Chapter 26, Problem 55AP

(a)

To determine

The capacitance of the three plate system ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}{\text{P}}_{\text{3}}$, when ${\text{P}}_{\text{3}}$ is connected to the negative terminal.

Answer to Problem 55AP

The capacitance of three plate system ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}{\text{P}}_{\text{3}}$ is $11.2\text{pF}$.

Explanation of Solution

Below figure shows the arrangements of plates ${\text{P}}_{\text{1}}$, ${\text{P}}_2$ and ${\text{P}}_3$ in the form of capacitors.

Figure-(1)

Write the expression for capacitance connected in parallel.

    $C=2\left(\frac{{\epsilon }_{0}A}{d}\right)$                                                                                                              (I)

Here, $A$ is the area of the plate, $d$ is the distance between two plates, ${\epsilon }_0$ is the permittivity of free space and $C$ is the capacitance.

Conclusion:

Substitute $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$, $7.5\times {10}^{-4}{\text{m}}^2$ for $A$ and $1.19\times {10}^{-3}\text{m}$ for $d$ in equation (I) to calculate $C$.

    $\begin{array}{c}C=2\left(\frac{\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)\left(7.5\times {10}^{-4}{\text{m}}^2\right)}{\left(1.19\times {10}^{-3}\text{m}\right)}\right)\\ =2\left(\frac{\left(66.375\times {10}^{-16}{\text{C}}^{\text{2}}/\text{N}\right)}{\left(1.19\times {10}^{-3}\text{m}\right)}\right)\\ =2\left(55.777\times {10}^{-13}{\text{C}}^{\text{2}}/\text{N}\cdot \text{m}\right)\left(\frac{\text{F}}{{\text{C}}^{\text{2}}/\text{N}\cdot \text{m}}\right)\\ =11.2\times {10}^{-12}\text{F}\left(\frac{1\text{pF}}{{10}^{-12}\text{F}}\right)\end{array}$

Further solve the above equation.

    $C=11.2\text{pF}$

Therefore, the capacitance of the three plate system ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}{\text{P}}_{\text{3}}$ is $11.2\text{pF}$.

(b)

To determine

The charge on ${\text{P}}_{\text{2}}$.

Answer to Problem 55AP

The charge on ${\text{P}}_{\text{2}}$ is $\text{134}\text{pC}$

Explanation of Solution

Charge on capacitor ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}$ is equal to charge on capacitor ${\text{P}}_2{\text{P}}_3$.

Write the expression for the charge on capacitors.

    $q=\frac{CV}{2}$                                                                                                                    (II)

Here, $q$ is the charge on the capacitor and $V$ is the potential difference.

Conclusion:

Substitute $11.2\text{pF}$ for $C$ and $12\text{V}$ for $V$ in Equation (II) to calculate $q$.

    $\begin{array}{c}q=\frac{\left(11.2\text{pF}\right)\left(12\text{V}\right)}{2}\left(\frac{{10}^{-12}\text{F}}{1\text{pF}}\right)\\ =\frac{\left(11.2\times {10}^{-12}\text{F}\right)\left(12\text{V}\right)}{2}\\ =\frac{134.4\times {10}^{-12}C}{2}\left(\frac{1\text{pC}}{{10}^{-12}C}\right)\\ =67.2\text{pC}\end{array}$

${\text{P}}_{\text{2}}$ is the common plate in the ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}{\text{P}}_{\text{3}}$ system, therefore net charge on ${\text{P}}_{\text{2}}$ is.

    $\begin{array}{c}{\text{P}}_{\text{2}}=2\times 67.2\text{pC}\\ =\text{134}\text{pC}\end{array}$

Therefore, the charge on ${\text{P}}_{\text{2}}$ is $\text{134}\text{pC}$.

(c)

To determine

The capacitance of the four plate system ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}{\text{P}}_{\text{3}}{\text{P}}_{\text{4}}$, when ${\text{P}}_{\text{4}}$ is connected to the positive terminal.

Answer to Problem 55AP

The capacitance of the four plate system ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}{\text{P}}_{\text{3}}{\text{P}}_{\text{4}}$ $16.7\text{pF}$.

Explanation of Solution

Write the expression for  net capacitance.

    $C^{\prime }=3\left(\frac{{\epsilon }_{0}A}{d}\right)$ (III)

Here, $C^{\prime }$ is the net capacitance.

Conclusion:

Substitute $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$, $7.5\times {10}^{-4}{\text{m}}^2$ for $A$ and $1.19\times {10}^{-3}\text{m}$ for $d$ in equation (III) to calculate $C$.

    $\begin{array}{c}{C}^{\prime }=3\left(\frac{\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)\left(7.5\times {10}^{-4}{\text{m}}^2\right)}{\left(1.19\times {10}^{-3}\text{m}\right)}\right)\\ =3\left(\frac{\left(66.375\times {10}^{-16}{\text{C}}^{\text{2}}/\text{N}\right)}{\left(1.19\times {10}^{-3}\text{m}\right)}\right)\\ =3\left(55.777\times {10}^{-13}{\text{C}}^{\text{2}}/\text{N}\cdot \text{m}\right)\left(\frac{\text{F}}{{\text{C}}^{\text{2}}/\text{N}\cdot \text{m}}\right)\\ =16.7\times {10}^{-12}\text{F}\left(\frac{1\text{pF}}{{10}^{-12}\text{F}}\right)\end{array}$

Further solving the above equation.

    $C^{\prime }=16.7\text{pF}$

Therefore, the capacitance of the four plate system ${\text{P}}_{\text{1}}{\text{P}}_{\text{2}}{\text{P}}_{\text{3}}{\text{P}}_{\text{4}}$ is $16.7\text{pF}$.

(d)

To determine

The charge on ${\text{P}}_{\text{4}}$.

Answer to Problem 55AP

The charge on ${\text{P}}_{\text{4}}$ is $67.0\text{pC}$.

Explanation of Solution

Write the expression for the net charge on capacitors.

    $Q=C^{\prime }V$                                                                                                                (IV)

Here, $Q$ is the charge on the capacitor.

Conclusion:

Substitute $16.7\text{pF}$ for $C^{\prime }$ and $12\text{V}$ for $V$ in Equation (IV) to calculate $Q$.

    $\begin{array}{c}Q=\left(16.7\text{pF}\right)\left(12\text{V}\right)\left(\frac{{10}^{-12}\text{F}}{1\text{pF}}\right)\\ =\left(16.7\times {10}^{-12}\text{F}\right)\left(12\text{V}\right)\\ =\left(200.4\times {10}^{-12}\text{C}\right)\left(\frac{1\text{pC}}{{10}^{-12}\text{C}}\right)\\ =200.4\text{pC}\end{array}$

The charge on ${\text{P}}_{\text{4}}$ alone is,

    $\begin{array}{c}{\text{P}}_{\text{4}}=\frac{200.4\text{pC}}{3}\\ =66.8\text{pC}\\ \simeq 67.0\text{pC}\end{array}$

Therefore, charge on ${\text{P}}_{\text{4}}$ is $67.0\text{pC}$.