Chapter 26, Problem 12OQ

(i)

To determine

The order of the capacitance from greatest to smallest.

Answer to Problem 12OQ

The order of the capacitance from greatest to smallest is $\left(\text{c}\right)>\left(\text{b}\right)>\left(\text{a}\right)>\left(\text{d}\right)>\left(\text{e}\right)$.

Explanation of Solution

Assume $C_1$, $C_2$, $C_3$, $C_4$ and $C_5$ as the capacitance in part $\left(\text{a}\right)$, $\left(\text{b}\right)$, $\left(\text{c}\right)$, $\left(\text{d}\right)$ and $\left(\text{e}\right)$ respectively.

Assume $Q_1$, $Q_2$, $Q_3$, $Q_4$ and $Q_5$ as the charge in part $\left(\text{a}\right)$, $\left(\text{b}\right)$, $\left(\text{c}\right)$, $\left(\text{d}\right)$ and $\left(\text{e}\right)$ respectively.

Assume $V_1$, $V_2$, $V_3$, $V_4$ and $V_5$ as the voltage in part $\left(\text{a}\right)$, $\left(\text{b}\right)$, $\left(\text{c}\right)$, $\left(\text{d}\right)$ and $\left(\text{e}\right)$ respectively.

Assume $U_1$, $U_2$, $U_3$, $U_4$ and $U_5$ as the stored energy in part $\left(\text{a}\right)$, $\left(\text{b}\right)$, $\left(\text{c}\right)$, $\left(\text{d}\right)$ and $\left(\text{e}\right)$ respectively.

Write the equation for the capacitance.

    $C=\frac{Q}{V}$                                                                                                                       (I)

Here, $C$ is the capacitance, $Q$ is the charge and $V$ is the voltage.

Write the equation for the relation between the capacitance and the energy stored.

    $\begin{array}{c}U=\frac{1}{2}C{V}^2\\ C=\frac{2U}{V^2}\end{array}$                                                                                                                (II)

Conclusion:

In the part $\left(\text{a}\right)$.

    $C_1=20\mu \text{F}$

In the part $\left(\text{b}\right)$.

    $C_2=30\mu \text{F}$

In the part $\left(\text{c}\right)$.

Substitute $2\text{-V}$ for $V$ and $80\mu \text{C}$ for $Q$ in the equation (I) to calculate the capacitance.

    $\begin{array}{c}{C}_3=\left(\frac{80\mu \text{C}}{2\text{-V}}\right)\\ =40\mu \text{F}\end{array}$

In the part $\left(\text{d}\right)$.

    $C_4=10\mu \text{F}$

In the part $\left(\text{e}\right)$.

Substitute $250\mu \text{J}$ for $U$ and $10\text{-V}$ for $V$ in the equation (II) to calculate the capacitance.

    $\begin{array}{c}{C}_5=\left(\frac{2\times 250\mu \text{J}}{{\left(10\text{-V}\right)}^2}\right)\\ =\left(\frac{500\mu \text{J}}{\left(100\text{-V}\right)}\right)\\ =5\mu \text{F}\end{array}$

Therefore, the order of the capacitance from greatest to smallest is $\left(\text{c}\right)>\left(\text{b}\right)>\left(\text{a}\right)>\left(\text{d}\right)>\left(\text{e}\right)$.

(ii)

To determine

The order of the potential difference from greatest to smallest.

Answer to Problem 12OQ

The order of the potential difference from greatest to smallest is $\left(\text{e}\right)>\left(\text{d}\right)>\left(\text{a}\right)>\left(\text{b}\right)>\left(\text{c}\right)$.

Explanation of Solution

Write the equation for the potential difference.

    $V=\frac{Q}{C}$                                                                                                                     (III)

Here, $C$ is the capacitance, $Q$ is the charge and $V$ is the voltage.

Write the equation for the relation between the potential difference and the energy stored.

    $\begin{array}{c}U=\frac{1}{2}C{V}^2\\ V=\sqrt{\frac{2U}{C}}\end{array}$                                                                                                             (IV)

Conclusion:

In the part $\left(\text{a}\right)$.

    $V_1=4\text{-V}$

In the part $\left(\text{b}\right)$.

Substitute $90\mu \text{C}$ for $Q$ and $30\mu \mathrm{F}$ for $C$ in the equation (III) to calculate the potential difference.

    $\begin{array}{c}{V}_2=\left(\frac{90\mu \text{C}}{30\mu \text{F}}\right)\\ =3\text{V}\end{array}$

In the part $\left(\text{c}\right)$.

    $V_3=2\text{V}$

In the part $\left(\text{d}\right)$.

Substitute $125\mu \text{J}$ for $U$ and $10\mu \text{F}$ for $C$ in the equation (IV) to calculate the potential difference.

    $\begin{array}{c}{V}_4=\left(\sqrt{\frac{2\times 125\mu \text{J}}{10\mu \text{F}}}\right)\\ =\left(\sqrt{25}\right)\text{V}\\ =5\text{V}\end{array}$

In the part $\left(\text{e}\right)$.

    $V_5=10\text{-V}$

Therefore, the order of the potential difference from greatest to smallest is $\left(\text{e}\right)>\left(\text{d}\right)>\left(\text{a}\right)>\left(\text{b}\right)>\left(\text{c}\right)$.

(iii)

To determine

The order of the magnitudes of the charges from greatest to smallest.

Answer to Problem 12OQ

The order of the magnitudes of the charges from greatest to smallest is $\left(\text{b}\right)>\left(\text{a}\right)=\left(\text{c}\right)>\left(\text{d}\right)=\left(\text{e}\right)$.

Explanation of Solution

Write the equation for the magnitude of the charge.

    $Q=CV$                                                                                                                    (V)

Here, $C$ is the capacitance, $Q$ is the charge and $V$ is the voltage.

Write the equation for the relation between the magnitude of the charge and the energy stored.

    $\begin{array}{c}Q=CV\\ =\frac{2U}{V^2}\cdot V\\ =\frac{2U}{V}\end{array}$                                                                                                              (VI)

Rewrite the above equation to calculate the magnitude of the charge.

    $\begin{array}{c}Q=\frac{2U}{\left(\frac{Q}{C}\right)}\\ Q^2=2UC\\ Q=\sqrt{2UC}\end{array}$                                                                                                           (VII)

Conclusion:

In the part $\left(\text{a}\right)$.

Substitute $20\mu \text{F}$ for $C$ and $4\text{-V}$ for $V$ in the equation (V) to calculate the magnitude of the charge.

    $\begin{array}{c}{Q}_1=\left[\left(20\mu \text{F}\right)\left(4\text{-V}\right)\right]\\ =80\mu \text{C}\end{array}$

In the part $\left(\text{b}\right)$.

    $Q_2=90\mu \text{C}$

In the part $\left(\text{c}\right)$.

    $Q_3=80\mu \text{C}$

In the part $\left(\text{d}\right)$.

Substitute $125\mu \text{J}$ for $U$ and $10\mu \text{F}$ for $C$ in the equation (VII) to calculate the magnitude of the charge.

    $\begin{array}{c}{Q}_4=\sqrt{2\left(125\mu \text{J}\right)10\mu \text{F}}\\ =\sqrt{2500}\mu \text{C}\\ =5\text{0}\mu \text{C}\end{array}$

In the part $\left(\text{e}\right)$.

Substitute $250\mu \text{J}$ for $U$ and $10\text{-V}$ for $V$ in the equation (VI) to calculate the magnitude of the charge.

    $\begin{array}{c}{Q}_5=\left[\frac{2\left(250\mu \text{J}\right)}{10\text{-V}}\right]\\ =5\text{0}\mu \text{C}\end{array}$

Therefore, the order of the magnitude of the charge from greatest to smallest is $\left(\text{b}\right)>\left(\text{a}\right)=\left(\text{c}\right)>\left(\text{d}\right)=\left(\text{e}\right)$.

(iv)

To determine

The order of the energy stored from greatest to smallest.

Answer to Problem 12OQ

The order of the energy stored from greatest to smallest is $\left(\text{e}\right)>\left(\text{a}\right)>\left(\text{b}\right)>\left(\text{d}\right)>\left(\text{c}\right)$.

Explanation of Solution

Write the equation for the energy stored.

    $U=\frac{1}{2}C{V}^2$                                                                                                           (VIII)

Write the equation for the relation between the energy stored and the charge.

    $U=\frac{Q^2}{2C}$                                                                                                                   (IX)

Write the equation for the relation between the energy stored, the charge and the potential.

    $U=\frac{QV}{2}$                                                                                                                   (X)

Conclusion:

In the part $\left(\text{a}\right)$.

Substitute $20\mu \text{F}$ for $C$ and $4\text{-V}$ for $V$ in the equation (VIII) to calculate the energy stored.

    $\begin{array}{c}{U}_1=\left[\frac{1}{2}\left(20\mu \text{F}\right){\left(4\text{-V}\right)}^2\right]\\ =160\mu \text{J}\end{array}$

In the part $\left(\text{b}\right)$.

Substitute $30\mu \text{F}$ for $C$ and $90\mu \text{C}$ for $Q$ in the equation (IX) to calculate the energy stored.

    $\begin{array}{c}{U}_2=\left[\frac{{\left(90\mu \text{C}\right)}^2}{2\left(30\mu \text{F}\right)}\right]\\ \text{=135}\mu \text{J}\end{array}$

In the part $\left(\text{c}\right)$.

Substitute $2\text{-V}$ for $V$ and $80\mu \text{C}$ for $Q$ in the equation (X) to calculate the energy stored

    $\begin{array}{c}{U}_3=\left[\frac{\left(80\mu \text{C}\right)\left(2\text{-V}\right)}{2}\right]\\ =8\text{0}\mu \text{J}\end{array}$

In the part $\left(\text{d}\right)$.

    $U_4=125\mu \text{J}$

In the part $\left(\text{e}\right)$.

    $U_5=250\mu \text{J}$

Therefore, the order of the energy stored from greatest to smallest is $\left(\text{e}\right)>\left(\text{a}\right)>\left(\text{b}\right)>\left(\text{d}\right)>\left(\text{c}\right)$.