Chapter 28, Problem 71AP

(a)

To determine

The charge on capacitor $C_1$ when the switch is closed.

Answer to Problem 71AP

The charge on capacitor $C_1$ when the switch is closed is $222\mu \text{C}$.

Explanation of Solution

Write the expression to calculate the power through resistor $R_2$.

    $P=I_2{}^{2}R$                                                                                                                    (I)

Here, $P$ is the power through the resistor $R_2$, $I_2$ is the current through the resistor $R_2$.

Resistor $R_1$ and $R_2$ are in series. So the current flowing through them will be of equal magnitude.

    $I_1=I_2$

Here, $I_1$ is the current flowing through resistor $R_1$.

Write the expression to calculate the  potential difference across $R_1$.

    $V_1=I_1{R}_1$                                                                                                                    (II)

Here, $V_1$ is the potential difference through resistor $R_1$.

Write the expression to calculate the charge on $C_1$.

    $q_1=C_1{V}_1$                                                                                                                  (III)

Here, $q_1$ is the charge on $C_1$ and $C_1$ is the capacitance.

Conclusion:

Substitute $2.40\text{W}$ for $P$ and $7.00\text{kΩ}$ for $R_2$ in equation (I) to solve for $I_2$.

    $\begin{array}{c}2.40\text{W}=I_2{}^2\left(7.00\text{kΩ}\times \frac{{10}^3\Omega }{1\text{kΩ}}\right)\\ I_2=\sqrt{\frac{2.40\text{W}}{\left(7.00\text{kΩ}\times \frac{{10}^3\Omega }{1\text{kΩ}}\right)}}\\ =0.0185\text{A}\times \frac{{10}^3\text{mA}}{1\text{A}}\\ =18.5\text{mA}\end{array}$

Substitute $18.5\text{mA}$ for $I_1$ and $4.00\text{kΩ}$ for $R_1$ in equation (II) to solve for $V_1$.

    $\begin{array}{c}{V}_1=\left(18.5\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(4.00\text{kΩ}\times \frac{{10}^3\text{Ω}}{\text{1}\text{kΩ}}\right)\\ =74\text{V}\end{array}$

Substitute $74\text{V}$ for $V_1$ and $3.00\mu \text{F}$ for $C_1$ in equation (III) to solve for $q_1$

    $\begin{array}{c}{q}_1=\left(74\text{V}\right)\left(3.00\mu \text{F}\times \frac{{10}^{-3}\text{F}}{1\mu \text{F}}\right)\\ =0.222\text{C}\times \frac{{10}^3\mu \text{C}}{1\text{C}}\\ =222\mu \text{C}\end{array}$

Therefore, the charge on the capacitor $C_1$ is $222\mu \text{C}$.

(b)

To determine

The change in charge of capacitor $C_2$ when the switch is open.

Answer to Problem 71AP

The change in charge of capacitor $C_2$ when the switch is open is $444\mu \text{C}$

Explanation of Solution

Write the expression to calculate the potential difference across $R_2$.

    $V_2=I_2{R}_2$                                                                                                                 (IV)

Here, $V_2$ is the potential difference across $R_2$.

Write the expression to calculate the charge on capacitor $C_2$.

    $q_2=V_2{C}_2$                                                                                                                 (V)

Here, $q_2$ is the charge on capacitor $C_2$.

Write the expression to calculate the emf of the battery.

    $\epsilon =V_1+V_2$                                                                                                               (VI)

Here, $\epsilon$ is the emf of the battery.

Write the expression to calculate the drop in the charge of the capacitor $C_2$ when the switch is opened.

    $Q_2=C_2\epsilon$                                                                                                               (VII)

Here, $Q_2$ is the drop in the charge of the capacitor $C_2$ when the switch is opened.

Write the expression to calculate the change in charge of capacitor $C_2$.

    $\Delta Q_2=Q_2-q_2$                                                                                                       (VIII)

Here, $\Delta Q_2$ is the change in charge of capacitor $C_2$.

Conclusion:

Substitute $18.5\text{mA}$ for $I_2$ and $7.00\text{kΩ}$ for $R_2$ in equation (IV) to solve for $V_2$.

    $\begin{array}{c}{V}_2=\left(18.5\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(7.00\text{kΩ}\times \frac{{10}^3\text{Ω}}{\text{1}\text{kΩ}}\right)\\ =129.5\text{V}\end{array}$

Substitute $129.5\text{V}$ for $V_2$ and $6.00\mu \text{F}$ for $C_2$ in equation (V) to solve for $q_2$

    $\begin{array}{c}{q}_2=\left(129.5\text{V}\right)\left(6.00\mu \text{F}\times \frac{{10}^{-3}\text{F}}{1\mu \text{F}}\right)\\ =0.777\text{C}\times \frac{{10}^3\mu \text{C}}{1\text{C}}\\ =777\mu \text{C}\end{array}$

Substitute $74\text{V}$ for $V_1$ and $129.5\text{V}$ for $V_2$ in equation (VI) to solve for $\epsilon$.

    $\begin{array}{c}\epsilon =74\text{V}+129.5\text{V}\\ \text{=203}\text{.5}\text{V}\end{array}$

Substitute $203.5\text{V}$ for $\epsilon$ and $6.00\mu \text{F}$ for $C_2$ in equation (VII) to solve for $Q_2$.

    $\begin{array}{c}{Q}_2=\left(6.00\mu \text{F}\times \frac{{10}^{-3}\text{F}}{1\mu \text{F}}\right)\left(203.5\text{V}\right)\\ =1.221\text{C}\times \frac{{10}^3\mu \text{C}}{1\text{C}}\\ =1221\mu \text{C}\end{array}$

Substitute $1221\mu \text{C}$ for $Q_2$ and $777\mu \text{C}$ for $q_2$ in equation (VIII) to solve for $\Delta Q_2$.

    $\begin{array}{c}\Delta Q_2=1221\mu \text{C}-777\mu \text{C}\\ =4\text{44}\mu \text{C}\end{array}$

Therefore, the change in charge of capacitor $C_2$ when the switch is open is $444\mu \text{C}$.