91Ó°ÊÓ

a) Area of the plates,$A=5.56×{10}^{-4}{\mathrm{m}}^2$

b) Separation of the plates,$d=5.56×{10}^{-3}\mathrm{m}$

c) The left half area of the capacitor is filled with dielectric,${\mathrm{k}}_1=7.0$

d) The right half area of the capacitor is filled with dielectric,${\mathrm{k}}_2=12.0$

The capacitor with two different dielectric constants can be considered as two capacitors connected in parallel but each has half the surface area of the original. So, we can find capacitances by considering two capacitors separately. Adding both we can get total capacitance.

Formula:

The capacitance of the plates due to dielectric substance,$C=\frac{\mathrm{kA}{\stackrel{Ë™}{\mathrm{o}}}_0}{2\mathrm{d}}$ …(¾±)

The equivalent capacitance of a parallel connection of capacitors,

$C_{\mathrm{requivalent}}=\underset{}{∑}{C}_i$ …(¾±¾±)

Where,

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

We can view these capacitors as capacitors in parallel.

The effective capacitance using equation (i) in equation (ii) can be written as follows:

$$\begin{gathered} C=C_1+C_2 \\ =\frac{A{\mathrm{Î\mu }}_0}{2d}.\left({\mathrm{k}}_1+{\mathrm{k}}_2\right)\left(âˆ\mu \mathrm{area}\mathrm{of}\mathrm{both}\mathrm{the}\mathrm{dielectric}\mathrm{substance}\mathrm{is}\mathrm{A}/2\right) \\ =\frac{\left(5.56×{10}^{-4}{\mathrm{m}}^2\right)×\left(8.85×{10}^{-12}\mathrm{F}.{\mathrm{m}}^{-1}\right)}{2×\left(5.56×{10}^{-3}\mathrm{m}\right)}\left(7.0+12.0\right) \\ =8.41×{10}^{-12}\mathrm{F} \end{gathered}$$

Hence, the value of the total capacitance is $8.41×{10}^{-12}\mathrm{F}$.