Chapter 18, Problem 129P

(a)

To determine

Find the total energy stored in two capacitor.

Answer to Problem 129P

The total energy stored in two capacitor is $175\text{ μJ}$.

Explanation of Solution

Write the equation for energy.

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

Here, $U$ is the energy, $C$ is the capacitance and $V$ is potential difference.

Conclusion:

For $2.00\text{ μF}$ capacitor, substitute $5.00\text{ V}$ for $V$ and $2.00\text{ μF}$ for $C$ in equation I.

$\begin{array}{c}{U}_1=\frac{1}{2}\left(2.00\text{ μF}\right){\left(5.00\text{ V}\right)}^2\\ =25\text{ μJ}\end{array}$

For $3.00\text{ μF}$ capacitor, substitute $10.0\text{ V}$ for $V$ and $3.00\text{ μF}$ for $C$ in equation I.

$\begin{array}{c}{U}_2=\frac{1}{2}\left(3.00\text{ μF}\right){\left(10.0\text{ V}\right)}^2\\ =150\text{ μJ}\end{array}$

Then the total energy stored is $\begin{array}{c}U=U_1+U_2\\ =25\text{ μJ}+150\text{ μJ}\\ =175\text{ μJ}\end{array}$

Thus, the total energy stored in two capacitor is $175\text{ μJ}$.

(b)

To determine

Find the total energy in the two capacitor and charge on each capacitor after disconnected from batteries.

Answer to Problem 129P

The total energy in the two capacitor is $160\text{ μJ}$ and charge on each capacitor after disconnected from batteries are $\text{16}\text{.0 μC}$ and $\text{24}\text{.0 μC}$.

Explanation of Solution

Write the equation for total charge.

$Q=C_2{V}_{2i}+C_3{V}_{3i}$ (II)

Here, $Q$ is the total charge, $C_2$ is the capacitance of $2.00\text{ μF}$ capacitor, $C_3$ is the capacitance of $3.00\text{ μF}$ capacitor, $V_{2i}$ is voltage across $2.00\text{ μF}$ capacitor and $V_{3i}$ is voltage across $3.00\text{ μF}$ capacitor.

Since the two capacitors are connected together the voltage across them is same, then the charge on each capacitor are,

$Q_{2f}=C_{2}V$ and $Q_{3f}=C_{3}V$

From the above two equation eliminate $V$.

$\frac{Q_{2f}}{C_2}=\frac{Q_{3f}}{C_3}$ (III)

The total charge is written as,

$Q=Q_{2f}+Q_{3f}$ (IV)

Substitute the value of $Q_{3f}$ from the above equation to equation III.

$\frac{Q_{2f}}{C_2}=\frac{Q_{3f}}{C_3}=\frac{Q-Q_{2f}}{C_3}$

Solve the above equation to get $Q_{2f}$.

$\begin{array}{c}{Q}_{2f}\left(\frac{1}{C_2}+\frac{1}{C_3}\right)=\frac{Q}{C_3}\\ Q_{2f}=\frac{Q}{C_3\left(\frac{1}{C_2}+\frac{1}{C_3}\right)}\end{array}$

$Q_{2f}=\frac{Q}{\frac{C_3}{C_2}+1}$ (V)

Write the equation for total charge.

$U_{total}=\frac{1}{2}{C}_2{V}^2+\frac{1}{2}{C}_3{V}^2$ (VI)

Conclusion:

Substitute $2.00\text{ μF}$ for $C_2$, $3.00\text{ μF}$ for $C_3$, $5.00\text{ V}$ for $V_{2i}$ and $10.0\text{ V}$ for $V_{3i}$  in equation II.

$\begin{array}{c}Q=\left(2.00\text{ μF}\right)\left(5.00\text{ V}\right)+\left(3.00\text{ μF}\right)\left(10.0\text{ V}\right)\\ =40.0\text{ μC}\end{array}$

Substitute $40.0\text{ μC}$ for $Q$, $2.00\text{ μF}$ for $C_2$ and $3.00\text{ μF}$ for $C_3$ in equation V.

$\begin{array}{c}{Q}_{2f}=\frac{40.0\text{ μC}}{\frac{3.00\text{ μF}}{2.00\text{ μF}}+1}\\ =16.0\text{ μC}\end{array}$

Substitute $16.0\text{ μC}$ for $Q_{2f}$ and $40.0\text{ μC}$ for $Q$ in equation IV.

$\begin{array}{c}{Q}_{3f}=40.0\text{ μC}-16.0\text{ μC}\\ =24.0\text{ μC}\end{array}$

Find the value of $V$.

$V=\frac{Q_{2f}}{C_2}=\frac{16.0\text{ μC}}{2.00\text{ μF}}=8.00\text{ V}$

Substitute $8.00\text{ V}$ for $V$, $2.00\text{ μF}$ for $C_2$ and $3.00\text{ μF}$ for $C_3$ in equation VI

$\begin{array}{c}{U}_{total}=\frac{1}{2}\left(2.00\text{ μF}\right){\left(8.00\text{ V}\right)}^2+\frac{1}{2}\left(3.00\text{ μF}\right){\left(8.00\text{ V}\right)}^2\\ =160\text{ μJ}\end{array}$

Therefore, the total energy in the two capacitor is $160\text{ μJ}$ and charge on each capacitor after disconnected from batteries are $\text{16}\text{.0 μC}$ and $\text{24}\text{.0 μC}$.

(c)

To determine

Explain the reason for missing energy.

Answer to Problem 129P

The missing energy is due to the appearance of internal energy in the wires.

Explanation of Solution

The missing energy is due to the internal energy caused by the wires connecting the capacitors during the current flowing time.

Conclusion:

Therefore, the missing energy is due to the appearance of internal energy in the wires.