Chapter 27, Problem 63AP

(a)

To determine

The resistance and resistivity of the wire in each case.

Answer to Problem 63AP

The potential difference between the plates of each capacitor is $\frac{Q}{4C}$.

Explanation of Solution

The c

Write the expression to obtain the total charge across both the capacitors when the switch is closed.

    $Q=Q_1+Q_2$                                                                                                              (I)

Here, $Q$ is the total charge across the both the capacitors, $Q_1$ is the charge on the left capacitor and $Q_2$ is the charge across the right capacitor.

Write the expression to obtain the potential difference across the capacitor.

    $\Delta V=\frac{Q}{C}$

Here, $\Delta V$ is the potential difference, $Q$ is the charge and $C$ is the capacitance of the capacitor.

Write the expression to obtain the voltage across left capacitor.

    $\Delta V_1=\frac{Q_1}{C}$                                                                                                                (II)

Here, $\Delta V_1$ is the voltage across left capacitor, $Q_1$ is the charge across left capacitor and $C$ is the capacitance of the left capacitor.

Write the expression to obtain the voltage across the right capacitor.

    $\Delta V_2=\frac{Q_2}{3C}$

Here, $\Delta V_2$ is the voltage across right capacitor, $Q_2$ is the charge across right capacitor and $3C$ is the capacitance of the right capacitor.

When switch is closed, the voltage across both the capacitor would be same.

    $\Delta V_1=\Delta V_2$

Substitute $\frac{Q_1}{C}$ for $\Delta V_1$ and $\frac{Q_2}{3C}$ for $\Delta V_2$ in the above expression.

    $\begin{array}{l}\frac{Q_1}{C}=\frac{Q_2}{3C}\\ Q_2=3Q_1\end{array}$

Substitute $Q-Q_1$ for $Q_2$ in the above expression.

    $\begin{array}{l}Q-Q_1=3Q_1\\ 4Q_1=Q\\ Q_1=\frac{Q}{4}\end{array}$                                                                                                           (III)

Substitute $\frac{Q}{4}$ for $Q_1$ in equation (II) to calculate $\Delta V_1$.

    $\begin{array}{c}\Delta V_1=\frac{\frac{Q}{4}}{C}\\ =\frac{Q}{4C}\end{array}$

The potential difference across each the capacitor is same.

Conclusion:

Therefore, the potential difference between the plates of each capacitor is $\frac{Q}{4C}$.

(b)

To determine

The charge on each capacitor.

Answer to Problem 63AP

The charge across capacitor whose capacitance is $\left(C\right)$ is $\frac{Q}{4}$ and charge across the capacitor whose capacitance is $\left(3C\right)$ is $\frac{3Q}{4}$.

Explanation of Solution

Consider equation (III),

    $Q_1=\frac{Q}{4}$

Substitute $\frac{Q}{4}$ for $Q_1$ in equation (I) to calculate $Q_2$.

    $\begin{array}{c}Q=\frac{Q}{4}+Q_2\\ Q_2=Q-\frac{Q}{4}\\ =\frac{3Q}{4}\end{array}$

Conclusion:

Therefore, the charge across capacitor whose capacitance is $\left(C\right)$ is $\frac{Q}{4}$ and charge across the capacitor whose capacitance is $\left(3C\right)$ is $\frac{3Q}{4}$.

(c)

To determine

The final energy stored in each capacitor.

Answer to Problem 63AP

The energy stored in capacitor whose capacitance is $\left(C\right)$ is $\frac{1}{32}\frac{Q^2}{C}$ and energy stored in capacitor whose capacitance is $\left(3C\right)$ is $\frac{3}{32}\frac{Q^2}{C}$.

Explanation of Solution

Write the expression to obtain the final energy stored in the capacitor.

    $U_{\text{f}}=\frac{1}{2}C{\left(\Delta V\right)}^2$                                                                                                     (IV)

Here, $U_{\text{f}}$ is the final energy stored in the capacitor, $C$ is the capacitance of the capacitor and $\Delta V$ is the potential difference across the capacitor.

Conclusion:

The energy stored across capacitor whose capacitance is $\left(C\right)$ is,

Substitute $\frac{Q}{4C}$ for $\Delta V$ in equation (IV) to calculate ${\left(U_{\text{f}}\right)}_{\text{C}}$.

    $\begin{array}{c}{\left(U_{\text{f}}\right)}_{\text{C}}=\frac{1}{2}C{\left(\frac{Q}{4C}\right)}^2\\ =\frac{1}{2}C\left(\frac{Q^2}{16C^2}\right)\\ =\frac{1}{32}\frac{Q^2}{C}\end{array}$

The energy stored across capacitor whose capacitance is $\left(3C\right)$ is,

Substitute $\frac{Q}{4C}$ for $\Delta V$ and $3C$ for $C$ in equation (IV) to calculate ${\left(U_{\text{f}}\right)}_{\text{3C}}$.

    $\begin{array}{c}{\left(U_{\text{f}}\right)}_{\text{3C}}=\frac{1}{2}3C{\left(\frac{Q}{4C}\right)}^2\\ =\frac{1}{2}3C\left(\frac{Q^2}{16C^2}\right)\\ =\frac{3}{32}\frac{Q^2}{C}\end{array}$

Conclusion:

Therefore, the energy stored in capacitor whose capacitance is $\left(C\right)$ is $\frac{1}{32}\frac{Q^2}{C}$ and energy stored in capacitor whose capacitance is $\left(3C\right)$ is $\frac{3}{32}\frac{Q^2}{C}$.

(d)

To determine

The internal energy in the resister.

Answer to Problem 63AP

The internal energy in the resister is $\frac{3}{8}\frac{Q^2}{C}$.

Explanation of Solution

Write the expression to obtain the internal energy in the resister.

    $E_{\mathrm{int}}=U-\left({\left(U_{\text{f}}\right)}_{\text{C}}+{\left(U_{\text{f}}\right)}_{\text{3C}}\right)$

Here, $E_{\mathrm{int}}$ is the internal energy in the resister, $U$ is the initial total energy across both the capacitor and ${\left(U_{\text{f}}\right)}_{\text{C}}$ is the final energy across capacitor whose capacitance $\left(C\right)$ and ${\left(U_{\text{f}}\right)}_{\text{3C}}$ is the final energy stored in the capacitor whose capacitance $\left(3C\right)$.

Substitute $\frac{1}{2}\frac{Q^2}{C}$ for $U$, $\frac{1}{32}\frac{Q^2}{C}$ for ${\left(U_{\text{f}}\right)}_{\text{C}}$ and $\frac{3}{32}\frac{Q^2}{C}$ for ${\left(U_{\text{f}}\right)}_{\text{3C}}$ in the above equation to calculate $E_{\mathrm{int}}$.

    $\begin{array}{c}{E}_{\mathrm{int}}=\frac{1}{2}\frac{Q^2}{C}-\left(\frac{1}{32}\frac{Q^2}{C}+\frac{3}{32}\frac{Q^2}{C}\right)\\ =\frac{1}{2}\frac{Q^2}{C}-\frac{4}{32}\frac{Q^2}{C}\\ =\frac{3}{8}\frac{Q^2}{C}\end{array}$

Conclusion:

Therefore, the internal energy in the resister is $\frac{3}{8}\frac{Q^2}{C}$.