Chapter 17, Problem 87QAP

To determine

(a)

Capacitance of the parallel plate capacitor

Answer to Problem 87QAP

Capacitance of the parallel plate capacitor is $\frac{{\epsilon }_{o}A}{2d}\left(\frac{3k_2{k}_3}{k_2+k_3}+\frac{k_1}{2}\right)$.

Explanation of Solution

Given:

The given capacitance diagram

  

Figure.1

Formula used:

The capacitance of the parallel plate capacitor is given as
$C=k\frac{{\epsilon }_{o}A}{d}$

Here,
$k$ is the dielectric constant
${\epsilon }_o$ is the permeability
$A$ is the area
$d$ is the distance between plates

Calculation:

It is clear from the figure.1 that capacitor with dielectric constant $k_2\text{and}{k}_3$ are in series connection, and the combination of these is in parallel connection with capacitor of dielectric constant of $k_1$

Therefor the equivalent capacitance of the arrangement is calculated as
$\begin{array}{c}{C}_{eq}=\left(\frac{1}{\frac{1}{k_2\frac{{\epsilon }_o\times 3A}{4\times \frac{d}{2}}}+\frac{1}{k_3\frac{{\epsilon }_o\times 3A}{4\times \frac{d}{2}}}}\right)+k_1\frac{{\epsilon }_{o}A}{4d}\\ =\frac{{\epsilon }_{o}A}{2d}\left(\frac{3k_2{k}_3}{k_2+k_3}+\frac{k_1}{2}\right)\end{array}$

Conclusion:

Capacitance of the parallel plate capacitor is $\frac{{\epsilon }_{o}A}{2d}\left(\frac{3k_2{k}_3}{k_2+k_3}+\frac{k_1}{2}\right)$.

To determine

(b)

Capacitance of the parallel plate capacitor

Answer to Problem 87QAP

Capacitance of the parallel plate capacitor is $\frac{{\epsilon }_{o}A}{2d}\left(\frac{3k_2}{k_2+1}+\frac{k_1}{2}\right)$.

Explanation of Solution

Given:

The given capacitance diagram

  

Figure.1

Formula used:

The capacitance of the parallel plate capacitor is given as
$C=k\frac{{\epsilon }_{o}A}{d}$

Here,
$k$ is the dielectric constant
${\epsilon }_o$ is the permeability
$A$ is the area
$d$ is the distance between plates

Calculation:

It is clear from the figure.1 that capacitor with dielectric constant $k_2\text{and}{k}_3$ are in series connection, and the combination of these is in parallel connection with capacitor of dielectric constant of $k_1$

If the dielectric with constant $k_3$ is replace by air, $k=1$ than the equivalent capacitance of the arrangement is calculated as
$\begin{array}{c}{C}_{eq}=\left(\frac{1}{\frac{1}{k_2\frac{{\epsilon }_o\times 3A}{4\times \frac{d}{2}}}+\frac{1}{1\times \frac{{\epsilon }_o\times 3A}{4\times \frac{d}{2}}}}\right)+k_1\frac{{\epsilon }_{o}A}{4d}\\ =\frac{{\epsilon }_{o}A}{2d}\left(\frac{3k_2}{k_2+1}+\frac{k_1}{2}\right)\end{array}$

Conclusion:

Capacitance of the parallel plate capacitor is $\frac{{\epsilon }_{o}A}{2d}\left(\frac{3k_2}{k_2+1}+\frac{k_1}{2}\right)$.