EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100461260
Author: SERWAY
Publisher: YUZU
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Chapter 26, Problem 26.56AP

For (he system of four capacitors shown in Figure P26.19. find (a) the total energy stored in the system and (b) the energy stored by each capacitor, (c) (Compare the sum of the answers in part (b) with your result to part (a) and explain your observation.

(a)

Expert Solution
Check Mark
To determine

The energy stored in the system.

Answer to Problem 26.56AP

The energy stored in the system is 1.35×102J .

Explanation of Solution

Given info: The electric potential in the system is 90.0V .

The Figure of the circuit diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 26, Problem 26.56AP

Figure (1)

Since, the capacitors C1 and C2 of first row are in series. So, the equivalent capacitance of first row for series connection is,

1Ceq=1C1+1C2

Here,

Ceq is the equivalent capacitance for series connection.

Substitute 3.00μF for C1 and 6.00μF for C2 in above equation to find Ceq .

1Ceq=13.00μF+16.00μFCeq=2.00μF

Thus, the equivalent capacitance of the first row is 2.00μF .

Since, the capacitors C3 and C4 of the second row are in series. So, the equivalent capacitance of the second row is,

1Ceq'=1C3+1C4

Here,

Ceq' is the equivalent capacitance of the second row.

Substitute 2.00μF for C3 and 4.00μF for C4 in above equation to find Ceq' .

1Ceq'=12.00μF+14.00μFCeq'=43μF

Thus, the equivalent capacitance of the second row is 43μF .

Since, the first and second rows are in parallel to each other. So, the equivalent capacitance of the system is,

Ceq''=Ceq+Ceq'

Here,

Ceq'' is the equivalent capacitance of the system.

Substitute 43μF for Ceq' and 2.00μF for Ceq in above equation to find Ceq'' .

Ceq''=2.00μF+43μF=3.33μF

Thus, the equivalent capacitance of the system is 3.33μF .

Formula to calculate the energy stored in the system is,

U=12Ceq''(ΔV)2

Here,

U is the energy stored in the system.

ΔV is the potential difference.

Substitute 3.33μF for Ceq'' and 90.0V for ΔV in above equation to find U .

U=12(3.33μF)(90.0V)2=1.348×102J1.35×102J

Conclusion:

Therefore, the energy stored in the system is 1.35×102J .

(b)

Expert Solution
Check Mark
To determine

The energy stored by each capacitor.

Answer to Problem 26.56AP

The energy stored in capacitors C1 , C2 , C3 and C4 are 5.4×103J , 2.7×103J , 3.6×103J and 1.8×103J respectively.

Explanation of Solution

Given info: The electric potential in the system is 90.0V .

Since, the electric potential through each capacitor is same.

Formula to calculate the charge on the first row of capacitor is,

Q1=CeqV

Here,

Q1 is the charge on the first row of capacitor.

Substitute 2.00μF for Ceq and 90.0V for V in above equation to find Q1 .

Q1=(2.00μF)(90.0V)=180μC

Thus, the charge on the first row of capacitor is 180μC .

Formula to calculate the energy stored is,

U=Q22C

Here,

C is the capacitance of the capacitor.

Q is the charge through first row of capacitor.

The energy stored in capacitor C1 is,

U1=Q122C1

Here,

U1 is the energy stored in capacitor C1 .

Substitute 180μC for Q1 and 3.00μF for C1 in above equation to find U1 .

U1=(180μC(106C1μC))22(3.00μF(106F1μF))=5.4×103J

Thus, the energy stored in capacitor C1 is 5.4×103J .

The energy stored in capacitor C2 is,

U2=Q122C2

Here,

U2 is the energy stored in capacitor C2 .

Substitute 180μC for Q1 and 3.00μF for C2 in above equation to find U2 .

U2=(180μC(106C1μC))22(6.00μF(106F1μF))=2.7×103J

Thus, the energy stored in capacitor C2 is 2.7×103J .

Formula to calculate the charge on the first row of capacitor is,

Q2=Ceq'V

Here,

Q2 is the charge on the second row of capacitor.

Substitute 43μF for Ceq' and 90.0V for V in above equation to find Q2 .

Q1=43μF(90.0V)=120μC

Thus, the charge on the first row of capacitor is 120μC .0

The energy stored in capacitor C3 is,

U3=Q222C3

Here,

U2 is the energy stored in capacitor C3 .

Substitute 180μC for Q2 and 3.00μF for C3 in above equation to find U3 .

U3=(120μC(106C1μC))22(2.00μF(106F1μF))=3.6×103J

Thus, the energy stored in capacitor C3 is 3.6×103J .

The energy stored in capacitor C4 is,

U4=Q222C4

Here,

U4 is the energy stored in capacitor C4 .

Substitute 180μC for Q2 and 4.00μF for C4 in above equation to find U4 .

U4=(120μC(106C1μC))22(4.00μF(106F1μF))=1.8×103J

Thus, the energy stored in capacitor C4 is 1.8×103J .

Conclusion:

Therefore, the energy stored in capacitors C1 , C2 , C3 and C4 are 5.4×103J , 2.7×103J , 3.6×103J and 1.8×103J respectively.

(c)

Expert Solution
Check Mark
To determine

The comparison of the sum of the energy in part (b) with energy of part (a) and explain it.

Answer to Problem 26.56AP

The sum of the energy of par (b) is same as the energy in part (a).

Explanation of Solution

Given info: The electric potential in the system is 90.0V .

The sum of the energy of the part (b) is,

U'=U1+U2+U3+U4

Substitute 5.4×103J fir U1 , 2.7×103J for U2 , 3.6×103J for U3 and 1.8×103J for U4 in above equation to find U' .

U'=(5.4×103J)+(2.7×103J)+(3.6×103J)+(1.8×103J)=1.35×102J

The sum of the energy of par (b) is same as the energy in part (a).

Conclusion:

Therefore, the sum of the energy of par (b) is same as the energy in part (a).

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Chapter 26 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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? 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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? 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