Chapter 26, Problem 26.12OQ

(i) Rank the following five capacitors from greatest to smallest capacitance, noting any cases of equality, (a) a 20-μF capacitor with a 4-V potential difference between its plates (b) a 30-μF capacitor with charges of magnitude 90 μC on each plate (c) a capacitor with charges of magnitude 80 μC on its plates, differing by 2 V in potential. (d) a 10-μF capacitor storing energy 125 μJ (e) a capacitor storing energy 250 μJ with a 10-V potential difference (ii) Rank the same capacitors in part (i) from largest to smallest according to the potential difference between the plates, (iii) Rank the capacitors in part (i) in the order of the magnitudes of the charges on their plates, (iv) Rank the capacitors in part (i) in the order of the energy they store.

To determine

The rank of the five capacitors from greatest to smallest capacitance.

Answer to Problem 26.12OQ

The rank of the five capacitors from greatest to smallest capacitance is $\text{c}>\text{b}>\text{a}>\text{d}>\text{e}$.

Explanation of Solution

Given info: The capacitances of the cases (a), (b) and (d) are $20\text{μF}$, $30\text{μF}$ and $10\text{μF}$, the electric potential for case (a) is $4.0\text{V}$, the charge on the capacitor for case (b) is $90\text{μC}$, the magnitude of charge and potential difference for case (c) are $80\text{μC}$ and $2\text{V}$ respectively, the storing energy for cases (d) and (e) are $125\text{μJ}$ and $250\text{μJ}$ respectively and the electric potential for case (e) is $10\text{V}$.

Formula to calculate the capacitance of the capacitor is,

$C=\frac{Q}{V}$

Here,

$Q$ is the charge on the capacitor.

$V$ is the electric potential.

$C$ is the capacitance of the capacitor.

For case (a):

The capacitance of the first capacitor is,

$\begin{array}{c}{C}_1=20\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\\ =20\times {10}^{-6}\text{F}\end{array}$

Here,

$C_1$ is the capacitance of the first capacitor.

For case (b):

The capacitance of the second capacitor is,

$\begin{array}{c}{C}_2=30\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\\ =30\times {10}^{-6}\text{F}\end{array}$

Here,

$C_2$ is the capacitance of the second capacitor.

For case (c):

The capacitance of the third capacitor is,

$C_3=\frac{Q}{V}$

Here,

$C_3$ is the capacitance of the third capacitor.

Substitute $2\text{V}$ for $V_3$ and $80\text{μC}$ for $Q_3$ in above equation to find $C_3$.

$\begin{array}{c}{C}_3=\frac{80\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{2\text{V}}\\ =40\times {10}^{-6}\text{C}\end{array}$

Thus, the capacitance of the third capacitor is $40\times {10}^{-6}\text{C}$.

For case (d):

The capacitance of the fourth capacitor is,

$\begin{array}{c}{C}_4=10\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\\ =10\times {10}^{-6}\text{F}\end{array}$

Here,

$C_4$ is the capacitance of the fourth capacitor.

For case (e):

Formula to calculate the energy stored in a capacitor is,

$U_5=\frac{1}{2}{C}_5{V}_5{}^2$

Here,

$C_5$ is the capacitance of the fifth capacitor.

$V_5$ is the voltage across the fifth capacitor.

Substitute $250\text{μJ}$ for $U_5$ and $10\text{V}$ for $V_5$ in above equation to find $C_5$.

$\begin{array}{c}250\text{μJ}\left(\frac{{10}^{-6}\text{J}}{\text{1}\text{μJ}}\right)=\frac{1}{2}{C}_5{\left(10\text{V}\right)}^2\\ C_5=5\times {10}^{-6}\text{F}\end{array}$

Thus, the capacitance of the fifth capacitor is $5\times {10}^{-6}\text{F}$.

The rank of the capacitor is,

$\begin{array}{c}{C}_3>C_2>C_1>C_4>C_5\\ \text{c}>\text{b}>\text{a}>\text{d}>\text{e}\end{array}$

Conclusion:

Therefore, the rank of the five capacitors from greatest to smallest capacitance is $\text{c}>\text{b}>\text{a}>\text{d}>\text{e}$.

To determine

The rank of the five capacitors from greatest to smallest capacitance according to the potential difference between plates.

Answer to Problem 26.12OQ

The rank of the five capacitors from greatest to smallest capacitance according to the potential difference between plates is $\text{c}>\text{d}>\text{a}>\text{d}>\text{e}$.

Explanation of Solution

Given info: The capacitances of the cases (a), (b) and (d) are $20\text{μF}$, $30\text{μF}$ and $10\text{μF}$, the electric potential for case (a) is $4.0\text{V}$, the charge on the capacitor for case (b) is $90\text{μC}$, the magnitude of charge and potential difference for case (c) are $80\text{μC}$ and $2\text{V}$, the storing energy for cases (d) and (e) are $125\text{μJ}$ and $250\text{μJ}$ respectively and the electric potential for case (e) is $10\text{V}$.

For case (a):

The electric potential across the first capacitor is,

$V_1=4\text{V}$

$V_1$ is the voltage across the first capacitor.

For case (b):

The electric potential for across the second capacitor is,

$V_2=\frac{Q_2}{C_2}$

$V_2$ is the voltage across the second capacitor.

$Q_2$ is the charge on the second capacitor.

Substitute $30\text{μF}$ for $C_2$ and $90\text{μC}$ for $Q_2$ in above equation to find $V_2$.

$\begin{array}{c}{V}_2=\frac{90\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{30\text{μF}\left(\frac{{10}^{-6}\text{F}}{\text{1}\text{μF}}\right)}\\ =3\text{V}\end{array}$

Thus, the electric potential for across the second capacitor is $3\text{V}$.

For case (c):

The electric potential for across the third capacitor is,

$V_3=2\text{V}$

Here,

$V_3$ is the voltage across the third capacitor.

For case (d):

Formula to calculate the energy stored in the fourth capacitor is,

$U_4=\frac{1}{2}{C}_4{V}_4{}^2$

Here,

$V_4$ is the voltage across the fourth capacitor.

Substitute $250\text{μJ}$ for $U_4$ and $10\text{μF}$ for $C_4$ in above equation to find $V_4$.

$\begin{array}{c}250\text{μJ}\left(\frac{{10}^{-6}\text{J}}{1\text{μJ}}\right)=\frac{1}{2}\left(10\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right)V_4{}^2\\ V_4=5\text{V}\end{array}$

Thus, the energy stored in the fourth capacitor $5\text{V}$.

For case (e):

The electric potential across the fifth capacitor is,

$V_5=10\text{V}$

Here,

$V_5$ is the voltage across the fifth capacitor.

The rank of the electric potential from highest to lowest is,

$\begin{array}{c}{V}_5>V_4>V_1>V_2>V_3\\ \text{e}>\text{d}>\text{a}>\text{b}>\text{c}\end{array}$

From the above expression, the capacitance of the capacitor is inversely proportional to the square of voltage. Hence, the rank of the five capacitors from greatest to smallest capacitance according to the potential difference between plates is $\text{c}>\text{d}>\text{a}>\text{d}>\text{e}$.

Conclusion:

Therefore, the rank of the five capacitors from greatest to smallest capacitance according to the potential difference between plates is $\text{c}>\text{d}>\text{a}>\text{d}>\text{e}$.

To determine

The rank of the five capacitors from greatest to smallest capacitance in the order of the magnitude of charge is $\text{b}>\text{a}=\text{c}>\text{d}=\text{e}$.

Answer to Problem 26.12OQ

The rank of the five capacitors from greatest to smallest capacitance in the order of the magnitude of charge is $\text{b}>\text{a}=\text{c}>\text{d}=\text{e}$.

Explanation of Solution

Given info: The capacitances of the cases (a), (b) and (d) are $20\text{μF}$, $30\text{μF}$ and $10\text{μF}$, the electric potential for case (a) is $4.0\text{V}$, the charge on the capacitor for case (b) is $90\text{μC}$, the magnitude of charge and potential difference for case (c) are $80\text{μC}$ and $2\text{V}$, the storing energy for cases (d) and (e) are $125\text{μJ}$ and $250\text{μJ}$ respectively and the electric potential for case (e) is $10\text{V}$.

For case (a):

The charge across the first capacitor is,

$Q_1=C_1{V}_1$

Here,

$Q_1$ is the charge on the first capacitor.

Substitute $4.0\text{V}$ for $V_1$ and $20\text{μF}$ for $C_1$ in above equation to find $Q_1$.

$\begin{array}{c}{Q}_1=\left(20\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right)\left(4.0\text{V}\right)\\ =80\times {10}^{-6}\text{C}\end{array}$

Thus, the charge across the first capacitor is $80\times {10}^{-6}\text{C}$.

For case (b):

The charge across the second capacitor is,

$\begin{array}{c}{Q}_2=90\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\\ =90\times {10}^{-6}\text{C}\end{array}$

Here,

$Q_2$ is the charge on the second capacitor.

Thus, the charge across the second capacitor is $90\times {10}^{-6}\text{C}$.

For case (c):

The charge across the third capacitor is,

$\begin{array}{c}{Q}_3=80\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\\ =80\times {10}^{-6}\text{C}\end{array}$

Here,

$Q_3$ is the charge on the third capacitor.

Thus, the charge across the second capacitor is $80\times {10}^{-6}\text{C}$.

For case (d):

Formula to calculate the energy stored in the fourth capacitor is,

$U_4=\frac{Q_4{}^2}{2C_4}$

Here,

$Q_4$ is the charge on the fourth capacitor.

Substitute $125\text{μJ}$ for $U_4$ and $10\text{μF}$ for $C_4$ in above equation to find $Q_4$.

$\begin{array}{c}125\text{μJ}\left(\frac{{10}^{-6}\text{J}}{1\text{μJ}}\right)=\frac{Q_4{}^2}{2\left(10\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right)}\\ Q_4=50\times {10}^{-6}\text{C}\end{array}$

Thus, the charge across the fourth capacitor is $50\times {10}^{-6}\text{C}$.

For case (e):

Formula to calculate the energy stored in the fifth capacitor is,

$U_5=\frac{1}{2}{C}_5{V}_5{}^2$

Substitute $250\text{μJ}$ for $U_5$ and $10\text{V}$ for $V_5$ in above equation to find $C_5$.

$\begin{array}{c}250\text{μJ}\left(\frac{{10}^{-6}\text{J}}{1\text{μJ}}\right)=\frac{1}{2}\left(10\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right){\left(10\text{V}\right)}^2\\ C_5=5\times {10}^{-6}\text{F}\end{array}$

Thus, the magnitude of the fifth capacitor is $5\times {10}^{-6}\text{F}$.

The charge across the fifth capacitor is,

$Q_5=V_5{C}_5$

Here,

$Q_5$ is the charge on the fifth capacitor.

Substitute $10\text{V}$ for $V_5$ and $5\times {10}^{-6}\text{F}$ for $C_5$ in above equation to find $Q_5$.

$\begin{array}{c}{Q}_5=\left(10\text{V}\right)\left(5\times {10}^{-6}\text{F}\right)\\ =50\times {10}^{-6}\text{C}\end{array}$

Thus, the charge across the fifth capacitor is $50\times {10}^{-6}\text{C}$.

The rank of the charge from highest to lowest is,

$\begin{array}{c}{Q}_2>Q_1=Q_3>Q_4=Q_5\\ \text{b}>\text{a}=\text{c}>\text{d}=\text{e}\end{array}$

Since, the capacitance of the capacitor is proportional to the charge. Hence, the rank of the five capacitors from greatest to smallest capacitance in the order of the magnitude of charge is $\text{b}>\text{a}=\text{c}>\text{d}=\text{e}$.

Conclusion:

Therefore, the rank of the five capacitors from greatest to smallest capacitance in the order of the magnitude of charge is $\text{b}>\text{a}=\text{c}>\text{d}=\text{e}$.

To determine

The rank of energy stored of the five capacitors from greatest to smallest capacitance.

Answer to Problem 26.12OQ

The rank of energy stored of the five capacitors from greatest to smallest capacitance is $\text{e}>\text{a}>\text{b}>\text{d}>\text{c}$.

Explanation of Solution

Given info: The capacitances of the cases (a), (b) and (d) are $20\text{μF}$, $30\text{μF}$ and $10\text{μF}$, the electric potential for case (a) is $4.0\text{V}$, the charge on the capacitor for case (b) is $90\text{μC}$, the magnitude of charge and potential difference for case (c) are $80\text{μC}$ and $2\text{V}$, the storing energy for cases (d) and (e) are $125\text{μJ}$ and $250\text{μJ}$ respectively and the electric potential for case (e) is $10\text{V}$.

For case (a):

The energy stored of the first capacitor is,

$\begin{array}{c}{U}_1=\frac{1}{2}\left(20\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right){\left(4\text{V}\right)}_1{}^2\\ =160\times {10}^{-6}\text{J}\end{array}$

Here,

$U_1$ is the energy stored of the first capacitor.

Thus, the energy stored of the first capacitor is $160\times {10}^{-6}\text{J}$.

For case (b):

The energy stored of the first capacitor is,

$U_2=\frac{Q_2{}^2}{2C_2}$

Here,

$U_2$ is the energy stored of the second capacitor.

Substitute $90\text{μC}$ for $Q_2$ and $30\text{μF}$ for $C_2$ in above equation to find $U_2$.

$\begin{array}{c}{U}_2=\frac{{\left(90\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}^2}{2\left(20\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right)}\\ =135\times {10}^{-6}\text{J}\end{array}$

Thus, the energy stored of the second capacitor is $135\times {10}^{-6}\text{J}$.

For case (c):

The energy stored of the third capacitor is,

$U_3=\frac{Q_3{}^2}{2C_3}$

Here,

$U_3$ is the energy stored of the third capacitor.

Substitute $80\text{μC}$ for $Q_3$ and $40\text{μF}$ for $C_3$ in above equation to find $U_3$.

$\begin{array}{c}{U}_3=\frac{{\left(80\text{μC}\left(\frac{{10}^{-6}\text{C}}{1\text{μC}}\right)\right)}^2}{2\left(40\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right)}\\ =80\times {10}^{-6}\text{J}\end{array}$

Thus, the energy stored of the third capacitor is $80\times {10}^{-6}\text{J}$.

For case (d):

The energy stored of the fourth capacitor is,

$\begin{array}{c}{U}_4=125\text{μJ}\left(\frac{{10}^{-6}\text{J}}{\text{1}\text{μJ}}\right)\\ =125\times {10}^{-6}\text{J}\end{array}$

Here,

$U_4$ is the energy stored of the fourth capacitor.

Thus, the energy stored of the fourth capacitor is $125\times {10}^{-6}\text{J}$.

For case (e):

The energy stored of the fifth capacitor is,

$\begin{array}{c}{U}_5=250\text{μJ}\left(\frac{{10}^{-6}\text{J}}{\text{1}\text{μJ}}\right)\\ =250\times {10}^{-6}\text{J}\end{array}$

Here,

$U_5$ is the energy stored of the fifth capacitor.

Thus, the energy stored of the fourth capacitor is $250\times {10}^{-6}\text{J}$.

The rank of the charge from highest to lowest is,

$\begin{array}{c}{U}_5>U_1>U_2>U_4>U_3\\ \text{e}>\text{a}>\text{b}>\text{d}>\text{c}\end{array}$

Conclusion:

Therefore, the rank of energy stored of the five capacitors from greatest to smallest capacitance is $\text{e}>\text{a}>\text{b}>\text{d}>\text{c}$.