Question

In figure, how much charge is stored on the parallel plate capacitors by the 12.0 Vbattery ? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathit{m}}^{\mathbf{2}}$and a plate separation of 2.00 mm.

Short answer

The total charge stored in the capacitors connected in parallel to the battery is

$1.06\times {10}^{-9}C\hspace{0.17em}$.

Step-by-step solution

The given data

a) Voltage of the battery, V = 12.0 V

b) There are two capacitors, one is filled with air and the other is dielectric.

c) Dielectric constant of the material, k = 3.0

d) Area of the plates, $A=5\times {10}^{-3}{m}^2$

e) Separation between the plates,$d=2\times {10}^{-3}m$

Understanding the concept of the capacitance with dielectric

Capacitors are connected in parallel to the battery, so they will cause the same potential drop across them. To find the charge supplied by the battery, we need to find the capacitance of both capacitors using the given relation and this will determine the value of the charge stored.

Formulae:

The capacitance due to presence of dielectric material,$C=\frac{k{\epsilon }_{0}A}{d}$ …(i)

The charge stored within the capacitor,$q=C\times V$ …(ii)

Where.

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

Calculation of the total stored charge

Using the given data in equation (i), the capacitance of the dielectric medium is given as follows:

$$\begin{gathered} C_1=\frac{3.0\times \left(8.85\times {10}^{-12}F.m^{-1}\right)\times \left(5\times {10}^{-3}{m}^2\right)}{\left(2\times {10}^{-3}m\right)} \\ =6.64\times {10}^{-11}F \end{gathered}$$

Also, using the same data for air filled capacitor in equation (i), we get the value of the capacitance as follows:

$$\begin{gathered} C_2=\frac{\left(8.85\times {10}^{-12}F.m^{-1}\right)\times \left(5\times {10}^{-3}{m}^2\right)}{\left(2\times {10}^{-3}m\right)} \\ =2.21\times {10}^{-11}F \end{gathered}$$

We have, the charge of the dielectric-filled capacitor can be given using the capacitance in equation (ii) as follows:

role="math" localid="1661344431796" $$\begin{gathered} q_1=C_1\times V=\left(6.64\times {10}^{-11}F\right)\times \left(12.0V\right) \\ =8.0\times {10}^{-10}C \end{gathered}$$

Similarly, the charge of the air-filled capacitor can be given using the capacitance in equation (ii) as follows:

$$\begin{gathered} q_2=\left(2.21\times {10}^{-11}F\right)\times \left(12.0V\right) \\ =2.6\times {10}^{-10}C \end{gathered}$$

Total charge can be found using the above charge values is given by,

$$\begin{gathered} q_{total}=q_1+q_2 \\ =1.06\times {10}^{-9}C \end{gathered}$$

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