Chapter 27, Problem 7PQ

In Figure P27.7, capacitor 1 (C₁ = 20.0 μF) initially has a potential difference of 50.0 V and capacitor 2 (C₂ = 5.00 μF) has none. The switches are then closed simultaneously. a. Find the final charge on each capacitor after a long time has passed. b. Calculate the percentage of the initial stored energy that was lost when the switches were closed.

FIGURE P27.7

To determine

The final charge on each capacitor.

Answer to Problem 7PQ

The final charges on each capacitor are $\underset{\_}{8.00\times {10}^2\text{μC , 2}.00\times {10}^2\text{μC}}$.

Explanation of Solution

Write the equation for the final charge in capacitor 1.

  $Q_1=C_1{V}_{\text{new}}$                                                                                                 (I)

Here, $Q_1$ is the charge on capacitor 1, $C_1$ is the capacitance of capacitor 1 and $V_{new}$ is the new potential difference.

Write the equation for the final charge in capacitor 2.

  $Q_2=C_2{V}_{\text{new}}$                                                                                                 (II)

Here, $Q_2$ is the charge on capacitor 2, $C_2$ is the capacitance of capacitor 2.

Write the expression for the equivalent capacitance.

$C_{eq}=C_1+C_2$ (III)

Here, $C_{eq}$ is the equivalent capacitance, $C_1$ is the capacitance 1 connected in the circuit and $C_2$ is the capacitance 2 connected in the circuit.

Write the expression for the new potential difference.

$V_{new}=\frac{Q}{C_{eq}}$ (IV)

Here, $Q$ is the initial charge.

Write the expression for the initial charge.

$Q=C_{1}V$ (V)

Here, $V$ is the battery emf.

Rewrite the expression for the final charge on capacitor 1.

$Q_1=C_1\left[\frac{\left(C_{1}V\right)}{\left(C_1+C_2\right)}\right]$ (VI)

Rewrite the expression for the final charge on capacitor 2.

$Q_2=C_2\left[\frac{\left(C_{1}V\right)}{\left(C_1+C_2\right)}\right]$ (VII)

Conclusion:

Substitute $20.0\text{μF}$ for $C_1$, $5.00\text{μF}$ for $C_2$, $50.0\text{V}$ for $V$ in Equation (VI) to find $Q_1$.

  $\begin{array}{c}{Q}_1=\left[\left(20.0\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\left[\frac{\left(\left[\left(20.0\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\left(50.0\text{V}\right)\right)}{\left(\left[\left(20.0\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]+\left[\left(5.00\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\right)}\right]\\ =\left[\left(20.0\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\left(40\text{V}\right)\\ =8.00\times {10}^2\text{μC}\end{array}$

Substitute $20.0\text{μF}$ for $C_1$, $5.00\text{μF}$ for $C_2$, $50.0\text{V}$ for $V$ in Equation (VII) to find $Q_2$.

$\begin{array}{c}{Q}_2=\left[\left(5.00\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\left[\frac{\left(\left[\left(20.0\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\left(50.0\text{V}\right)\right)}{\left(\left[\left(20.0\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]+\left[\left(5.00\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\right)}\right]\\ =\left[\left(5.00\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]\left(40\text{V}\right)\\ =2.00\times {10}^2\text{μC}\end{array}$

Thus, the final charges on each capacitor are $\underset{\_}{8.00\times {10}^2\text{μC , 2}.00\times {10}^2\text{μC}}$.

To determine

The percentage of the initial stored energy that was lost when the switches were closed.

Answer to Problem 7PQ

The percentage of the initial stored energy that was lost when the switches were closed is $\underset{\_}{20\%}$.

Explanation of Solution

Write the equation for the fraction of loss of energy.

  $f=\left[\frac{U_i-U_f}{U_i}\right]\times 100\%$ (VIII)

Here, $f$ is the fraction of

Write the expression for the initial energy,

$U_i=\frac{1}{2}{C}_1{V}^2$ (IX)

Write the expression for the final energy,

$U_f=\frac{1}{2}{C}_{eq}{V}^2$ (X)

Rewrite the expression for the fraction of loss of energy.

$\begin{array}{c}f=\left[\frac{\left[\frac{1}{2}{C}_1{V}^2\right]-\left[\frac{1}{2}\left(C_1+C_2\right)V^2\right]}{\left[\frac{1}{2}{C}_1{V}^2\right]}\right]\times 100\%\\ =\left[\frac{\left[C_1\right]-\left[\left(C_1+C_2\right)\right]}{\left[C_1\right]}\right]\times 100\%\end{array}$ (X)

Conclusion:

Substitute $20.0\text{μF}$ for $C_1$, $5.00\text{μF}$ for $C_2$ in Equation (X) to find $f$.

  $\begin{array}{c}f=\left[\frac{\left[\left(20.0\text{μF}\right)\right]-\left[\left(20.0\text{μF}+5.00\text{μF}\right)\right]}{\left[\left(20.0\text{μF}\right)\right]}\right]\times 100\%\\ =\left(0.200\right)\times 100\%\\ =20\%\end{array}$

Thus, the percentage of the initial stored energy that was lost when the switches were closed is $\underset{\_}{20\%}$.