91Ó°ÊÓ

The total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity, as per Gauss Law. The electric flux in a specified locality is calculated by adding the electric field by area of the surface projected in a plane perpendicular to the field.

a) We can begin with Gauss's law for areas as a beginning point. $1$and $5$.We can see a left Gaussian cylinder in the picture and write Gauss's initials on it. The above is the law for the top, bottom, and sides of both the cylinder:

$\begin{array}{r}\text{∮}\stackrel{→}{E}â‹\dots d\stackrel{→}{A}=\frac{Q_{in}}{{\mathrm{Ï\mu }}_0}\\ \text{∮}\stackrel{→}{E}â‹\dots d\stackrel{→}{A}={∫}_{\text{bottom}} \stackrel{→}{E}â‹\dots d\stackrel{→}{A}+{∫}_{\text{top}} \stackrel{→}{E}â‹\dots d\stackrel{→}{A}+{∫}_{\text{sides}} \stackrel{→}{E}â‹\dots d\stackrel{→}{A}=E_{1}s+E_{5}s+0\\ \frac{Q_{up}}{{\mathrm{Ï\mu }}_0}+\frac{Q_{\text{down}}}{{\mathrm{Ï\mu }}_0}=E_{1}s+E_{5}s\end{array}$

Because the electric field vector is perpendicular to the vector field of the surface s, which is the top and bottom face of the Probabilistic cylinder, the integration by sides is zero.

We can already use expression for determine charge density.

The up slab's charge density is:

localid="1648833873919" ${\mathrm{σ}}_{up}=\frac{Q}{A}$

The following charges are associated with the up slab cylinder:

$Q_{up}=\mathrm{σ}s$

Substitute ${\mathrm{σ}}_{up}:$

$Q_{up}=\frac{Q}{A}s$

The low slab's charge density is:

${\mathrm{σ}}_{low}=\frac{2Q}{A}$

Substitute ${\mathrm{σ}}_{low}:$

$Q_{low}=\frac{2Q}{A}s$

Gauss's law can be replaced with a prior phrase.

$\begin{array}{r}{E}_{1}s+E_{5}s=\frac{Q}{A{\mathrm{Ï\mu }}_0}s+\frac{2Q}{A{\mathrm{Ï\mu }}_0}s\\ E_{1}s+E_{5}s=\frac{3Q}{A{\mathrm{Ï\mu }}_0}\end{array}$

We have electrostatic stability inside metal because two slabs are formed of metal${\stackrel{→}{E}}_2={\stackrel{→}{E}}_4=0$. The filed of a charge plane is unaffected by distance from the aircraft and because of that $E_1=E_5$.We can now determine $E_1$and $E_5$.

$\begin{array}{r}{E}_1+E_1=\frac{3Q}{A{\mathrm{Ï\mu }}_0}\\ E_1=\frac{3Q}{2A{\mathrm{Ï\mu }}_0}\\ E_1=E_5=\frac{3Q}{2A{\mathrm{Ï\mu }}_0}\end{array}$

On the right side, we can now consider the Gaussian cylinder. The lower slab provides a more powerful field than the upper slab, so electric field in region $3$is upward.

$\begin{array}{r}\text{∮}\stackrel{→}{E}â‹\dots d\stackrel{→}{A}={∫}_{bottom} \stackrel{→}{E}â‹\dots d\stackrel{→}{A}+{∫}_{top} \stackrel{→}{E}â‹\dots d\stackrel{→}{A}+{∫}_{sides} \stackrel{→}{E}â‹\dots d\stackrel{→}{A}\\ =E_{1}s−E_{3}s+0\end{array}$

$\left(−E_{3}s\right)$because there is an angle between the vector of the electric field and the normal vector on the surface $\mathrm{π}$.

Solving for $E_3$:

$\begin{array}{r}\text{∮}\stackrel{→}{E}â‹\dots d\stackrel{→}{A}=\frac{Q_{in}}{{\mathrm{Ï\mu }}_0}\\ Q_{in}=Q_{up}=\frac{Q}{A}s\\ E_{1}s−E_{3}s=\frac{Qs}{A{\mathrm{Ï\mu }}_0}\\ E_3=E_1−\frac{Q}{A{\mathrm{Ï\mu }}_0}\\ E_3=\frac{3Q}{2A{\mathrm{Ï\mu }}_0}−\frac{Q}{A{\mathrm{Ï\mu }}_0}\\ E_3=\frac{Q}{2A{\mathrm{Ï\mu }}_0}\end{array}$

b) Now let us start with an expression for the electric field at the conductor's surface of conductor by expressing $\mathrm{η}:$

$\begin{array}{r}E=\frac{\mathrm{η}}{{\mathrm{Ï\mu }}_0}\\ \mathrm{η}=E{\mathrm{Ï\mu }}_0\end{array}$

For each surface, we can now substitute formulas for electric fields:

It should be noted that (+) sign because $E_1$the surface is directed outward.

$\begin{array}{r}{\mathrm{η}}_a=+E_1{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_a=\frac{3Q}{2A{\mathrm{Ï\mu }}_0}{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_a=+\frac{3Q}{2A}\end{array}$

It should be noted that (-) sign because $E_3$the surface is directed toward.

$\begin{array}{r}{\mathrm{η}}_b=−E_3{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_b=−\frac{Q}{2A{\mathrm{Ï\mu }}_0}{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_b=−\frac{Q}{2A}\end{array}$

It should be noted that (+) sign that shows$E_3$that the surface is directed outward.

$\begin{array}{r}{\mathrm{η}}_c=+E_3{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_c=+\frac{Q}{2A{\mathrm{Ï\mu }}_0}{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_c=+\frac{Q}{2A}\end{array}$

It should be noted that (+) sign denotes localid="1648885087744" $E_5$that the surface is directed outward.

$\begin{array}{r}{\mathrm{η}}_d=+E_5{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_d=\frac{3Q}{2A{\mathrm{Ï\mu }}_0}{\mathrm{Ï\mu }}_0\\ {\mathrm{η}}_d=\frac{3Q}{2A}\end{array}$