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Assignment 8: Capacitance
Due: 8:00am on Monday, February 6, 2012
Note: To understand how points are awarded, read your instructor's Grading Policy.
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Introduction to Capacitance
Learning Goal: To understand the meaning of capacitance and ways of calculating capacitance
When a positive charge is placed on a conductor that is insulated from ground, an electric field emanates from
the conductor to ground, and the conductor will have a nonzero potential relative to ground. If more charge is
placed on the conductor, this voltage will increase proportionately. The ratio of charge to voltage is called the
capacitance of this conductor: .
Capacitance is one of the central concepts in electrostatics, and specially constructed devices called capacitors are
essential elements of electronic circuits. In a capacitor, a second conducting surface is placed near the first (they
are often called electrodes), and the relevant voltage is the voltage between these two electrodes.
This tutorial is designed to help you understand capacitance by assisting you in calculating the capacitance of a
parallel-plate capacitor, which consists of two plates each of area separated by a small distance with air or
vacuum in between. In figuring out the capacitance of this configuration of conductors, it is important to keep in
mind that the voltage difference is the line integral of the electric field between the plates.
Part A
What property of objects is best measured by their capacitance?
ANSWER:
ability to conduct electric current
ability to distort an external electrostatic field
ability to store charge
Correct
Capacitance is a measure of the ability of a system of two conductors to store electric charge and energy.
Capacitance is defined as . This ratio remains constant as long as the system retains its geometry and the
amount of dielectric does not change. Capacitors are special devices designed to combine a large capacitance
with a small size. However, any pair of conductors separated by a dielectric (or vacuum) has some capacitance.
Even an isolated electrode has a small capacitance. That is, if a charge is placed on it, its potential with
respect to ground will change, and the ratio is its capacitance .
Part B
Assume that charge is placed on the top plate, and is placed on the bottom plate. What is the magnitude
of the electric field between the plates?
Hint B.1 How do you find the magnitude of the electric field?
Hint not displayed
Hint B.2 What is the electric flux integral due to the electric field?
Hint not displayed
Hint B.3 Find the enclosed charge
Hint not displayed
Hint B.4 Recall Gauss's law
Hint not displayed
Express in terms of and other quantities given in the introduction, in addition to and any other constants
needed.
ANSWER:
= Correct
Part C
What is the voltage between the plates of the capacitor?
Hint C.1 The electric field is the derivative of the potential
Hint not displayed
Express in terms of the quantities given in the introduction and any required physical constants.
ANSWER:
= Correct
Part D
Now find the capacitance of the parallel-plate capacitor.
Express in terms of quantities given in the introduction and constants like .
ANSWER:
= Correct
You have derived the general formula for the capacitance of a parallel-plate capacitor with plate area and plate
separation . It is worth remembering.
Part E
Consider an air-filled charged capacitor. How can its capacitance be increased?
Hint E.1 What does capacitance depend on?
Hint not displayed
ANSWER:
Increase the charge on the capacitor.
Decrease the charge on the capacitor.
Increase the spacing between the plates of the capacitor.
Decrease the spacing between the plates of the capacitor.
Increase the length of the wires leading to the capacitor plates.
Correct
Part F
Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance?
ANSWER:
Double the charge and double the plate area.
Double the charge and double the plate separation.
Halve the charge and double the plate separation.
Halve the charge and double the plate area.
Halve the plate separation; double the plate area.
Double the plate separation; halve the plate area.
Correct
A Capacitor with Thick Plates
Consider a capacitor that consists of two metal plates of nonzero thickness separated by a positive distance .
When such a capacitor is connected across the terminals of a battery with emf , two things happen
simultaneously: Charge flows from one plate to another, and a potential difference appears between the two plates.
In this problem, you will determine the amount and distribution of charge on each metallic plate for an ideal
(infinite) parallel-plate capacitor. There is competition between the repulsive forces from like charges on a plate
and the attractive forces from the opposite charge on the other plate. You will determine which force "wins."
Let , , , and be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge
densities to be positive for now. Some of them are in fact negative, and this will be revealed at the end of the
calculation.
Part A
If we assume that the metallic plates are perfect conductors, the electric field in their interiors must vanish. Given
that the electric field due to a charged sheet with surface charge is given by
,
and that it points away from the plane of the sheet, how can the condition that the electric field in plate I vanishes
be written?
Hint A.1 How to approach the problem
Hint not displayed
ANSWER:
Correct
Part B
Similarly, how can the condition that the electric field in plate II vanishes be written?
Hint B.1 How to approach the problem
Hint not displayed
ANSWER:
Correct
Part C
There is an electric field in the region between the two plates. The magnitude of this electric field is . This
imposes another condition on the charge densities on the surfaces of the plates. How can this condition be
expressed?
Hint C.1 How to approach the problem
Hint not displayed
ANSWER:
Correct
Part D
Finally, how can the condition that the total charge of the system is conserved be expressed?
Hint D.1 How to approach the problem
Hint not displayed
ANSWER:
Correct
You now have four equations in four variables. If you solve the four equations simultaneously you obtain
,
,
which is the familiar expression for the surface charge density on a parallel-plate capacitor. However, you have
now also established that all the charge resides on the insides of the capacitor plates.
Capacitors in Parallel
Learning Goal: To understand how to calculate capacitance, voltage, and charge for a parallel combination of
capacitors.
Frequently, several capacitors are connected together to form a collection of capacitors. We may be interested in
determining the overall capacitance of such a collection. The simplest configuration to analyze involves capacitors
connected in series or in parallel. More complicated setups can often (though not always!) be treated by combining
the rules for these two cases. Consider the example of a parallel combination of capacitors: Three capacitors are
connected to each other and to a battery as shown in the figure.
The individual capacitances are , , and , and the battery's voltage is .
Part A
If the potential of plate 1 is , then, in equilibrium, what are the potentials of plates 3 and 6? Assume that the
negative terminal of the battery is at zero potential.
Hint A.1 Electrostatic equilibrium
Hint not displayed
ANSWER:
and
and
and
and
Correct
Part B
If the charge of the first capacitor (the one with capacitance ) is , then what are the charges of the second and
third capacitors?
Hint B.1 Definition of capacitance
Hint not displayed
Hint B.2 Voltages across the capacitors
Hint not displayed
ANSWER:
and
and
and
and
Correct
Part C
Suppose we consider the system of the three capacitors as a single "equivalent" capacitor. Given the charges of
the three individual capacitors calculated in the previous part, find the total charge for this equivalent
capacitor.
Express your answer in terms of and .
ANSWER:
=
Correct
Part D
Using the value of , find the equivalent capacitance for this combination of capacitors.
Hint D.1 Using the definition of capacitance
Hint not displayed
Express your answer in terms of .
ANSWER:
=
Correct
The formula for combining three capacitors in parallel is
.
How do you think this formula may be generalized to capacitors?
Capacitors in Series
Learning Goal: To understand how to calculate capacitance, voltage, and charge for a combination of capacitors
connected in series.
Consider the combination of capacitors shown in the figure.
Three capacitors are connected to each other in series, and then to the battery. The values of the capacitances are
, , and , and the applied voltage is . Initially, all of the capacitors are completely discharged; after the
battery is connected, the charge on plate 1 is .
Part A
What are the charges on plates 3 and 6?
Hint A.1 Charges on capacitors connected in series
Hint not displayed
Hint A.2 The charges on a capacitor's plates
Hint not displayed
ANSWER:
and
and
and
and
0 and
0 and
Correct
Part B
If the voltage across the first capacitor (the one with capacitance ) is , then what are the voltages across the
second and third capacitors?
Hint B.1 Definition of capacitance
Hint not displayed
Hint B.2 Charges on the capacitors
Hint not displayed
ANSWER:
and
and
and
0 and
Correct
Part C
Find the voltage across the first capacitor.
Hint C.1 How to analyze voltages
Hint not displayed
Express your answer in terms of .
ANSWER:
= Correct
Part D
Find the charge on the first capacitor.
Express your answer in terms of and .
ANSWER:
= Correct
Part E
Using the value of just calculated, find the equivalent capacitance for this combination of capacitors in
series.
Hint E.1 Using the definition of capacitance
Hint not displayed
Express your answer in terms of .
ANSWER:
= Correct
The formula for combining three capacitors in series is
.
How do you think this formula may be generalized to capacitors?
Exercise 24.22
The Figure below shows a system of four capacitors, where the
potential difference across is 50.0 .
Part A
Find the equivalent capacitance of this system between and .
ANSWER:
= 3.47
Correct
Part B
How much charge is stored by this combination of capacitors?
ANSWER:
= 174
Correct
Part C
How much charge is stored in the 10.0- capacitor?
ANSWER:
= 174
Correct
Part D
How much charge is stored in the 9.0- capacitor?
ANSWER:
= 174
Correct
Problem 24.60
The capacitors in the Figure are initially uncharged and are
connected, as in the diagram, with switch open. The applied potential difference is = + 210 .
Part A
What is the potential difference ?
Express your answer using two significant figures.
ANSWER:
= 70
Correct
Part B
What is the potential difference after switch is closed?
ANSWER:
= 105
Correct
Part C
What is the potential difference after switch is closed?
ANSWER:
= 105
Correct
Part D
What is the potential difference after switch is closed?
ANSWER:
= 105
Correct
Part E
What is the potential difference after switch is closed?
ANSWER:
= 105
Correct
Part F
How much charge flowed through the switch when it was closed?
ANSWER:
= 315
Correct
± Capacitance and Electric Field of a Spherical Capacitor
A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner
sphere has a radius of = 12.5 , and the outer sphere has a radius of = 14.9 . A potential difference of
120 is applied to the capacitor.
Part A
What is the capacitance of the capacitor?
Hint A.1 How to approach the problem
Hint not displayed
Hint A.2 Which Gaussian surface to use
Hint not displayed
Hint A.3 Evaluate the integral in Gauss's law
Hint not displayed
Hint A.4 Find an expression for the potential difference
Hint not displayed
Hint A.5 Definiton of capacitance
Hint not displayed
Use = 8.85×10−12
for the permittivity of free space.
ANSWER:
= 8.63×10
−11
Correct
Part B
What is the magnitude of the electric field at radius 12.6 , just outside the inner sphere?
Hint B.1 How to approach the problem
Hint not displayed
Hint B.2 Calculate the charge on the inner sphere
Hint not displayed
ANSWER:
= 5870
Correct
Part C
What is the magnitude of at 14.6 , just inside the outer sphere?
Hint C.1 How to approach the problem
Hint not displayed
ANSWER:
= 4370
Correct
In a parallel-plate capacitor, the field between the plates is uniform. In a spherical capacitor, however, the field is
a function of the radial distance from the center, so the field is not uniform between the spheres.
In problem 24.48 the square plates have side lengths of 16 cm.
Problem 24.48
A parallel-plate air capacitor is made by using two plates 16 square, spaced 3.7 apart. It is connected to a
12- battery.
Part A
What is the capacitance?
Express your answer using two significant figures.
ANSWER:
= 6.1×10
−11
Correct
Part B
What is the charge on each plate?
Express your answer using two significant figures.
ANSWER:
= 7.4×10
−10
Correct
Part C
What is the electric field between the plates?
Express your answer using two significant figures.
ANSWER:
= 3200
Correct
Part D
What is the energy stored in the capacitor?
Express your answer using two significant figures.
ANSWER:
= 4.4×10
−9
Correct
Part E
The battery is disconnected and then the plates are pulled apart to a separation of 7.4 .
What is the capacitance in this case?
Express your answer using two significant figures.
ANSWER:
= 3.1×10
−11
Correct
Part F
What is the charge on each plate in this case?
Express your answer using two significant figures.
ANSWER:
= 7.4×10
−10
Correct
Part G
What is the electric field between the plates in this case?
Express your answer using two significant figures.
ANSWER:
= 3200
Correct
Part H
What is the energy stored in the capacitor?
Express your answer using two significant figures.
ANSWER:
= 8.8×10
−9
Correct