Physics for Scientists and Engineers (AP Edition)
Physics for Scientists and Engineers (AP Edition)
9th Edition
ISBN: 9781133953951
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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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.

(i)

Expert Solution
Check Mark
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 c>b>a>d>e .

Explanation of Solution

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

Formula to calculate the capacitance of the capacitor is,

C=QV

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,

C1=20μF(106F1μF)=20×106F

Here,

C1 is the capacitance of the first capacitor.

For case (b):

The capacitance of the second capacitor is,

C2=30μF(106F1μF)=30×106F

Here,

C2 is the capacitance of the second capacitor.

For case (c):

The capacitance of the third capacitor is,

C3=QV

Here,

C3 is the capacitance of the third capacitor.

Substitute 2V for V3 and 80μC for Q3 in above equation to find C3 .

C3=80μC(106C1μC)2V=40×106C

Thus, the capacitance of the third capacitor is 40×106C .

For case (d):

The capacitance of the fourth capacitor is,

C4=10μF(106F1μF)=10×106F

Here,

C4 is the capacitance of the fourth capacitor.

For case (e):

Formula to calculate the energy stored in a capacitor is,

U5=12C5V52

Here,

C5 is the capacitance of the fifth capacitor.

V5 is the voltage across the fifth capacitor.

Substitute 250μJ for U5 and 10V for V5 in above equation to find C5 .

250μJ(106J1μJ)=12C5(10V)2C5=5×106F

Thus, the capacitance of the fifth capacitor is 5×106F .

The rank of the capacitor is,

C3>C2>C1>C4>C5c>b>a>d>e

Conclusion:

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

(ii)

Expert Solution
Check Mark
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 c>d>a>d>e .

Explanation of Solution

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

For case (a):

The electric potential across the first capacitor is,

V1=4V

V1 is the voltage across the first capacitor.

For case (b):

The electric potential for across the second capacitor is,

V2=Q2C2

V2 is the voltage across the second capacitor.

Q2 is the charge on the second capacitor.

Substitute 30μF for C2 and 90μC for Q2 in above equation to find V2 .

V2=90μC(106C1μC)30μF(106F1μF)=3V

Thus, the electric potential for across the second capacitor is 3V .

For case (c):

The electric potential for across the third capacitor is,

V3=2V

Here,

V3 is the voltage across the third capacitor.

For case (d):

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

U4=12C4V42

Here,

V4 is the voltage across the fourth capacitor.

Substitute 250μJ for U4 and 10μF for C4 in above equation to find V4 .

250μJ(106J1μJ)=12(10μF(106F1μF))V42V4=5V

Thus, the energy stored in the fourth capacitor 5V .

For case (e):

The electric potential across the fifth capacitor is,

V5=10V

Here,

V5 is the voltage across the fifth capacitor.

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

V5>V4>V1>V2>V3e>d>a>b>c

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 c>d>a>d>e .

Conclusion:

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

(iii)

Expert Solution
Check Mark
To determine

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

Explanation of Solution

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

For case (a):

The charge across the first capacitor is,

Q1=C1V1

Here,

Q1 is the charge on the first capacitor.

Substitute 4.0V for V1 and 20μF for C1 in above equation to find Q1 .

Q1=(20μF(106F1μF))(4.0V)=80×106C

Thus, the charge across the first capacitor is 80×106C .

For case (b):

The charge across the second capacitor is,

Q2=90μC(106C1μC)=90×106C

Here,

Q2 is the charge on the second capacitor.

Thus, the charge across the second capacitor is 90×106C .

For case (c):

The charge across the third capacitor is,

Q3=80μC(106C1μC)=80×106C

Here,

Q3 is the charge on the third capacitor.

Thus, the charge across the second capacitor is 80×106C .

For case (d):

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

U4=Q422C4

Here,

Q4 is the charge on the fourth capacitor.

Substitute 125μJ for U4 and 10μF for C4 in above equation to find Q4 .

125μJ(106J1μJ)=Q422(10μF(106F1μF))Q4=50×106C

Thus, the charge across the fourth capacitor is 50×106C .

For case (e):

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

U5=12C5V52

Substitute 250μJ for U5 and 10V for V5 in above equation to find C5 .

250μJ(106J1μJ)=12(10μF(106F1μF))(10V)2C5=5×106F

Thus, the magnitude of the fifth capacitor is 5×106F .

The charge across the fifth capacitor is,

Q5=V5C5

Here,

Q5 is the charge on the fifth capacitor.

Substitute 10V for V5 and 5×106F for C5 in above equation to find Q5 .

Q5=(10V)(5×106F)=50×106C

Thus, the charge across the fifth capacitor is 50×106C .

The rank of the charge from highest to lowest is,

Q2>Q1=Q3>Q4=Q5b>a=c>d=e

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 b>a=c>d=e .

Conclusion:

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

(iv)

Expert Solution
Check Mark
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 e>a>b>d>c .

Explanation of Solution

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

For case (a):

The energy stored of the first capacitor is,

U1=12(20μF(106F1μF))(4V)12=160×106J

Here,

U1 is the energy stored of the first capacitor.

Thus, the energy stored of the first capacitor is 160×106J .

For case (b):

The energy stored of the first capacitor is,

U2=Q222C2

Here,

U2 is the energy stored of the second capacitor.

Substitute 90μC for Q2 and 30μF for C2 in above equation to find U2 .

U2=(90μC(106C1μC))22(20μF(106F1μF))=135×106J

Thus, the energy stored of the second capacitor is 135×106J .

For case (c):

The energy stored of the third capacitor is,

U3=Q322C3

Here,

U3 is the energy stored of the third capacitor.

Substitute 80μC for Q3 and 40μF for C3 in above equation to find U3 .

U3=(80μC(106C1μC))22(40μF(106F1μF))=80×106J

Thus, the energy stored of the third capacitor is 80×106J .

For case (d):

The energy stored of the fourth capacitor is,

U4=125μJ(106J1μJ)=125×106J

Here,

U4 is the energy stored of the fourth capacitor.

Thus, the energy stored of the fourth capacitor is 125×106J .

For case (e):

The energy stored of the fifth capacitor is,

U5=250μJ(106J1μJ)=250×106J

Here,

U5 is the energy stored of the fifth capacitor.

Thus, the energy stored of the fourth capacitor is 250×106J .

The rank of the charge from highest to lowest is,

U5>U1>U2>U4>U3e>a>b>d>c

Conclusion:

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

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