UNIT II - Mr. Greenberg Physics · Web viewPHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 26 FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed.,

Unit IV – CAPACITORS & DIELECTRICS

References:PHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 26

FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed., Ch. 26

Unit ObjectivesWhen you have completed Unit IV you should be able to:

1. Define the terms CAPACITOR and CAPACITANCE and use these definitions to relate the capacitance, difference in potential (aka: voltage), and the charge of a capacitor.

2. Derive and apply the expressions for the capacitance of capacitors having planar, cylindrical or spherical symmetry.

3. Determine the equivalent capacitance of a set of capacitors connected together, and determine the charge stored on each and the voltage across each capacitor of the set.

4. Determine the energy stored in a capacitor or a combination of capacitors.

5. Describe the effect on a capacitor's capacitance, the charge stored, the voltage across the capacitor, the energy stored, as well as the electric field in the capacitor, if the space between its conductors contains a dielectric material.

IV-1


Capacitors are important devices in electronic circuits, electrical machinery, etc...., since they are devices which store charge and electrical energy. The charge and energy may be stored for a short period of time in an alternating current circuit or for a longer time until it is needed, as in the circuitry that operates the strobe light for a camera. You at one time have probably been a capacitor by storing up enough charge on your body so that when you touch another conductor you get “zapped”. In this unit, we will cover the basic definitions and physics involved in capacitors.

Imagine two isolated conductors having equal but opposite charges. The difference in potential between the two objects, V, is directly proportional to the charge on the objects. This should be rather obvious since doubling the charge also doubles the electric field in the region around the objects and hence, the work-per-unit-charge (that is, the potential difference, V) required to move a test charge between two points in the field also doubles. We can write then, that

or,

Where C is a constant of proportionality. Our two equal but oppositely charged objects together are what we call a CAPACITOR, since as long as they are isolated they have a capacity for storing charge. The constant C is called the CAPACITANCE of our capacitor where

.As can be seen from the equation, the capacitance is a measure of how much charge can be stored in the capacitor per unit potential difference between the objects. Note that Q is not the net charge on the capacitor (since Qnet = 0) but is the magnitude of the charge stored on either conductor. Therefore, since Q is (+) and we will take V to mean the magnitude of the potential difference between the conductors, hence the capacitance, C, will always be a (+) number.

As can be seen from the relation C = Q/V, the MKSA unit of capacitance is the coulomb/volt or C/V. The C/V has been given a special name:

A capacitance of 1 F is sort of uncommon as we’ll see later. More common units are:

IV-2


Now for a brief recap:

1. A capacitor consists of two conducting hunks of stuff having equal but opposite charges.

2. If the magnitude of the charge on either hunk is Q and the potential difference between the hunks (taken positively) is V, then the charge stored on either hunk is proportional to V. This means that if the charge stored on each hunk is +2 and –2 coulombs when the voltage between them is 1 volts, then at 2 volts it will have +4 and –4 coulombs stored at 5 volts

+10 and –10 coulombs, etc...3. The capacitance = C = Q/V and is a measure of a particular capacitor’s ability to store

charge.

Example: Given a 2 F and a 5 F capacitor. With a 5 volt potential difference across each capacitor the 2 F will be storing

of charge on each “hunk” and the 5 F will be storing 25 C, etc. Thus you see that the capacitance (C) tells you how much charge will be stored per volt of potential difference between the terminals of the capacitor.

=============================================

Given the two identical conducting spheres shown at the right having equal but opposite charges. Their centers are separated by a distance R. Find the capacitance of the system.

For answers & solutions see “In-Text Fill-in Answers” beginning on page IV-19

Solution:

V1 = potential at the center of #1 = (IV-3 #1)

(don’t forget to include the potential of #1 itself)

V2 = Potential at the center of #2 = (IV-3 #2)

(don’t forget to include the potential of #2 itself)

|∆V| = (IV-3 #3)

C = Q/∆V = (IV-3 #4)

The MKSA units of you answer should be . Show that these units reduce to farads.

(IV-3 #5)

IV-3


The Parallel Plate CapacitorInstead of two “hunks” of conducting stuff, let’s now go to the specific. Let our “hunks” be two parallel conducting sheets. We can insure that they have equal but opposite charges by connecting a battery across the plates. When the battery is connected charge will flow onto the plates until the ∆V across the plates is the same as the ∆V of the battery. In doing this, we have stored some charge on the plates and thus have stored some electrical potential energy. Therefore, the system qualifies as a capacitor.

Let’s determine the capacitance of our parallel plate capacitor.

In terms of the between the plates and the plate separation, d, the potential difference between parallel oppositely charged sheets is given by:

∆V = (Hint: See Unit III, p. III-27 problem 6b) (IV-4 #1)

The -field between parallel oppositely charged plates in terms of , the magnitude of the charge on either plate, Q, and the area of each plate A is given by:

E = (IV-4 #2) Hint: Remember = charge/area, see Unit III, p. III-25 problem 3(iii). (You should be able to quickly derive

this with Gauss’ Law using a “pillbox.” You may have to do it from scratch on an exam!)

Combining the two expressions above and eliminating E, we get:

∆V = (IV-4 #3)

Since capacitance is C = Q/∆V, the capacitance of a parallel plate capacitor is

C = _____________ (IV-4 #4)

This relation is, I hope, as you would expect since if A is increased, the larger area enables more charge to be stored, or if the plates are moved closer together (d made smaller), the increased attraction between the charges enables more charge to be stored.

IV-4


It was mentioned earlier in this unit that a 1 F capacitor was kinda rare. Let’s see why.

Suppose we had two square parallel plates separated by a distance of 1 mm. If the capacitance of this system is 1 Farad, find the length of one side of the plates.

Solution - For the parallel plates, C is given by

C = _____________ (IV-5 #1)

For a square of side L the area, A = ________ (IV-5 #2). Combining these two relations and

solving for L in terms of , C, and d:

L = _____________ (IV-5 #3)

Substituting and calculating L:

L = _____________ meters or _____________ miles (IV-5 #4)

See why a 1 F parallel plate capacitor is kinda uncommon?

The Coaxial CapacitorA coaxial capacitor is composed of a solid conducting cylinder of length L surrounded by a cylindrical conducting shell also of length L as shown. Find the capacitance.

Let’s work this one out in gory detail from scratch.

a. The battery is connected as shown resulting in the cylinder being given a charge of ____Q and the outer

(+, –)shell being given a charge of ____Q. As a result, the (+, –)

-field between the cylinder and the shell is directed ____________. (IV-5 #5)

b. In the space below use Gauss’ Law to find the magnitude of at a distance r from the axis where a < r < b. (If you get stuck, see Unit III, p. III-29

problem 13)

E = ________________ (IV-5 #6)

IV-5


c. In the space below use to find the difference in potential between the cylinder and the shell. Do the integration in the direction that makes ∆V come out (+). (Again, if you get stuck, see your Unit III, p. III-29 problem 13)

∆V = ________________ (IV-6 #1)

d. Use C = Q/∆V and calculate the capacitance.

C = ________________ (IV-6 #2)

Again, just as we would expect:

1) As L increases, the area of the cylinder and outer shell increases, thereby ________________ the capacitor’s charge storing ability. Hence, C gets (increasing, decreasing) __________ as L increases. (IV-6 #3)

(larger/smaller)

2) As a and b approach one another in size (the cylinder-shell gap decreases) the charge storing ability of the capacitor _______________ and hence C gets ___________

(increases, decreases) (larger/smaller) as (b – a) decreases. (IV-6 #4)

For more examples of this type, see Examples 26.1, 26.2, and 26.3 on p. 807, 808 and 809 in Serway & Beichner.

At this point try problems 1 through 3 at the back of this unit.

IV-6


Capacitor CircuitsFirst let’s take the simplest circuit possible involving a capacitor; that of a capacitor with each side connected to the terminals of a battery. Initially as the battery is connected (+) charge will flow from the (+) terminal of the battery to the upper side of the capacitor and (–) charge from the (–) terminal to the lower side. Eventually, when the battery can no longer supply enough energy to move any more charges to the plates of the capacitor, the charges come to rest. This means that the upper plate, wire a-e-b, and the (+) terminal of the battery form an equipotential surface as does the lower plate, wire d-f-c, and the (–) terminal of the battery. Hence, the potential differences ∆Vca, ∆Vfb, ∆Vde, ∆Vda, and ∆Vcb are all equal. In other words, the potential difference across the battery is the same as the potential difference across the capacitor.

Since all parts of the circuit have a charge [upper part (+), lower part (–)] there must be an -field in the region around the circuit as shown in the drawing at the right. Since the

-field is a conservative field then the work done moving a charge around any closed loop is zero. That is:

Remember integrate around any arbitrary closed loop.

Now since , then the sum of all the ∆Vs around the closed loop equals

. Or in shorter notation: .

Let’s show what this means in the circuit at the top of the page.

∆Vda is an increase in potential and is therefore (+).∆Vaeb = 0 since the wire is an equipotential surface.∆Vbc is a decrease in potential and is therefore (–).∆Vcfd = 0 since the wire is an equipotential surface.

Now if these are summed, we will be adding all the ∆Vs around a closed loop. Doing this we get:

or∆Vda = ∆Vbc which is the same result arrived at previously.

Remember this idea:

, it will be very useful.

IV-7


Capacitors Connected In SeriesSuppose we have three capacitors C1, C2, and C3 connected IN SERIES as shown in the drawing at the right. When the battery is connected, charge flows onto plates A and B. Since plates G and D and the wire connecting them are electrically isolated from the rest of the circuit, no charge can flow into or out of the dotted loop. The battery then just separates the charge on this part of the circuit. The charge on A induces an equal but opposite charge on G, leaving a charge on D that is equal and opposite to the charge on G. D then induces on equal but opposite charge on E… and so on. The point is that no matter what the capacitances, the charge on each capacitor is the same. That is,

Q1 = Q2 = Q3 (since they’re the same, call ‘em Q).

Now recalling that :

or

From this we can write

And since , we can replace the three capacitors with one that has a capacitance given

by and thereby simplify the circuit from to

Generally then, given n capacitors connected in SERIES can be replaced by an equivalent capacitance, Ceq, given by:

IV-8


Sample Exercise IV-1: Three capacitors whose capacitances are 5, 10, and 50 µF are connected in series across a 12 V battery as shown below.

a. Find the capacitance of a single equivalent capacitor that when connected to the battery would store the same charge.

Solution:

(IV-9 #1)

__________ F (IV-9 #2)

b. Find the charge on each of the three capacitors.

Solution: From the arguments on the previous page, the all have the same charge, Q,which can be calculated from:

Q = __________ C (IV-9 #3)

c. Find the ∆V across each capacitor.

Solution: ∆V is given by Q/C for the capacitor.

(IV-9 #4)

Note that ∆Vbattery = ∆V1 + ∆V2 + ∆V3.

Capacitors Connected In Parallel The three capacitors sketched at the right are connected IN PARALLEL with a battery of voltage ∆Vb across its terminals. Using the same ideas as we did for the capacitors connected in series, namely, that the sum of the ∆Vs around a closed loop as zero, we can see that in loop 1: ∆Vb = ∆V1 where ∆V1 is the potential difference across capacitor C1. Similarly, in loop 2: ∆V1 = ∆V2 and in loop 3: ∆V2 = ∆V3 where ∆V2 and ∆V3 are the potential differences across C2 and C3, respectively. Thus we see that the potential difference (or voltage) across a bunch of capacitors connected in parallel is the same.

IV-9


Note that since the capacitance is given by C = Q/∆V, then if then the charges

. Here we have the main difference between capacitors connected in series and those connected in parallel.

For n capacitors in series: Q1 = Q2 = Q3 = . . . = Qn

For n capacitors in parallel: ∆V1 = ∆V2 = ∆V3 = . . . = ∆Vn

In our circuit above, note that after the battery has been connected and the situation has become static, the total charge, Q, that has been supplied by the battery is

Q = Q1 + Q2 + Q3

Where Q1, Q2, and Q3 are the charges on each capacitor.

Now since C = Q/∆V we can write

Q = C1∆V1 + C2∆V2 + C3∆V3

And since ∆V1 = ∆V2 = ∆V3 = ∆Vb

Q = ∆Vb(C1 + C2 + C3)

or

This expression indicates that the tree capacitors can be replaced by one equivalent capacitor, Ceq with a capacitance given by

Ceq = C1 + C2 + C3.

Generally then, if we have n capacitors all connected in parallel, we can replace them with a single capacitor whose capacitance is given by

Sample Exercise IV-2: Suppose the same three capacitors used in the previous example (5 µF, 10 µF, and 50 µF) are connected in parallel as shown at the right.

a) Find the capacitance of a single equivalent capacitor that when connected to the battery would store the same charge as all three capacitors.

Solution:

Ceq = C1 + C2 + C3 = ( _______ ) + ( _______ ) + ( _______ )

__________ F (IV-10 #1)

IV-10

Unit IV – CAPACITORS & DIELECTRICSIV-11


b) Find the voltage across each capacitor.

Solution: The voltage across each is the same as ∆Vbatt. Therefore

∆V1 = ∆V2 = ∆V3 = ∆Vbatt = __________ (IV-11 #1)

c) Find the charge stored on each capacitor.

Solution: Q = C∆V thus:

Q1 = C1∆V1 = ( _______ ) ( _______ ) = __________ C

Q2 = C2∆V2 = ( _______ ) ( _______ ) = __________ C

Q3 = C3∆V3 = ( _______ ) ( _______ ) = __________ C (IV-11 #2)

Sample Exercise IV-3: A 300 V battery is connected in series with a 2 µF and an 8 µF capacitor.

a) In the space at the right, sketch the circuit, label each circuitelement and indicate which plate of each capacitor is (+) and which is (–). (IV-11 #3)

b) Find the charge stored and potential difference acrosseach capacitor.

Solution: Since the capacitors are in series, their

__________ must be the same. (IV-11 #4) (charge, ∆V)

Knowing Ceq and Ceq = Q/∆Vbatt we can find Q.

Ceq = _________ and Q = Ceq∆Vbatt = __________ (IV-11 #5)

where Q is also the charge on each capacitor. Now calculate ∆V1 and ∆V2:

∆V1 = Q/C1 = __________ ∆V2 = Q/C2 = __________ (IV-11 #6)

c) Now imagine that we disconnect the capacitors, being careful not to discharge them, and then connect the (+) plates together (connect A to B) and the (–) plates together (connect C to D). What will be the resulting charge and ∆V for each capacitor now?

Solution: The charge on the upper plates will redistribute itself so that the upper portion of the circuit is an equipotential. The same will happen to the (–) charge with the lower portion becoming an equipotential. This being true, then we know that the __________

(charge, ∆V)will be the same for both capacitors. (IV-11 #7)

IV-12


Before we connected them, the charge on the 2 µF was _________ C and on the 8 µF

_________ C. Now since charge cannot be created or destroyed (i.e. charge is

conserved) when they are connected (+) to (+) and (–) to (–) the total amount of (+)

charge on the upper portion of the circuit will be _________ C and the lower

_________ C. So now if Q2 and Q8 are the magnitude of the new charges on each

capacitor, we know that (IV-12 #1)

Q2 + Q8 = _________ C. (IV-12 #2)

AlsoQ2 = C2∆V2 and Q8 = C8∆V8

Using these equations and the fact that ∆V2 = ∆V8 and going through an orgy of algebra, we find that

Q2 = _________ Q8 = _________and

∆V2 = ∆V8 = _________ (IV-12 #3)

For an additional example of this type of problem, see Example 26.5 on p. 816 of Serway & Beichner. Also, there are other examples in the other references.

At this point try problems 4 & 8 at the back of this unit.

IV-13


Dielectrics An insulator or dielectric is a material that does not conduct electricity. That is, the number of electrons that are free to move

within a dielectric when it is placed in an -field and is super teensy weensy in comparison to a conducting material. The molecules of a

dielectric are, however, influenced by the -field. The molecules become polarized as illustrated in sketch (a) & (b) at the right. The (+) nucleus is displaced in the direction of the -field and the negative electrons are displaced in the opposite direction, resulting in a polarized molecule. That is, one that is a little more (+) on one end and a little more (–) on the other. The amount of polarization is, of course, a property of the molecule itself and is related to how tightly the electrons are bound to the nucleus.

Now if a hunk of dielectric material is placed in an -field, its molecules become polarized with the result that the left end acquires a small (+) charge and the right hand end an equal (–) charge. Note that the effect of this polarization is to create an -

field inside the dielectric, , which is in the opposite direction to the externally applied -field. Therefore, the NET -field in the

region containing the dielectric is = + is less than the -field outside the dielectric.

OK- so bid deal! What does this stuff have to do with capacitors?

What if we put the hunk of dielectric between the plates of a charged capacitor. Since the -field is reduced in the region containing the dielectric, the work required to move a (+) charge from the lower to the upper plate is less. This means that the difference in potential across the plates decreases when the dielectric is introduced even though we still have the same amount of charge on each plate. The result is we have increased the capacitance of our by inserting the dielectric! (Since C = Q/∆V, if Q is the same and ∆V decreases, then C increases.)

IV-14


Let’s back up a second. It wasn’t pointed out before but the capacitance developed previously assumed that nothing was between the plates. So let’s have C mean the capacitance when the plates have a vacuum between them and Cd mean the capacitance when the space between the plates contains a dielectric. Now it turns out that Cd and C are directly proportional and thus we can write:

where (the Greek letter kappa) is called the dielectric constant of the dielectric material. It is a measure of the polarizability of the material; the easier it is to polarize the molecules, the bigger the . Since Cd is always greater than C, is always greater than 1. (It is equal to 1 when the gap is empty). The greater , the greater the polarization the smaller ∆V is between the plates the larger Cd is. Note that all this is consistent with:

,

IF and only IF the capacitor is charged and the battery disconnected so that the charge Q on the capacitor cannot change.

What if the battery remains connected when the dielectric is inserted. In this situation the battery maintains a constant ∆V

across the capacitor. If ∆V remains constant the -field must remain the same also, hence, upon the insertion of the dielectric, more charge must flow onto the capacitor plates to maintain this

constant -field. If the new charge on the plates is Qd, then since the capacitances before and after the insertion of the dielectric are still proportional,

The table below lists the dielectric constants for a number of dielectric materials. In addition, the “dielectric strength” of the materials is given. Here is the meaning of this quantity: A dielectric material insulates one plate of a capacitor from the other. In cannot serve this purpose for any ∆V applied across the plates. At some value of ∆V and, therefore, at some maximum value of the -field between the plates, the dielectric loses its insulating qualities and arcing between the plates occurs. This max is called the dielectric strength of the material.

Material

DielectricConstant

Dielectric StrengthEmax at Room Temp.

(V/m)Air (1 atm)Air (100 atm)Pyrex glassQuartzParafined paperMicaBarium titanate

1.000591.0548

5.63.825

1200

30 x 106

15 x 106

8 x 106

40 x 106

200 x 106

IV-15


Sample Exercise IV-4: Given the situation sketched at the right where a piece of mica fills the space between the plates of the capacitor.

a. Before the mica is inserted find:

i. the capacitance of the empty capacitor

(IV-15 #1)

ii. the magnitude of the charge on each plate

(IV-15 #2)

iii. the -field in the gap between the plates

in terms of Q & A, or in terms of V & d

= __________ , __________ (IV-15 #3) (magnitude) (direction)

b. After the mica is inserted find:

i. the capacitance with the mica inserted

(IV-15 #4)

ii. the magnitude of the charge on each plate (Careful! - Remember it’s still connected to the battery)

(IV-15 #5)

iii. the net -field in the gap between the plates

= the same as in part (a) iii because __________________________

__________________________________________________ (IV-15 #6)

Now suppose we connect the battery, charge up the capacitor giving it a charge |Q| on each plate, then disconnect the battery. All the answers to part (a) before the mica is inserted are the same. Let’s see what they are after the mica is inserted.

c. i. , so Cd is still _________. (IV-15 #7)

ii. Here’s where the difference comes in. Once we disconnect the battery, the chargeon each plate can’t change if the capacitor is not hooked up to anything else. Thus

Qd still equals _________. (IV-15 #8)

IV-16

Unit IV – CAPACITORS & DIELECTRICSIV-17


Sample Exercise IV-4, (Cont.)

ii. (Cont.) The thing that does change however, is ∆V.

(IV-16 #1)

iii. The net -field in the gap now

(IV-16 #2)

Now for a couple of ideas that will help on the problems. Suppose you are given a problem with this type of capacitor in it:

this is nothing more than two capacitors in series because the boundary between the dielectrics is an equipotential… so treat them as such!)

Suppose you are given a problem with this type of capacitor in it:

This is nothing more than

two capacitors in parallel… so treat them as such!

At this point try problems 9 - 13 at the back of this unit.

IV-18


Energy Stored In A Capacitor

In the situation sketched at the right, suppose the net charge on either plate is initially zero. If (+) charges are repeatedly removed from side a [thus leaving it with a net (–) charge] and transferred to side b, an ever increasing amount of work is required since we have to work against the increasing repulsion of the (+) side and against the attraction of the (–) side. In transferring this charge, the energy we put into the system is stored in the capacitor as electric potential energy. Let’s calculate how much work is done (and therefore how much energy is stored) when we transfer a total amount of charge Q. At some instant let the potential differences between the plates be ∆V and the charge on the plates be q. If the amount of work required to transfer the next teensy weensy bit of charge dq is dW, then

dW = ∆Vdq.

But ∆V = q/C where C is the capacitance. Hence,

If we want to calculate the total work done charging the capacitor from zero charge to charge Q, we just add up all the dWs (integrate):

Thus the total electric potential energy stored in a capacitor is:

The last two come from using C=Q/∆V and doing a little algebraic messing around

Sample Exercise IV-5A rule of thumb for capacitor discharge circuits is that 20 J of electrical energy can stop the human heart (which results in a usually undesirable condition technically referred to as “death”). In many old radio and TV circuits it was not uncommon to have voltages as high as 500 V. Assume that in such a circuit there is an undischarged capacitor with 500 V across its terminals.

a) What would the capacitance be if the capacitor was super duper dangerous?

Solution: Known quantities - Energy stored in an dangerous capacitor = ________ J

∆V across the capacitor = ________ VFrom the box above choose the form of equation involving the knowns and theunknown, C, and solve for C.

(IV-17 #1) Equation Substitution Answer

IV-19


Sample Exercise IV-5 (cont)

b) How much charge does this involve?

Solution: Same knowns as in (a).

Choose the form of the equation in the box above involving the knowns and theunknown, Q, and solve it for Q.

(IV-18 #1)

At this point try problems 14 through 17 at the back of this unit.

– End Unit IV –

IV-20


In-Text Fill-In Answers

Page Fill-In # Fill-In Answer

IV-3 1

2

3

4

5

IV-4 1 V = Ed

2 E = /o = Q/(oA)

3 V = o = Qd/(oA)

4

IV-51

2 A = L2

IV-21


3

4 L ~ 104 m or ~ 6.2 miles

5 Cylinder: +Q, outer shell: –Q, -field directed outward.

6


Page Fill-In # Fill-In Answer

IV-6 1 Since is outward

2

3 increasing, larger

4 increases, larger

IV-9 1

2 Ceq ~ 3.1 F

3 Q = ( 3.1 F )( 12 V ) ~ 37 C

4 ∆V1 = 7.5 V, ∆V2 = 3.7 V, ∆V3 = 0.75 V

IV-10 1 Ceq = 65 F

IV-11 1 ∆V1 = ∆V2 = ∆V3 = ∆Vbatt = 12 V

2 Q1 = 60 C, Q2 = 120 C, Q3 = 600 C

IV-22


3

4 Their charge is the same.

5 Ceq = 8/5 F and Q = Ceq∆Vbatt = 480 C

6 ∆V1 = Q/C1 = (480 C)/(2 F) = 240 V∆V2 = Q/C2 = (480 C)/(8 F) = 60 V

7 ∆V will be the same

IV-23



IV-12 1 Q2 = 480 C, Q8 = 480 C, Qupper = 960 C, Qlower = – 960 C

2 Q2 + Q8 = 960 C

3 Q2 = 192 C, Q8 = 768 C,∆V2 = ∆V8 = 96 V

IV-15 1 C ~ 89 pF

2 Q ~ 4.5 nC

3 = 50 x 103 (N/C or V/m), downward

4 Cd = 445 pF

5 Qd = 22.3 nC

6 = the same as in part a) iii because the battery is stil connected.

7 Cd = 445 pF

IV-16 1

2 104 V/m, downward

IV-17 1 Known – Udangerous = 20 J,V = 500 V

IV-18 1

IV-24


End of Unit Problems1. You are given a parallel plate capacitor with square plates of area A and separation d, in a

vacuum. What is the qualitative effect of each of the following on its capacitance?a. Reduce d.b. Put a slab of copper between the plates, touching neither plate.c. Double the area between both plates.d. Double the potential difference between the plates.e. Tilt one place so that the separation remains d at one end but it is d/2 at the other.

2. A parallel plate capacitor has circular plates of 8 cm radius and 1 mm separation.a. Find its capacitance. (Ans: 56.6 pF)b. If 100 V battery is connected across the capacitor, what charge will appear on the

plates? [Ans: 5.7 nC]

3. Given to concentric conducting spherical shells of radii a and b. The inner shell has a charge of -Q and the outer positive +Q.a. Use Gauss’ law to find the magnitude and direction of the

electric field and regions I, II, and III in the sketch of the

right. [Ans: , radially inward]

b. Use to show that the potential difference between the inner and outer

shells is given by .

c. What is the capacitance of the shells?

4. Determine the equivalent capacitance of each of the networks of capacitors shown below.

Assume all capacitors have a capacitance C. [Ans: ]

5. What single capacitor is equivalent to the capacitor network shown at the right? [Ans: 5 F]

a. If VAB is 12 volts, what is the V across each capacitor? [Ans: 6 V]

b. What is the charge stored on each capacitor? [Ans: Q4 = 24 C, Q6 = 36 C, Q10 = 60 C]

IV-25


End of Unit Problems6. A 35 F capacitor is required in certain circuit but only 10 F capacitors are available. How

should the minimum number of 10 F capacitors be connected so that the combination has an equivalent capacitance of 35 F?

7. In the circuit at the right, the values of the capacitors are C1 = 12 F, C2 = 10 F, C3 = 15 F, and C4 = 6 F.a. What is the equivalent capacitance between terminals A and

B? [Ans: 6 F]b. What is the V across C4 if a 12 V battery is connected

between terminals A and B? [Ans: 6 V]

8. In the circuit diagrammed that the right, the battery provides a constant V of 50 volts. C1 and C2 are initially uncharged. Switch S1 is closed charging the 100 F capacitor C1. Then S1 is opened disconnecting the battery from the circuit. Following this S2 is closed.a. Before S2 is closed, what charge is stored on C1? [Ans: 5 mC]b. After S2 is closed the V across C2 is measured to be 35 V. Determine the capacitance

of C2. [Ans: 43 F]

9. For making a capacitor you have available two copper plates, a sheet of mica (thickness = 0.1 mm), a sheet of pyrex glass (thickness = 1.0 mm), and a sheet of paraffined paper (thickness = 0.2 mm). Each of these sheets are alternately placed between the copper plates to form a capacitor. Which material will result in the greatest capacitance? . . . the least? [Ans: mica, pyrex glass]

10. A paper capacitor is made by placing two metallic foils, each 50 cm² in area, on each side of a piece of paraffin paper having a thickness of 0.02 mm. What is the rating of the capacitor? That is, what is its (a) capacitance, and (b) the maximum V that can be applied across it? [Ans: 4.4 nF, 800 V]

11. A parallel plate capacitor is filled with two dielectrics of equal thickness as shown at the right. Show that the capacitance is given by:

where A is the area of the plates and d is the thickness of each slab of dielectric material.

12. Consider a cylindrical capacitor with two layers of dielectric material between the inner and outer cylinder as shown at the right. Show that the capacitance of the capacitor in terms of the given parameters is given by:

End of Unit Problems

IV-26


13. Suppose the volume between the shells in problem #3 is completely filled with a dielectric material having a = 4. a. Find the new capacitance. [Ans: Cd = 4Co]b. With the same V between the shells in both cases, find the ratio of the charge stored

in this situation to that stored in problem #3. [Ans: 4:1]

c. Suppose our dielectric is a liquid and we only fill the volume between the shells halfway as shown at the right. Find the ratio of the capacitance in this situation to the capacitance when empty.

[Ans: 5:2]

14. a. A 10 F capacitor can withstand a maximum V of 4 kV. What is the maximum energythis capacitor can store? [Ans: 80 J]

b. If charged to this voltage, what will be the charge stored on each plate of the capacitor? [Ans: 40 mC]

15. Two capacitors are connected in parallel as shown in the sketch below. The plates for C1 have an area A, are separated by distance d, and the space between them is evacuated. The plates for C2 have area 2A, are separated by a distance 2d, with the space between them being filled by a sheet of mica.a. Show that the equivalent capacitance of this system is

.b. If C1 = 10pF, what is C2? [Ans: 50 pF]c. If d equals 1 mm what is the maximum voltage that could be applied to C2?

[Ans: 400 kV]d. At Vmax, what is the charge stored on C2? [Ans: 20 C]e. At Vmax, what is the ratio of the energy stored in C2 to that stored in C1? [Ans: 5:1]

16. In Problem #12 let: the inner conducting shell have a radius of 5 mm, the inner dielectric be a 2 mm thick cylinder of mica, and the outer dielectric be a cylinder of paraffined paper having a thickness of 3 mm. The shells have a length of 2 cm.a. Calculate the capacitance. [Ans: 4.5 pF]b. If V between the inner and outer conducting shells is 200 V, find the energy stored in

the capacitor. [Ans: 90 nJ]c. If V is 200 V, find the charge on either plate. [Ans: 0.9 nC]

17. A spherical “capacitor” has an inner radius of 0.50 m and an outer radius of 0.51 m.a. Calculate its capacitance. [Ans: 2.8 nF]b. If the V between the inner and outer shells is 100 V, how much energy is stored in the

capacitor? [Ans: 14 J]

IV-27

Documents

UNIT II - Mr. Greenberg Physics · Web viewPHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 26 FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed.,