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 26.26P

Find (a) the equivalent capacitance of the capacitors in Figure P26.26, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

Chapter 26, Problem 26.26P, Find (a) the equivalent capacitance of the capacitors in Figure P26.26, (b) the charge on each

(a)

Expert Solution
Check Mark
To determine

The equivalent capacitance of the capacitors.

Answer to Problem 26.26P

The equivalent capacitance of the capacitors in the system is 2.67μF .

Explanation of Solution

Given: The figure given below shows the arrangement of capacitors.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 26, Problem 26.26P

Figure 1

Write the equation for equivalent capacitance in the series combination.

1Ceq=1C1+1C2+C3+1C4

Here,

Ceq is the equivalent capacitance.

C1 is the capacitance of the first capacitor.

C2 is the capacitance of the second capacitor in the upper branch.

C3 is the capacitance of the third capacitor in the lower branch.

C4 is the capacitance of the fourth capacitor.

Substitute 8.00μF for C1 and 6.00μF for C2 , 2.00μF for C3 and 8.00μF for C4 in above equation to find Ceq .

1Ceq=1(8.00μF)+1(6.00μF)+(2.00μF)+1(8.00μF)Ceq=2.67μF

Conclusion:

Therefore, the equivalent capacitance of the capacitors in the system is 2.67μF .

(b)

Expert Solution
Check Mark
To determine

The charge on each capacitor.

Answer to Problem 26.26P

The charge on 8.00μF capacitor to the left, 6.00μF capacitor, 2.00μF capacitor and 8.00μF capacitor to the right are 24.0μC , 18.0μC , 6.00μC and 24.0μC respectively.

Explanation of Solution

Write the equation for charge on capacitor C1 .

Q1=Ceq(ΔV)

Here,

Q1 is the charge on capacitor C1 .

ΔV is the potential difference.

Substitute 2.67μF for Ceq and 9.00V for ΔV in above equation to find Q1 .

Q1=(2.67μF)(9.00V)=24.0μC

Thus, the charge on 8.00μF capacitor to the left is 24.0μC .

The charge on 8.00μF capacitor to the left and 8.00μF capacitor to the right is same due to the same potential difference.

The charge on capacitor C4 is,

Q4=24.0μC

Thus, the charge on 8.00μF capacitor to the right is 24.0μC .

Write the equation for charge on capacitor C2 .

Q2=C2(Q1C2+C3)

Here,

Q2 is the charge on capacitor C2 .

Substitute 6.00μF for C2 , 24.0μC for Q1 and 2.00μF for C3 in above equation to find Q2 .

Q2=(6.00μF)((24.0μC)(6.00μF)+(2.00μF))=18.0μC

Thus, the charge on 6.00μF capacitor is 18.0μC .

Write the equation for charge on capacitor C3 .

Q3=C3(Q1C2+C3)

Here,

Q3 is the charge on capacitor C3 .

Substitute 2.00μF for C3 , 24.0μC for Q1 and 6.00μF for C2 in above equation to find Q3 .

Q3=(2.00μF)((24.0μC)(6.00μF)+(2.00μF))=6.00μC

The charge on 2.00μF capacitor is 6.00μC .

Conclusion:

Therefore, the charge on 8.00μF capacitor to the left, 6.00μF capacitor, 2.00μF capacitor and 8.00μF capacitor to the right are 24.0μC , 18.0μC , 6.00μC and 24.0μC respectively.

(c)

Expert Solution
Check Mark
To determine

The potential difference across each capacitor.

Answer to Problem 26.26P

The potential difference across all the capacitors are same and equal to 3.00V .

Explanation of Solution

Write the equation for potential difference across capacitor C1 .

ΔV1=Q1C1

Here,

ΔV1 is the potential difference across 8.00μF to the left.

Substitute 24.0μC for Q1 and 8.00μF for C1 in above equation to find ΔV1 .

ΔV1=(24.0μC)(8.00μF)=3.00V

Thus, the potential difference across 8.00μF to the left is 3.00V .

The potential difference across 8.00μF capacitor to the left and 8.00μF capacitor to the right is same due to the same charge and capacitance.

The potential difference across capacitor C4 is,

ΔV4=3.00V

Thus, the potential difference across 8.00μF to the right is 3.00V .

Write the equation for potential difference across capacitor C2 .

ΔV2=Q1C2+C3

Here,

ΔV2 is the charge on capacitor C2 .

Substitute 24.0μC for Q1 , 6.00μF for C2 and 2.00μF for C3 in above equation to find ΔV2 .

ΔV2=(24.0μC)(6.00μF)+(2.00μF)=3.00V

Thus, the potential difference across 6.00μF is 3.00V .

The potential difference across the 2.00μF capacitor is same as the potential difference across 6.00μF due to the parallel combination.

The potential difference across capacitor C3 is,

ΔV3=3.00V

The potential difference across 2.00μF is 3.00V .

Conclusion:

Therefore, the potential difference across all the capacitors are same and equal to 3.00V .

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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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