Figure 25-48 shows a parallel plate capacitor with a plate area $\mathbf{A}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{89}\mathbf{}{\mathbf{cm}}^{\mathbf{2}}$and plate separation d = 4.62 mm. The top half of the gap is filled with material of dielectric constant${\mathbf{k}}_{\mathbf{1}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{0}$; the bottom half is filled with material of dielectric constant ${\mathbf{k}}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}$. What is the capacitance?

a) Area of the plates,$\mathrm{A}=7.89\times {10}^{-4}{\mathrm{m}}^2$

b) Separation of the plates,$d=4.62\times {10}^{-3}\mathrm{m}$

c) Dielectric constant of material in the top half gap,${\mathrm{k}}_1=11.0$

d) Dielectric constant of material in the bottom half gap,${\mathrm{k}}_2=12.0$

The capacitor with two different dielectric materials can be considered as the two capacitors connected in series, each with the same plate area, the thickness of each dielectric is d/2. The potential drop across the plates is calculated by adding the potential drops across each dielectric material; we assume that the electric field across the capacitor plates is uniform.

Formulae:

The potential drop between the capacitor plates, V =E.d …(i)

The electric field due to flux between the capacitor plates, $E=\frac{\mathrm{\sigma }}{{\mathrm{k\epsilon }}_0}\mathrm{or}\frac{\mathrm{\sigma }}{{\mathrm{k\epsilon }}_0\mathrm{A}}$ …(ii)

The capacitance of the capacitor plates due to stored charge, C = Q/V …(iii)

Where,

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

Total potential is equal to the sum of potential across each capacitor as the capacitors are in series..

Using the equation for potential difference from equation (i) and the given data, we can write the equation as follows:

$$\begin{gathered} V=E_1.\frac{d}{2}+E_2.\frac{d}{2} \\ =\frac{Q}{A.{\epsilon }_0{k}_1}.\frac{d}{2}+\frac{Q}{A.{\epsilon }_0{k}_2}.\frac{d}{2}\left(\mathrm{The}\mathrm{electric}\mathrm{field}\mathrm{values}\mathrm{are}\mathrm{sunstituted}\mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right)\right) \\ =\frac{Q}{2{\epsilon }_{0}A}.\left(\frac{1}{{\mathrm{k}}_1}+\frac{1}{{\mathrm{k}}_2}\right) \\ =\frac{Qd}{2{\epsilon }_{0}A}.\left(\frac{k_1+k_2}{{\mathrm{k}}_1{\mathrm{k}}_2}\right) \end{gathered}$$

Now, using this potential value in equation (iii), we can get the capacitance value between the capacitors using the given data as follows:

$$\begin{gathered} C=\frac{2{\epsilon }_{0}A}{d\left(k_1+k_2\right)}{k}_1{k}_2 \\ =\frac{2\times \left(8.85\times {10}^{-12}\mathrm{F}.{\mathrm{m}}^{-1}\right)\times \left(7.89\times {10}^{-4}{\mathrm{m}}^2\right)\times \left(11\times 12\right)}{\left(4.62\times {10}^{-3}\mathrm{m}\right)\times \left(11\times 12\right)} \\ =17.3\times {10}^{-12}\mathrm{F} \\ =17.3\mathrm{pF} \end{gathered}$$

Hence, the value of the capacitance is 17.3 pF.