Chapter 25, Problem 58SP

A 2.0- $\mu \text{F}$ capacitor is charged to 50 V and then connected in parallel (positive plate to positive plate) with a 4.0- $\mu \text{F}$ capacitor charged to 100 V. (a) What are the final charges on the capacitors? (b) What is the potential difference across each?

To determine

The final charges on the capacitors when the $2.0\mu \text{F}$ capacitor and $4.0\mu \text{F}$ capacitor are joined in parallel.

Answer to Problem 58SP

Solution:

$0.17\text{ mC}$, $\text{0}\text{.33 mC}$

Explanation of Solution

Given data:

The $2.0\mu \text{F}$ capacitor, which is charged to $50\text{ V}$, is connected in parallel to $4.0\mu \text{F}$ capacitor, which is charged to $100\text{ V}$.

Formula used:

Write the expression for equivalent capacitance when the capacitors are connected in parallel:

$C_{eq}=C_1+C_2$

Here, $C_{eq}$ is the equivalent capacitance;$C_1$ and $C_2$ are the individual capacitances.

Write the expression for total charge when capacitors are connected in parallel:

$q=q_1+q_2$

Here, $q$ is the total charge;$q_1$ and $q_2$ are the charges on individual capacitances.

Write the expression for charge on capacitor:

$q=CV$

Here, $q$ is the charge on capacitor, $C$ is the capacitance, and $V$ is the voltage.

Explanation:

Recall expression for charge on capacitor 1:

$q_1=C_1{V}_1$

Here, $C_1$ is the capacitance of the capacitor 1 and $V_1$ is the voltage across the capacitor 1.

Substitute $2.0\mu \text{F}$ for $C_1$ and $50\text{ V}$ for $V_1$

$\begin{array}{c}{q}_1=\left(2.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)\right)\left(50\text{ V}\right)\\ =1\times {10}^{-4}\text{ C}\left(\frac{1\text{ mC}}{{10}^{-3}\text{ C}}\right)\\ =0.1\text{ mC}\end{array}$

Again, recall expression for charge on capacitor 2:

$q_2=C_2{V}_2$

Here, $C_2$ is the capacitance of the capacitor 2 and $V_2$ is the voltage across the capacitor 2.

Substitute $4.0\mu \text{F}$ for $C_2$ and $100\text{ V}$ for $V_2$

$\begin{array}{c}{q}_2=\left(4.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)\right)\left(100\text{ V}\right)\\ =4\times {10}^{-4}\text{ C}\left(\frac{1\text{ mC}}{{10}^{-3}\text{ C}}\right)\\ =0.4\text{ mC}\end{array}$

Recall the expression for equivalent capacitance when capacitors are connected in parallel:

$C_{eq}{}^{\prime }=C_1+C_2$

Here, $C_{eq}\text{'}$ is the equivalent capacitancewhen capacitors are connected in parallel.

Substitute $2.0\mu \text{F}$ for $C_1$ and $4.0\mu \text{F}$ for $C_2$

$\begin{array}{c}{C}_{eq}{}^{\prime }=\left(2\mu \text{F}\right)+\left(4\mu \text{F}\right)\\ =6\mu \text{F}\end{array}$

Recall the expression for total charge when capacitors are connected in parallel:

$q^{\prime }=q_1+q_2$

Here, $q\text{'}$ is the charge on capacitorwhen capacitors are connected in parallel.

Substitute $0.1\text{ mC}$ for $q_1$ and $0.4\text{ mC}$ for $q_2$

$\begin{array}{c}{q}^{\prime }=\left(0.1\text{ mC}\right)+\left(0.4\text{ mC}\right)\\ =0.5\text{ mC}\end{array}$

When the capacitors are joined in parallel, the voltage acrosseach capacitor is equal.

Recall the expression for charge on the system of capacitors in parallel:

$q_1{}^{\prime }=C_{eq}{}^{\prime }{V}^{\prime }$

Here, $V^{\prime }$ is the voltage across the capacitor.

Substitute $0.5\text{ mC}$ for $q^{\prime }$ and $6.0\mu \text{F}$ for $C_{eq}{}^{\prime }$

$0.5\text{ mC}=\left(6.0\mu \text{F}\right)V^{\prime }$

Solve for $V^{\prime }$

$\begin{array}{c}{V}^{\prime }=\frac{0.5\text{ mC}\left(\frac{{10}^{-3}\text{ C}}{1\text{ mC}}\right)}{6.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)}\\ =\frac{0.5\times {10}^{-3}\text{ C}}{6.0\times {10}^{-6}\text{ F}}\\ =0.0833\times {10}^3\text{ V}\\ =\text{83}\text{.3 V}\end{array}$

Recall the expression for charge on capacitor 1 when the connection is parallel:

$q_1{}^{\prime }=C_1{V}^{\prime }$

Substitute $2.0\mu \text{F}$ for $C_1$ and $83.33\text{ V}$ for $V^{\prime }$

$\begin{array}{c}{q}_1\text{'}=\left(2.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)\right)\left(83.33\text{V}\right)\\ =1.7\times {10}^{-4}\text{ C}\\ =\text{0}\text{.17}\times {\text{10}}^{-3}\text{ C}\left(\frac{1\text{ mC}}{{10}^{-3}\text{ C}}\right)\\ =\text{0}\text{.17 mC}\end{array}$

Recall the expression for charge on capacitor 2 when the connection is parallel:

$q_2{}^{\prime }=C_2{V}^{\prime }$

Substitute $4.0\mu \text{F}$ for $C_2$ and $83.33\text{ V}$ for $V^{\prime }$

$\begin{array}{c}{q}_2\text{'}=\left(4.0\mu \text{F}\left(\frac{{10}^{-6}\text{ F}}{1\mu \text{F}}\right)\right)\left(83.33\text{V}\right)\\ =3.3\times {10}^{-4}\text{ C}\\ =\text{0}\text{.33}\times {\text{10}}^{-3}\text{ C}\left(\frac{1\text{ mC}}{{10}^{-3}\text{ C}}\right)\\ =\text{0}\text{.33 mC}\end{array}$

Conclusion:

The final charges on the capacitors are $0.17\text{ mC}$ and $0.33\text{ mC}$.

To determine

The potential difference across each capacitor when the $2.0\mu \text{F}$ capacitor and $4.0\mu \text{F}$ capacitor are joined in parallel.

Answer to Problem 58SP

Solution:

$\text{83 V}$

Explanation of Solution

Given data:

The $2.0\mu \text{F}$ capacitor, which is charged to $50\text{ V}$, is connected in parallel to the $4.0\mu \text{F}$ capacitor, which is charged to $100\text{ V}$.

Formula used:

Explanation:

Recall that the voltage across system of capacitors when capacitors are connected in parallel combination from subpart (a):

$V\text{'}=83.3\text{V}$

From Fig 1, infer that when the capacitors are connected in parallel, the voltage across each capacitor is equal.

So, the voltage across each capacitor is same and equal to the voltage across the system of capacitors connected in parallel.

The potential difference across each capacitor is approximated as

$V\text{'}\simeq 83\text{V}$

Conclusion:

The potential difference across each capacitor is $83\text{ V}$.