculation of the electric field below the sheets

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

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ =\frac{1.77\times {10}^{-22}\mathrm{C}/{\mathrm{m}}^2}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2} \\ =2.00\times {10}^{-11}\mathrm{N}/\mathrm{C} \end{gathered}$$

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