Question

In Fig. 23-44, two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge densities of opposite signs and magnitude$\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{22}}\mathbf{C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. In unit-vector notation, what is the electric field at points (a) to the left of the plates, (b) to the right of them, and (c) between them?

Short answer

a) The electric field at points to the left of the plates is 0 N/C .

b) The electric field at points to the right of the plates is 0 N/C .

c) The electric field at points between the plates is-$\left(7.91\times {10}^{-11}\mathrm{N}/\mathrm{C}\right)\stackrel{⏜}{\mathrm{i}}$ .

Step-by-step solution

The given data

The inner faces of the plates have the magnitude of surface densities $\mathrm{\sigma }=7.{010}^{-22}\mathrm{C}/{\mathrm{m}}^2$with opposite signs.

Understanding the concept of Gauss law-planar symmetry

Using the concept of Gauss law and the planar symmetry, we can get the required values of the electric field at the left of the plates, right of the plates, and between the plates.

Formula:

The electric field of a non-conducting sheet,$\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_0}$ (1)

a) Calculation of the electric field at the left of the plates

To the left of the plates, the net electric field using equation (1) is given as follows:

$\mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(\stackrel{⏜}{\mathrm{i}}\right)\left(\mathrm{from}\mathrm{the}\mathrm{right}\mathrm{plate}\right)+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(+\stackrel{⏜}{\mathrm{i}}\right)\left(\mathrm{from}\mathrm{the}\mathrm{left}\mathrm{plate}\right)$

Hence, the value of the electric field is 0 N/C .

b) Calculation of the electric field at the right of the plates

To the right of the plates, the net electric field using equation (1) is given as follows:

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(-\stackrel{⏜}{\mathrm{i}}\right)\left(\mathrm{from}\mathrm{the}\mathrm{right}\mathrm{plate}\right)+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(+\stackrel{⏜}{\mathrm{i}}\right)\left(\mathrm{from}\mathrm{the}\mathrm{left}\mathrm{plate}\right) \\ =0\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is 0 N/C .

c) Calculation of the electric field between the plates

Between the plates, the net electric from both the plates is given using equation (1) as follows:

$$\begin{gathered} \overrightarrow{\mathrm{E}}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(\stackrel{⏜}{\mathrm{i}}\right)+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left(-\stackrel{⏜}{\mathrm{i}}\right) \\ =\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0}\left(-\stackrel{⏜}{\mathrm{i}}\right) \\ =\left(\frac{7.0\times {10}^{-22}\mathrm{C}/{\mathrm{m}}^2}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2}\right)\stackrel{⏜}{\mathrm{i}} \\ =-\left(7.91\times {10}^{-11}\mathrm{N}/\mathrm{C}\right)\stackrel{⏜}{\mathrm{i}} \end{gathered}$$

Hence, the value of the electric field is$-\left(7.91\times {10}^{-11}\mathrm{N}/\mathrm{C}\right)\stackrel{⏜}{\mathrm{i}}$ .