Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
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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Chapter 26 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 26 - Assume a device is designed to obtain a large...Ch. 26 - (i) What happens to the magnitude of the charge...Ch. 26 - A capacitor with very large capacitance is in...Ch. 26 - A parallel-plate capacitor filled with air carries...Ch. 26 - (i) A battery is attached to several different...Ch. 26 - A parallel-plate capacitor is charged and then is...Ch. 26 - (i) Rank the following five capacitors from...Ch. 26 - True or False? (a) From the definition of...Ch. 26 - You charge a parallel-plate capacitor, remove it...Ch. 26 - (a) Why is it dangerous to touch the terminals of...Ch. 26 - Assume you want to increase the maximum operating...Ch. 26 - If you were asked to design a capacitor in which...Ch. 26 - Prob. 26.4CQCh. 26 - Explain why the work needed to move a particle...Ch. 26 - An air-filled capacitor is charged, then...Ch. 26 - The sum of the charges on both plates of a...Ch. 26 - Because the charges on the plates of a...Ch. 26 - (a) When a battery is connected to the plates of a...Ch. 26 - Two conductors having net charges of +10.0 C and...Ch. 26 - (a) How much charge is on each plate of a 4.00-F...Ch. 26 - An air-filled parallel-plate capacitor has plates...Ch. 26 - A 50.0-in length of coaxial cable has an inner...Ch. 26 - (a) Regarding (lie Earth and a cloud layer 800 m...Ch. 26 - When a potential difference of 150 V is applied to...Ch. 26 - Prob. 26.8PCh. 26 - An air-filled capacitor consists of two parallel...Ch. 26 - A variable air capacitor used in a radio tuning...Ch. 26 - An isolated, charged conducting sphere of radius...Ch. 26 - Review. A small object of mass m carries a charge...Ch. 26 - Two capacitors, C1 = 5.00 F and C2 = 12.0 F, are...Ch. 26 - What If? The two capacitors of Problem 13 (C1 =...Ch. 26 - Find the equivalent capacitance of a 4.20-F...Ch. 26 - Prob. 26.16PCh. 26 - According to its design specification, the timer...Ch. 26 - Why is the following situation impossible? A...Ch. 26 - For the system of four capacitors shown in Figure...Ch. 26 - Three capacitors are connected to a battery as...Ch. 26 - A group of identical capacitors is connected first...Ch. 26 - (a) Find the equivalent capacitance between points...Ch. 26 - Four capacitors are connected as shown in Figure...Ch. 26 - Consider the circuit shown in Figure P26.24, where...Ch. 26 - Find the equivalent capacitance between points a...Ch. 26 - Find (a) the equivalent capacitance of the...Ch. 26 - Two capacitors give an equivalent capacitance of...Ch. 26 - Two capacitors give an equivalent capacitance of...Ch. 26 - Consider three capacitors C1, C2. and C3 and a...Ch. 26 - The immediate cause of many deaths is ventricular...Ch. 26 - A 12.0-V battery is connected to a capacitor,...Ch. 26 - A 3.00-F capacitor is connected to a 12.0-V...Ch. 26 - As a person moves about in a dry environment,...Ch. 26 - Two capacitors, C1 = 18.0 F and C2 = 36.0 F, are...Ch. 26 - Two identical parallel-plate capacitors, each with...Ch. 26 - Two identical parallel-plate capacitors, each with...Ch. 26 - Two capacitors, C1 = 25.0 F and C2 = 5.00 F, are...Ch. 26 - A parallel-plate capacitor has a charge Q and...Ch. 26 - Review. A storm cloud and the ground represent the...Ch. 26 - Consider two conducting spheres with radii R1 and...Ch. 26 - Review. The circuit in Figure P26.41 (page 804)...Ch. 26 - A supermarket sells rolls of aluminum foil,...Ch. 26 - (a) How much charge can be placed 011 a capacitor...Ch. 26 - The voltage across an air-filled parallel-plate...Ch. 26 - Determine (a) the capacitance and (b) the maximum...Ch. 26 - A commercial capacitor is to be constructed as...Ch. 26 - A parallel-plate capacitor in air has a plate...Ch. 26 - Each capacitor in the combination shown in Figure...Ch. 26 - A 2.00-nF parallel-plate capacitor is charged to...Ch. 26 - A small rigid object carries positive and negative...Ch. 26 - An infinite line of positive charge lies along the...Ch. 26 - A small object with electric dipole moment p is...Ch. 26 - The general form of Gausss law describes how a...Ch. 26 - Find the equivalent capacitance of' the group of...Ch. 26 - Four parallel metal plates P1, P2, P3, and P4,...Ch. 26 - For (he system of four capacitors shown in Figure...Ch. 26 - A uniform electric field E = 3 000 V/m exists...Ch. 26 - Two large, parallel metal plates, each of area A,...Ch. 26 - A parallel-plate capacitor is constructed using a...Ch. 26 - Why is the following situation impossible? A...Ch. 26 - Prob. 26.61APCh. 26 - A parallel-plate capacitor with vacuum between its...Ch. 26 - A 10.0-F capacitor is charged to 15.0 V. It is...Ch. 26 - Assume that the internal diameter of the...Ch. 26 - Two square plates of sides are placed parallel to...Ch. 26 - (a) Two spheres have radii a and b, and their...Ch. 26 - A capacitor of unknown capacitance has been...Ch. 26 - A parallel-plate capacitor of plate separation d...Ch. 26 - Prob. 26.69APCh. 26 - Example 25.1 explored a cylindrical capacitor of...Ch. 26 - To repair a power supply for a stereo amplifier,...Ch. 26 - The inner conductor of a coaxial cable has a...Ch. 26 - Some physical systems possessing capacitance...Ch. 26 - Consider two long, parallel, and oppositely...Ch. 26 - Determine the equivalent capacitance of the...Ch. 26 - A parallel-plate capacitor with plates of area LW...Ch. 26 - Calculate the equivalent capacitance between...Ch. 26 - A capacitor is constructed from two square,...
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