91Ó°ÊÓ

a) Area of the plates,$A=10.5×{10}^{-4}{\mathrm{m}}^2$

b) Separation of the plates,$2d=7.12×{10}^{-3}\mathrm{m}$

c) Dielectric constant of material in half of the left area,${\mathrm{k}}_1=21.0$

d) Dielectric constant of material in the top half of the half of right area,${\mathrm{k}}_2=42.0$

e) Dielectric constant of material in the bottom half of the half of right area,${\mathrm{k}}_3=58.0$

The given capacitor is considered as a combination of three different capacitors, the first half substance is connected in parallel with the other two capacitors in the other half, and later these two are connected in series with a plate area, A/2. For the first capacitor, plate separation is d. Later two have the plate separation, d/2.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{equivatent}}=\underset{}{∑{\displaystyle \frac{1}{C_i}}}$ …(¾±).

The equivalent capacitance of a parallel connection of capacitors,

$\frac{1}{C_{equivatent}}=\underset{}{∑{\displaystyle C_i}}$ …(¾±¾±)

The capacitance between the two plates due to stored charge, …(iii)

Where,

${\mathrm{Î\mu }}_0=8.85×{10}^{-12}\mathrm{F}.{\mathrm{m}}^{-1}$ is permittivity of free space

We find the capacitor for the three different dielectric materials using the formula of equation (iii) as follows:

$$\begin{gathered} C_1=\frac{{\stackrel{Ë™}{o}}_0\mathrm{A}}{2}\frac{{\mathrm{k}}_1}{2d} \\ {C}_2=\frac{{\stackrel{Ë™}{o}}_0\mathrm{A}}{2}\frac{{\mathrm{k}}_2}{d} \\ {C}_3=\frac{{\stackrel{Ë™}{o}}_0{\mathrm{Ak}}_3}{2d} \end{gathered}$$

Capacitor $C_2$and $C_3$are in series, the equivalent capacitance of $C_2\mathrm{and}{C}_3$are connected in parallel with $C_1$,

Thus, the equivalent capacitance of the three materials between the plates is given using equations (i) and (ii) as:

$$\begin{gathered} C=C_1+\frac{C_2{C}_3}{C_2+C_3} \\ =\frac{{\stackrel{˙}{o}}_{0}A}{2}\frac{{\mathrm{k}}_1}{2d}+\frac{{\stackrel{˙}{o}}_{0}A}{2d}\frac{{\mathrm{k}}_2{\mathrm{k}}_3}{{\mathrm{k}}_2+{\mathrm{k}}_3} \\ =\frac{{\stackrel{˙}{o}}_{0}A}{4d}\left({\mathrm{k}}_1+\frac{2{\mathrm{k}}_2{\mathrm{k}}_3}{\left({\mathrm{k}}_2+{\mathrm{k}}_3\right)}\right) \\ =\frac{\left(8.85×{10}^{-12}\mathrm{F}.{\mathrm{m}}^{-1}\right)×\left[10.5×{10}^{-4}{\mathrm{m}}^2\right]}{4×\left(3.56×{10}^{-3}\mathrm{m}\right)}.\left(21.0+\frac{2×42×58}{42+58}\right) \\ =4.55×{10}^{-11}\mathrm{F} \end{gathered}$$

Hence, the value of the capacitance is $4.55×{10}^{-11}\mathrm{F}$.