Chapter 19, Problem 74GP

(a)

To determine

The configuration of the capacitors.

Answer to Problem 74GP

The seven capacitors are connected in the parallel connection.

Explanation of Solution

Introduction:

The equivalent capacitance of the capacitors in the parallel connection increases with addition of more capacitors, so if new capacitors are added to capacitors in parallel connection then new equivalent capacitors also increase.

The area of the variable capacitor changes from $2{\text{cm}}^2$ to $9.5{\text{cm}}^2$ . So, the capacitance of the variable capacitor increases. The equivalent capacitance of the capacitor can only increases in the parallel connection. So, the seven capacitors are connected in the parallel connection.

Conclusion:

Thus, the seven capacitors are connected in the parallel connection.

(b)

To determine

Range of the capacitance values.

Answer to Problem 74GP

The range of the capacitances is from $1.18\times {10}^{-12}\text{F}$ to $5.61\times {10}^{-12}\text{F}$ .

Explanation of Solution

Given info:

The minimum area of the variable capacitor is, $A_{\min }=2{\text{cm}}^2=2{\text{cm}}^2\times \frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}=2\times {10}^{-4}{\text{m}}^2$ .

The maximum area of the variable capacitor is, $A_{\max }=9.5{\text{cm}}^2=9.5{\text{cm}}^2\times \frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}=9.5\times {10}^{-4}{\text{m}}^2$ .

The separation between the plates of the capacitors is, $d=1.5\text{mm}=1.5\text{mm}\times \frac{1\text{m}}{1000\text{mm}}=1.5\times {10}^{-3}\text{m}$ .

Formula Used:

The expression to calculate the minimum capacitance is,

  $C_{\min }=\frac{{\epsilon }_0{A}_{\min }}{d}$

Here,

  ${\epsilon }_0$ is the permittivity of free space and its value is $8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ .

The expression to calculate the maximum capacitance is,

  $C_{\max }=\frac{{\epsilon }_0{A}_{\max }}{d}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_{\min }=\frac{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(2\times {10}^{-4}{\text{m}}^2\right)}{1.5\times {10}^{-3}\text{m}}\\ =1.18\times {10}^{-12}\text{F}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{C}_{\max }=\frac{\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(9.5\times {10}^{-4}{\text{m}}^2\right)}{1.5\times {10}^{-3}\text{m}}\\ =5.61\times {10}^{-12}\text{F}\end{array}$

Conclusion:

Thus, the range of the capacitances is from $1.18\times {10}^{-12}\text{F}$ to $5.61\times {10}^{-12}\text{F}$ .