The conducting box in FIGURE EX24.26 has been given an excess negative charge. The surface density of excess electrons at the center of the top surface is $5.0\times {10}^{10}electrons/m^2$ . What are the electric field strengths $E_1$to $E_3$at points 1 to 3?

Given : The surface density at the center of the top surface is :$5.0\times {10}^{10}electrons/m^2$

Theory used :

The electric field inside a conductor is zero at all times when it is in electrostatic equilibrium. However, all surplus charges on the conductor accumulate on the outside surface, and as further charges are added, they spread out on the outer surface until they reach the electrostatic equilibrium points.

The electric field at the surface of a charged conductor is given by the equation role="math" localid="1649668425589" $E_{surface}=\frac{\eta }{{\epsilon }_0}$ (1)

where$\eta$ is the surface charge density, which is a physical parameter that relies on the conductor's form.

Because the excess charge is expressed in $electrons/m^2$, we must convert the surface charge density to$C/m^2$using

$$\begin{gathered} \eta =(5\times {10}^{10}electrons/m^2)\left(\frac{-1.6x{10}^{-19}C}{1electron}\right) \\ =\underline{-8x{10}^{-9}C/m^2} \end{gathered}$$

There is an electric field at point 1, thus we apply

$$\begin{gathered} E_1=E_{surface}=\frac{\eta }{{\epsilon }_0} \\ =\frac{(-8x{10}^{-9}C/m^2)}{(8.85x{10}^{-12}{C}^2/N{m}^2)} \\ =\boxed{\mathbf{-}\mathbf{904}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}} \end{gathered}$$

As previously stated, there is no electric field inside the wire, hence the electric field at point 2 is zero. That is $E_2=\boxed{\mathbf{0}}$.

Because all charges are concentrated at the outer surface at point 1, there are no net charges at point 3, and the electric field is zero. That is, $E_3=\boxed{\mathbf{0}}$.