Chapter 21, Problem 69P

(a)

To determine

The charge on the capacitor $C_1$ in the electric circuit.

Answer to Problem 69P

The charge on the capacitor $C_1$ in the electric circuit is $\underset{\_}{222\mu \text{C}}$.

Explanation of Solution

The switch is closed in the electric circuit and the current exists in a simple series circuit as shown in figure $1$.

Write the expression for the power delivered to the resistor.

    $P=I^2{R}_2$        (I)

Here, $P$ is the power delivered to the resistor, $I$ is the current, $R_2$ is the resistance.

Use equation (I) to solve for $I$.

    $I=\sqrt{\frac{P}{R_2}}$        (II)

Write the expression for the potential difference across the resistor $R_1$.

    $\Delta V_1=I{R}_1$        (III)

Here, $\Delta V_1$ is the potential difference across the resistor, $R_1$ is the resistance.

Write the expression for the charge on the capacitor $C_1$.

    $Q_1=C_1\Delta V_1$        (IV)

Here, $C_1$ is the capacitance, $Q_1$ is the charge on the $C_1$ .

Conclusion:

Substitute $2.40\text{W}$ for $P$, $7.00\text{k}\Omega$ for $R_2$ in equation (II) to find $I$.

    $\begin{array}{c}I=\sqrt{\frac{2.40\text{W}}{7.00\text{k}\Omega \times \frac{{10}^3\Omega }{1\text{k}\Omega }}}\\ =18.5\times {10}^{-3}\text{A}\\ \text{=18}\text{.5}\text{mA}\end{array}$

Substitute $18.5\text{mA}$ for $I$, $4.00\text{k}\Omega$ for $R_1$ in equation (III) to find $\Delta V_1$.

    $\begin{array}{c}\Delta V_1=\left(18.5\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(4.00\text{k}\Omega \times \frac{{10}^3\Omega }{1\text{k}\Omega }\right)\\ =74.1\text{V}\end{array}$

Substitute $3.00\mu \text{F}$ for $C_1$, $74.1\text{V}$ for $\Delta V_1$ in equation (IV) to find $Q_1$.

    $\begin{array}{c}{Q}_1=\left(3.00\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\left(74.1\text{V}\right)\\ =222\times {10}^{-6}\text{C}\\ \text{=222}\mu \text{C}\end{array}$

Therefore, the charge on the capacitor $C_1$ in the electric circuit is $\underset{\_}{222\mu \text{C}}$.

(b)

To determine

The amount of charge on the capacitor $C_2$ is changed.

Answer to Problem 69P

The amount of charge on the capacitor $C_2$ is changed is $\underset{\_}{444\mu \text{C}}$.

Explanation of Solution

Consider the switch is closed to find the emf of the battery and the charge in the capacitor $C_2$.

Write the expression for the potential difference across $R_2$.

    $\Delta V_2=I{R}_2$        (V)

Write the expression for the charge on the capacitor $C_2$.

    $Q_2=C_2\Delta V_2$        (VI)

Write the expression for the emf of the battery.

    $V=I{R}_{\text{eq}}$        (VII)

Here, $V$ is the emf of the battery, $R_{\text{eq}}$ is the equivalent resistance.

Write the expression for $R_{\text{eq}}$.

    $R_{\text{eq}}=R_1+R_2$        (VIII)

Use equation (VIII) in (VII) to solve for $V$.

    $V=I\left(R_1+R_2\right)$        (IX)

Here, $V$ is the emf of the battery.

After the switch is opened, no current exists. The potential difference across each resistor is zero. The emf of the battery appears across both capacitors.

Write the expression for the new charge on the $C_2$.

    $Q=C_{2}V$        (X)

Write the expression for the amount of the charge on the capacitor is changed $C_2$.

    $\Delta Q=Q-Q_2$        (XI)

Here, $\Delta Q$ is the change in the charge on the $C_2$.

Conclusion:

Substitute $18.5\text{mA}$ for $I$, $7.0\text{k}\Omega$ for $R_2$ in equation (V) to find $\Delta V_2$.

    $\begin{array}{c}\Delta V_2=\left(18.5\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left(7.00\text{k}\Omega \times \frac{{10}^3\Omega }{1\text{k}\Omega }\right)\\ =130\text{V}\end{array}$

Substitute $6.00\mu \text{F}$ for $C_2$, $130\text{V}$ for $\Delta V_2$ in equation (VI) to find $Q_2$.

    $\begin{array}{c}{Q}_2=\left(6.00\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\left(130\text{V}\right)\\ =778\times {10}^{-6}\text{C}\\ \text{=778}\mu \text{C}\end{array}$

Substitute $18.5\text{mA}$ for $I$, $4.00\text{k}\Omega$ for $R_1$, $7.00\text{k}\Omega$ in equation (IX) to find $V$.

    $\begin{array}{c}V=\left(18.5\text{mA}\times \frac{{10}^{-3}\text{A}}{1\text{mA}}\right)\left[\left(4.00\text{k}\Omega \times \frac{{10}^3\Omega }{1\text{k}\Omega }\right)+\left(7.00\text{k}\Omega \times \frac{{10}^3\Omega }{1\text{k}\Omega }\right)\right]\\ =204\text{V}\end{array}$

Substitute $6.00\mu \text{F}$ for $C_2$, $204\text{V}$ for $V$ in equation (X) to find $Q$.

    $\begin{array}{c}Q=\left(6.00\mu \text{F}\times \frac{{10}^{-6}\text{F}}{1\mu \text{F}}\right)\left(204\text{V}\right)\\ =1222\times {10}^{-6}\text{C}\\ \text{=1222}\mu \text{C}\end{array}$

Substitute $1222\mu \text{C}$ for $Q$, $778\mu \text{C}$ for $Q_2$ in equation (XI) to find $\Delta Q$.

    $\begin{array}{c}\Delta Q=\left(1222\mu \text{C}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)-\left(778\mu \text{C}\times \frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\\ =444\times {10}^{-6}\text{C}\\ \text{=444}\mu \text{C}\end{array}$

Therefore, the amount of charge on the capacitor $C_2$ is changed is $\underset{\_}{444\mu \text{C}}$.