In Fig. 25-56, the parallel-plate capacitor of plate area $\mathbf{2}\mathbf{.}\mathbf{00}\mathit{x}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$is filled with two dielectric slabs, each with thickness. One slab has dielectric constant 3.00, and the other, 4.00. How much charge does the 7.00 Vbattery store on the capacitor?

• Plate area$A=2.00\times {10}^{-2}{m}^2$
• Thickness, d = 4.00 mm
• Dielectric constant$k_1=3.00$
• Dielectric constant$k_2=4.00$
• Potential across the battery V = 7.00 V

If the space between the plates of a capacitor is completely filled with a dielectric material, the capacitance C is increased by a factor, called the dielectric constant, which is characteristic of the material. In a region that is completely filled by a dielectric, all electrostatic equations containing must be modified by replacing${\mathbf{\in }}_{\mathbf{0}}$ with$\mathit{k}{\mathbf{\in }}_{\mathbf{0}}$.

When the capacitors${\mathit{c}}_{\mathbf{1}}$ and${\mathit{c}}_2$ are connected in series, the equivalent capacitance${\mathit{c}}_{eq}$ is given as,

$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{e}\mathbf{q}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}$

When the capacitors${\mathit{c}}_{\mathbf{1}}$ and${\mathit{c}}_2$ are connected in parallel, the equivalent capacitance${\mathit{c}}_{eq}$ is given as,

${\mathit{C}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{C}}_{\mathbf{1}}\mathbf{+}{\mathit{C}}_{\mathbf{2}}$

We use the formula of the capacitance of the parallel plate capacitor to find the capacitance of the parallel plate capacitor. And using the relation of charge and capacitance, we can find the charge on the capacitor.

Formulae:

The capacitance between the plates with dielectric, $C=\frac{k{\in }_{0}A}{d}$ ...(i)

Here, C is capacitance, k is dielectric constant of the material,${\in }_0$is the permittivity of the free space, A is the area of cross-section, and d is the separation between the two plates.

The charge stored between the plates, q=CV ...(ii)

Here, q is electric charge, C is capacitance and V is the potential difference across the two plates.

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{equivalent}}=\underset{}{\sum {\displaystyle \frac{1}{C_i}}}$ ...(iii).

We may think that there are two capacitors in series of different dielectric constant$k_1=3and{k}_2=4$

We assume that the two capacitors are in series and their equivalent capacitance is given using equation (iii) by,

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{d}{k_1{\in }_{0}A}+\frac{d}{k_2{\in }_{0}A} \\ {C}_{eq}=\left(\frac{k_1{k}_2}{k_1+k_2}\right)\frac{{\in }_{0}A}{d} \\ =\left(\frac{3.00\times 4.00}{3.00+4.00}\right)\frac{\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)\left(2.00\times {10}^{-2}{m}^2\right)}{\left(2.00\times {10}^{-3}m\right)} \\ =1.52\times {10}^{-10}F \end{gathered}$$

And the charge is given using equation (iii) by,

$$\begin{gathered} \mathrm{q}=1.52\times {10}^{-10}\mathrm{F}\times 7\mathrm{V} \\ =1.06\times {10}^{-9}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $1.06\times {10}^{-9}\mathrm{C}$.