91Ó°ÊÓ

Consider the formula for the electric field as

$\mathbf{E}\mathbf{=}\frac{\mathrm{σ}}{\mathbf{2}{\mathbf{Î\mu }}_0}$

Consider the formula for the dielectric constant as

$\mathbf{D}\mathbf{=}{\mathbf{°ìÎ\mu }}_0\mathbf{E}$

Consider the formula for the polarization constant as

$\mathbf{P}\mathbf{=}{\mathbf{Î\mu }}_0{\mathbf{χ}}_e\mathbf{E}$

Consider the formula for the electric potential as

$\mathbf{V}\mathbf{=}\mathbf{Ed}$

(a)

Determine the formula for the electrical displacement in each of the slab as

$âˆ\circledR \mathbf{D}â‹\dots \mathbf{da}=q_{encl}$

Consider the infinite slab has the surface charge density as $+s$at the top surface. Since the electric field is perpendicular to the charge at the top of the surface. The slab is conducting and all the charges accumulate at the top surface. Since there are not charges inside the slab, then the formula becomes:

$âˆ\circledR \mathbf{D}â‹\dots \mathbf{da}=0$

Consider the plates with the negative charges moving from the positive to negative side.This shows that the electric field inside the space between the two plates is two times of the field because of each sheet. Then, the formula for the electric sheet is

$E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$

Consider the direction of the electric field E for every plate moves in positive x direction and is

$\begin{array}{c}E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_1}+\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_2}\\ =\frac{\mathrm{σ}}{2\left(k_1{\mathrm{Î\mu }}_0\right)}+\frac{\mathrm{σ}}{2\left(k_2{\mathrm{Î\mu }}_0\right)}\end{array}$

Substitute and solve for the electric field as

$\begin{array}{c}E=\frac{\mathrm{σ}}{2\left(2{\mathrm{Î\mu }}_0\right)}+\frac{\mathrm{σ}}{2\left(1.5{\mathrm{Î\mu }}_0\right)}\\ =\frac{7\mathrm{σ}}{12{\mathrm{Î\mu }}_0}\end{array}$

Consider the formula for the dielectric constant for the slab 1 as

$D_{slab1}=k_1{\mathrm{Î\mu }}_{0}E$

Substitute the values and solve further as

$D_{slab1}=k_1{\mathrm{Î\mu }}_{0}E$

Substitute the values and solve as

$\begin{array}{c}{D}_{slab1}=2{\mathrm{Î\mu }}_0\left(\frac{7\mathrm{σ}}{12{\mathrm{Î\mu }}_0}\right)\\ =\frac{7\mathrm{σ}}{6}\end{array}$

Solve for the dielectric constant for the slab 2 as

$\begin{array}{c}{D}_{slab1}=\left(1.5\right){\mathrm{Î\mu }}_0\left(\frac{7\mathrm{σ}}{12{\mathrm{Î\mu }}_0}\right)\\ =\frac{21\mathrm{σ}}{24}\end{array}$

Therefore, the dielectric constant for the slab 1 is $\frac{7\mathrm{σ}}{6}$and for slab 2 is $\frac{21\mathrm{σ}}{24}$.

(b)

Consider the formula for the electric field in terms of the surface charge density as

$E=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_{r1}}$

Solve for the electric field in the first slab as

$E_1=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$

Solve for the electric field in the second slab as

$E_2=\frac{\mathrm{σ}}{1.5{\mathrm{Î\mu }}_0}$

Therefore, the electric field for the first vector is$\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$ and the second vector is $\frac{\mathrm{σ}}{1.5{\mathrm{Î\mu }}_0}$.

(c)

Consider the formula for the polarization as

$\begin{array}{l}P={\mathrm{Î\mu }}_0{\mathrm{χ}}_e\mathbf{E}\\ P={\mathrm{Î\mu }}_0{\mathrm{χ}}_e\frac{\mathrm{σ}}{\mathrm{Î\mu }}\end{array}$

Here, the susceptibility of the medium is${\mathrm{χ}}_e$ .

Resolve the equation as

$\begin{array}{c}\mathbf{P}=\frac{{\mathrm{Î\mu }}_0{\mathrm{χ}}_e\mathrm{σ}}{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}\\ =\frac{{\mathrm{χ}}_e\mathrm{σ}}{{\mathrm{Î\mu }}_r}\\ =\frac{({\mathrm{Î\mu }}_r−1)\mathrm{σ}}{{\mathrm{Î\mu }}_r}\\ =(1−{\mathrm{Î\mu }}_r^{−1})\mathrm{σ}\end{array}$

Substitute the values and solve for polarization constant of the slab 1 as

$\begin{array}{c}\mathbf{P}=(1−2^{−1})\mathrm{σ}\\ =\frac{1}{2}\mathrm{σ}\end{array}$

Substitute the values and solve for polarization constant of the slab 2 as

$\begin{array}{c}\mathbf{P}=(1−{1.5}^{−1})\mathrm{σ}\\ =\frac{1}{3}\mathrm{σ}\end{array}$

Therefore, the polarization for the slab 1 is$\frac{1}{2}\mathrm{σ}$ and for slab 2 is $\frac{1}{3}\mathrm{σ}$.

(d)

Consider the formula for the potential between the slabs if the distance between them is d is as follows:

$V=Ed$

Resolve the equations as

$\begin{array}{c}V=E_{1}a+E_{2}a\\ =a(\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}+\frac{\mathrm{σ}}{3{\mathrm{Î\mu }}_0})\end{array}$

Substitute the values and solve as

$\begin{array}{c}V=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}[\frac{1}{2}+\frac{2}{3}]a\\ =\frac{7\mathrm{σ}}{6{\mathrm{Î\mu }}_0}a\end{array}$

Therefore, the potential between the two slabs is $\frac{7\mathrm{σ}}{6{\mathrm{Î\mu }}_0}a$.

(e)

Consider the formula for the surface bound charge at the top of the slab 1 is as follows:

$\mathrm{σ}=P_1â‹\dots \widehat{\mathbf{n}}$

Consider at the top of the surface the slab 1 is $\widehat{\mathbf{n}}$in the upward direction and $P_1$points in the downward direction. Then, the formula is

$\begin{array}{c}{\mathrm{σ}}_b=−P_1\\ =−\frac{\mathrm{σ}}{2}\end{array}$

Consider the expression for the surface bound charge at the bottom of the slab 2 is as follows:

$\begin{array}{c}{\mathrm{σ}}_{b2}=P_2â‹\dots \widehat{\mathbf{n}}\\ =P_2\\ =\frac{\mathrm{σ}}{3}\end{array}$

Consider the formula for the surface bound charge at the top of the slab 2 is

$\begin{array}{c}{\mathrm{σ}}_2=P_2â‹\dots \widehat{\mathbf{n}}\\ =−P_2\\ =−\frac{\mathrm{σ}}{3}\end{array}$

Solve for the total volume for the bound charges is as follows:

role="math" localid="1658735677296" $\begin{array}{c}{\mathrm{ÒÏ}}_b=−\frac{\mathrm{σ}}{2}+\frac{\mathrm{σ}}{2}+\frac{\mathrm{σ}}{3}−\frac{\mathrm{σ}}{3}\\ =0\end{array}$

Therefore, the total amount of the bound charge is 0.

(f)

Consider the formula diagrams for the condition with the electric fields as

Consider the total charge above the slab 1 is

$\mathrm{σ}−\frac{\mathrm{σ}}{2}=\frac{\mathrm{σ}}{2}$

Consider the surface charge below the slab 1 is

$\frac{\mathrm{σ}}{2}−\frac{\mathrm{σ}}{2}+\frac{\mathrm{σ}}{3}−\mathrm{σ}=−\frac{\mathrm{σ}}{2}$

Solve for the total surface charge above the slab 1 is

$\begin{array}{c}{\mathrm{σ}}_1=\frac{\mathrm{σ}}{2}−(−\frac{\mathrm{σ}}{2})\\ =\frac{\mathrm{σ}}{2}+\frac{\mathrm{σ}}{2}\\ =\mathrm{σ}\end{array}$

Solve for the magnitude of the electric field slab is

$E_1=\frac{{\mathrm{σ}}_1}{2{\mathrm{Î\mu }}_0}$

Consider the formula for the total slab charge for slab 2 above it is

$\begin{array}{c}\mathrm{σ}=\frac{\mathrm{σ}}{2}+\frac{\mathrm{σ}}{2}−\frac{\mathrm{σ}}{3}\\ =\frac{2\mathrm{σ}}{3}\end{array}$

Consider the total slab charge below the slab 2 is

$\begin{array}{c}{\mathrm{σ}}_2=\frac{2\mathrm{σ}}{3}−(−\frac{2\mathrm{σ}}{3})\\ =\frac{4\mathrm{σ}}{3}\end{array}$

Solve for the electric field for the slab 2 as

$\begin{array}{c}{E}_2=\frac{\left(\frac{4\mathrm{σ}}{3}\right)}{2{\mathrm{Î\mu }}_0}\\ =\frac{2\mathrm{σ}}{3{\mathrm{Î\mu }}_0}\end{array}$

Therefore, the magnitude of the electric field for the slab 1 is$\frac{{\mathrm{σ}}_1}{2{\mathrm{Î\mu }}_0}$ and for the slab 2 is$\frac{2\mathrm{σ}}{3{\mathrm{Î\mu }}_0}$ .