2. Chapter 26 capacitance 26-1 Definition of Capacitance
Calculating Capacitance
Combinations of Capacitors
Energy Stored in a Charged Capacitor
Capacitors with Dielectrics
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3. Units: Coulombs/Volt or Farads
Week 04, Day 1
26.1 definition of capacitance
Consider two conductors carrying charges of equal magnitude and opposite sign,
as shown in the Figure. Such a combination of two conductors is called a capacitor.
The conductors are called plates. A potential difference Δ V exists between the conductors due to the presence of the charges
Capacitor:Two isolated conductors Equal and opposite charges ±Q Potential difference DV between them.
Experiments show the quantity of electric charge Q on a capacitor is linearly proportional to the potential difference between the conductors, that is Q ~ DV. Or we write Q = C DV
The capacitance C of a capacitor is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between them:
Units: Coulombs/Volt or Farads
C is Always Positive
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4. Example
A storage capacitor on a random access memory (RAM) chip has a capacitance of 55 nF. If the capacitor is charged to 5.3 V, how many excess electrons are on the negative plate?
We can find the number of excess electrons on the negative plate if we know the total charge q on the plate. Then, the number of electrons n=q/e, where e is the electron charge in coulomb.
:The charge q of the plate is related to the voltage V to which the capacitor is charged: q=CV.
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5. V = 16 V, C = 0.75 mF = 0.75 x 10-6 F C = Q/V Q = CV
Example
A 0.75 mF capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?
V = 16 V, C = 0.75 mF = 0.75 x 10-6 F
C = Q/V Q = CV
Q = (0.75 x 10-6)(16) = 1.2 x 10-5 C
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6. Capacitance of an isolated sphere
26.2 Calculating Capacitance
Capacitance of an isolated sphere
Calculate the capacitance of an isolated spherical conductor of radius R and charge Q by assuming that the second conductor making up the capacitor is a concentric hollow sphere of infinite radius.
Electric potential of the sphere of radius R is
kQ/R and V= 0 at infinity, we have Q
C = Q DV =
kQ/R R k
= 4peo R
C is proportional to its radius and independent of both the charge on the sphere and the potential difference.
The capacitance of a pair of conductors depends on the geometry of the conductors.
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7. Example
(a) If a drop of liquid has capacitance 1.00 pF, what is its radius ? (b) If another drop has radius 2.00 mm, what is its capacitance ? (c) What is the charge on the smaller drop if its potential is 100V ?
C = 4πeo R
R = (9 x 109 N · m2/C2)(1.00 x 10–12 F) = 9 mm
C = 4 π (8.85 x 10-12) x 2.0x10-3 = pF
Q =CV = pF x 100 V = 2.22 x C
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8. Example What is the capacitance of the Earth ?
Think of Earth spherical conductor and the outer conductor of the “spherical capacitor” may be considered as a conducting sphere at infinity where V approaches zero.
C= 7.08 x 10-4 F
A large capacitor !
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9. Example V = 16 V, C = 0.75 mF = 0.75 x 10-6 F C = Q/V or Q = CV
: A 0.75 mF capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor?
V = 16 V, C = 0.75 mF = 0.75 x 10-6 F
C = Q/V or Q = CV
Q = (0.75 x 10-6)(16) = 1.2 x 10-5 C
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10. Parallel - Plate Capacitors
A parallel-plate capacitor consists of two parallel conducting plates, each of area A, separated by a distance d. When the capacitor is charged, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge. And surface charge density of each plate is s = Q/A
The plates are charged by connection to a battery..
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11. Top Sheet: + + + + + + + + + + + + + + Bottom Sheet:
Week 04, Day 1
Top Sheet:
Bottom Sheet:
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12. Variation with A If plates are large, then charges can distribute themselves over a substantial area, and the amount of charge that can be stored on a plate for a given potential diff increases as A is increased. Thus we expect C to be proportional to A  C ~ A
Variation with d Potential difference DV constant across, E field increases as d decreases. Imagine d decreases and consider situation before any charges have had a chance to move in response to this change. Because no charge move  E the same but over a shorter distance. DV = Ed means that DV decreases
The difference between this new capacitor voltage and the terminal voltage of the battery now exists as a potential difference across the wires connecting the battery to the capacitor.
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13. More charges has accumulated at the capacitor as a result.
A E field result in the wires that drives more charge onto the plates, increasing the potential diff. DV until it matches that of the battery.  potential diff. Across wire = 0  flow of charges stop.
More charges has accumulated at the capacitor as a result.
We have d decrease, Q increases. Similarly d increases  Q decreases.
Capacitance inversely proportional to d.
 C ~ 1/d
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14. Assume electric field uniform between the plates, we have
(see lecture on Gauss’s Law)
The general properties of a parallel-plate capacitor – that the capacitance increases as the plates become larger and decreases as the separation increases – are common to all capacitors.
Charge density:
Electric field:
Potential diff.:
Only the geometry of the plates (A and d) affect the capacitance.
Capacitance:
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15. The electric field between the plates of a parallel-plate capacitor is uniform near the center but nonuniform near the edges.
Electric field pattern of two oppositely charged conducting parallel plates.
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16. Cylindrical capacitor
Capacitance configurations
Spherical Capacitance
Cylindrical capacitor

17. Example
A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm
What is the capacitance?
C = e0A/d
= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)
= 2.21 x 10–9 F
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18. Example
We have a parallel plate capacitor constructed of two parallel plates, each with area 625 cm2 separated by a distance of 1.00 mm.
What is the capacitance of this parallel plate capacitor?
C = nF
Result: A parallel plate capacitor constructed out of square conducting plates 25 cm x 25 cm separated by 1 mm produces a capacitor with a capacitance of about 0.5 nF.
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19. Example
We have a parallel plate capacitor constructed of two parallel plates separated by a distance of 1.00 mm.
What area is required to produce a capacitance of 1.00 F?
Result: A parallel plate capacitor constructed out of square conducting plates 10.6 km x 10.6 km (6 miles x 6 miles) separated by 1 mm produces a capacitor with a capacitance of 1 F.
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20. Example.
What is the AREA of a 1 F capacitor that has a plate separation of 1 mm?
d = 1 mm = m, eo = 8.85 x C2/(Nm2)
1.13 x 108 m2
10629 m
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21. example : Changing Dimensions
Week 04, Day 2
example
: Changing Dimensions
A parallel-plate capacitor has plates with equal and opposite charges ±Q, separated by a distance d, and is connected to a battery. The plates are pulled apart to a distance D > d. What happens?
V increases, Q increases
V decreases, Q increases
V is the same, Q increases
V increases, Q is the same
V decreases, Q is the same
V is the same, Q is the same
V increases, Q decreases
V decreases, Q decreases
V is the same, Q decreases
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22. Answer: 9. V is the same, Q decreases
Week 04, Day 2
Answer: 9. V is the same, Q decreases
With a battery connected to the plates the potential V between them is held constant
In this situation, since
V = E d
As d increases, E must decrease.
Since the electric field is proportional to the charge on the plates, Q must decrease as well.
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23. Example.
A parallel plate capacitor is constructed with plate of an area of m2 and a separation of 0.55 mm. Find the magnitude of the charge of this capacitor when the potential difference between the plate is 20.1 V.
A = m2 d
= 0.55 mm = m,
V = 20.1 V
Q = CV
C = eo A/d
C = (8.85 x )(0.028) /( )= 4.51 x F
Q = (4.51 x )(20.1) = 9.06 x 10-9 C
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24. Example C= eoA/d = 8.85x10-12 x (1000)2/800 = 11.1 nF
Regarding the Earth and a cloud layer 800 m above the Earth as the “plates” of a capacitor, calculate the capacitance if the cloud layer has an area of 1.00 x 1.00 km2. Assume that the air between the cloud and the ground is pure and dry.
Assume that charge builds up on the cloud and on the ground until a uniform electric field of 3.00 x 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold?
C= eoA/d = 8.85x10-12 x (1000)2/800 = 11.1 nF
Potential between ground and cloud is
DV = Ed = 3.0 x106 x 800 = 2.4 x 109 V
Q = C(DV) = 26.6 C
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25. 26.3 Combinations of Capacitors
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26. Parallel Combination For parallel capacitors
When a potential difference V is applied across several capacitors connected in parallel, that potential difference V is applied across each capacitor.
The total charge q stored on the capacitors is the sum of the charges stored on all the capacitors.
Capacitors connected in parallel can be replaced with an equivalent capacitor that has the same total charge q and the same potential difference V as the actual capacitors.
For parallel capacitors
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27. Capacitors in Series
When a potential difference V is applied across several capacitors connected in series, the capacitors have identical charge q.
The sum of the potential differences across all the capacitors is equal to the applied potential difference V.
Capacitors that are connected in series can be replaced with an equivalent capacitor that has the same charge q and the same total potential difference V as the actual series capacitors.
SERIES:
Q is same for all capacitors
Total potential difference = sum of V
For series capacitors
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28. Example: Equivalent Capacitance
In series use 1/C=1/C1+1/C2
2.50 mF
20.00 mF
6.00 mF
In series use 1/C=1/C1+1/C2
8.50 mF
In parallel use C=C1+C2
5.965 mF
20.00 mF
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29. Example: Equivalent Capacitance
In parallel use C=C1+C2
In parallel use C=C1+C2
In series use 1/C=1/C1+1/C2
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30. Example: Equivalent Capacitance
In parallel use Ceq=C+C/2+C/3
In series use 1/CA=1/C+1/C C
C/2
C/3
In series use 1/CB=1/C+1/C+1/C
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31. Example: 1/Cs=1/C3+1/Cp Step 1: Cp=C1+C2 Cp=0.10 mF+0.20 mF

32. Find the Charge stored in the capacitor.
Example
Find the Charge stored in the capacitor.
12V
20μF
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33. Example Find the total charge stored in each the capacitor.
First calculate the total capacitance.
12V
10μF
20μF
30μF
Then find the total charge
In a series circuit the charge stored on each capacitor is the same and the voltage is split.
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34. Example Find the total charge stored in all the capacitors.
First find the total capacitance.
12V
10μF
20μF
30μF
calculate the total charge.
In a parallel circuit the charge stored on each capacitor can be different and the voltage must be the same.
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35. Example Find the total capacitance of all the capacitors.
First find the total capacitance of the parallel capacitors
12V
12μF
8μF
4μF
combine this value in series with 12μF capacitor.
Energy Stored in a Capacitor
Units: Joules (J)
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36. Example.
½ C
2 C
1/3 C
3 C C
2/5 C
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37. Example C3
C12
Step 1 V
C123
Step 2 V C1 C2
parallel V
series C3
C1 = 12.0 mF, C2 = 5.3 mF, C3 = 4.5 mF
C123 = ( )4.5/( ) mF = 3.57 mF

38. Example
parallel C6
C45
series C4 C1 C6 C5 V
parallel C2 C3 C3 C3

39. Example
series
C1456 V
C23
Complete solution

40. 26.4 Energy Stored in Capacitor
Week 04, Day 1
26.4 Energy Stored in Capacitor
Suppose that, at a given instant, a charge q′ has been transferred from one plate of a capacitor to the other. The potential difference DV between the plates at that instant will be q / C.
If an extra increment of charge dq is then transferred, the increment of work required will be,
dW = DV dq
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41. Energy To Charge Capacitor
Week 04, Day 1
Energy To Charge Capacitor
1. Capacitor starts uncharged.
2. Carry +dq from bottom to top.
Now top has charge q = +dq, bottom -dq
3. Repeat
4. Finish when top has charge q = +Q, bottom -Q +q -q
At some point top plate has +q, bottom has –q
Potential difference is DV = q / C
Work done lifting another dq is dW = dq DV
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42. So work done to move dq is:
Week 04, Day 1
So work done to move dq is:
Total energy to charge to q = Q:
A plot of potential difference versus charge for a capacitor is a straight line having a slope 1/C. The work required to move charge dq through the potential difference DV across the capacitor plates is given by the area of the shaded rectangle. The total work required to charge the capacitor to a final charge Q is the triangular area under the straight line, W = QDV/2.
1V = 1 J/C hence the unit for the area is joule J.
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43. Energy Stored in Capacitor
Week 04, Day 1
Energy Stored in Capacitor
Work done in charging the capacitor = electric potential energy U stored in the capacitor.
Since
Where is the energy stored???
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44. Energy Stored in Capacitor
Week 04, Day 1
Energy Stored in Capacitor
Energy stored in the E field!
Parallel-plate capacitor:
Ad = Volume occupied by the E field. This lead to a new quantity known as Energy Density uE = U/Volume = U/Ad
The energy stored in a capacitor can be put to a number of uses: a camera flash; a cardiac defibrillator; and others. In addition, capacitors form an essential part of most electrical devices used today.
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45. Example
Suppose the capacitor shown here is charge to Q and then the battery is disconnected.
Now suppose I pull the plates further apart so that the final separation is d1.
How do the quantities Q, C, E, V, U change? d A Q: C: E: V: U:
remains the same.. no way for charge to leave.
decreases.. since capacitance depends on geometry
remains the same... depends only on charge density
increases.. since C ¯, but Q remains same (or d  but E the same)
increases.. add energy to system by separating
How much do these quantities change?
Answers:
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46. Example. A + + + + V d - - - - - C: V: Q: E: U:
Suppose the battery (V) is kept attached to the capacitor. A V d
Again pull the plates apart from d to d1.
Now what changes? C: V: Q: E: U:
decreases (capacitance depends only on geometry)
must stay the same - the battery forces it to be V
must decrease, Q=CV charge flows off the plate
must decrease ( , )
must decrease ( )
How much do these quantities change?.. exercise for student!!
Answers:
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47. example The final stored energy is smaller
Week 04, Day 2
example
A parallel-plate capacitor, disconnected from a battery, has plates with equal and opposite charges, separated by a distance d.
Suppose the plates are pulled apart until separated by a distance D > d.
How does the final electrostatic energy stored in the capacitor compare to the initial energy?
The final stored energy is smaller
The final stored energy is larger
Stored energy does not change.
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48. Answer: 2. The stored energy increases
Week 04, Day 2
Answer: 2. The stored energy increases
As you pull apart the capacitor plates you increase the amount of space in which the E field is non-zero and hence increase the stored energy. Where does the extra energy come from? From the work you do pulling the plates apart.
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49. Example
A parallel-plate capacitor is charged and then disconnected from a battery. By what fraction does the stored energy change (increase or decrease) when the plate separation is doubled ?
U = Q2/2C and C = eoA/d and d2 = 2 d1 then
C1= C2/ and the energy stored doubles.
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50. example B.) Find the energy stored in the charged up defibrillator.
In a typical defibrillator, a 175 mF, is charged until the potential difference between the plates is 2240 V. A.) What is the charge on each plate? V = 2240 V, C = 175 mF = 175 x 10-6 F Q = CV = (175 x 10-6)(2240) = C
example
B.) Find the energy stored in the charged up defibrillator.
U = ½ CV2 U = Q2/(2C)
U = ½ QV = ½ (0.392)(2240)
U = 439 J
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51. Example
Consider the circuit as shown, where C1 = 6.00mF and C2= 3.00 mF and DV =20.0V. Capacitor C1 is first charged by closing of switch S1. Switch S1 is then opened and the charged capacitor is connected to the uncharged capacitor by the closing of S2. Calculate the initial charge acquired by C1 and the final charge on each.
S1 close, S2 open 
C = Q/V  Q = mC
After S1 open, S2 close 
Q1 + Q2 = mC
Same potential  Q1 /C1 = Q2 / C2  (120-Q2)/C1= Q2/C2
( Q2)/6 = Q2/ 3  Q2 = 40 mC
 Q 1= 80 mC
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52. example
--Find the charge on (Q) and potential difference (V) across each capacitor. What is the total energy stored in the system?

54. 26.5 Capacitors with Dielectrics
A dielectric is a nonconducting material, such as rubber, glass, or waxed paper. Dielectric constant is a property of a material and varies from one material to another.
A charged capacitor (a) before and (b) after insertion of a dielectric between the plates. The charge on the plates remains unchanged, but the potential difference decreases from DVo to DV = DVo/k Thus the capacitance increases from Co to k Co.
Note no battery is involved in this example.
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55. When a dielectric is inserted between the plates of a capacitor, the capacitance increases.
If the dielectric completely fills the space between the plates, the capacitance increases by a dimensionless factor k , which is called the dielectric constant.
Capacitance increases by the factor k when dielectric completely fills the region between the plates.
If the dielectric is introduced while the potential difference is being maintained constant by a battery, the charge increases to a value
Q = k Qo . The additional charge is supplied by the battery and the capacitance again increases by the factor k.
For parallel plate capacitor:
eoA
C = k d
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56. Dielectrics Definition:
The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it:
k values are always > 1 (e.g., glass = 5.6; water = 78)
They INCREASE the capacitance of a capacitor (generally good, since it is hard to make “big” capacitors)
They permit more energy to be stored on a given capacitor than otherwise with vacuum (i.e., air):
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57. Dielectric Strength
If the electric field in a dielectric becomes too large, it can tear the electrons off the atoms, thereby enabling the material to conduct. This is called dielectric breakdown; the field at which this happens is called the dielectric strength. For any given separation d, the maximum voltage that can be applied to a capacitor without causing a discharge depends on the dielectric strength (maximum electric field) of the dielectric.
. If magnitude of the electric field in the dielectric exceeds the dielectric strength, then the insulating properties break down and the dielectric begins to conduct.
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58. the advantages of dielectric material in a capacitor?
Increase the capacitance
Increase the maximum operating voltage
Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C.
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59. Example. Compare the voltages of the two capacitors.
Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected it is filled with a dielectric.
Compare the voltages of the two capacitors.
a) V1 > V b) V1 = V c) V1 < V2
Compare the electric fields between the plates of both capacitors.
a) E1 > E b) E1 = E c) E1 < E2
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60. Example. When we insert the dielectric into the capacitor C2 we do:
a) positive work
b) negative work
c) no work
Recall the meaning of negative work is the energy of the system is reduced. The dielectric is sucked into the capacitor. When the charge is constant, the total energy of the capacitor decreases because the presence of the dielectric increases the capacitance . It turns out that the dielectric is pulled in even if the voltage is held constant, e.g., via a battery. On a microscopic scale the force on the dielectric arises due to “fringing fields” at the edges of the capacitor.
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61. The capacitance of the capacitor?
Examples:
(a) A parallel-plate capacitor is connected to a power supply that maintains a constant potential difference V0 between the two plates. If the gap between the plates is first filled with air (≈ vacuum) and then a dielectric with dielectric constant κ=3 is inserted, how do the following quantities change:
The capacitance of the capacitor?
The charge on the positive plate of the capacitor?
The electric field between the capacitor plates?
The energy stored in the capacitor?
If A and d don’t change then: if κ ↑3  C↑3
If V doesn’t change then: if C↑3  Q↑3
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62. Example.
Two identical parallel plate capacitors are connected in series to a battery as shown below. If a dielectric is inserted in the lower capacitor, which of the following increase for that capacitor?
I and III.
I, II and IV.
I, II and III.
All except II.
All increase. I.
Capacitance of capacitor
II.
Voltage across capacitor
III.
Charge on capacitor
IV.
Energy stored on capacitor V C k

63. A Closer Look Insert dielectric Capacitance goes up by k q’ q
Charge increases C V
Charge on upper plate comes from upper capacitor, so its charge also increases. V
Since q’ = CV1 increases on upper capacitor, V1 must increase on upper capacitor. q q’ k C kC V
Since total V = V1 + V2 = constant, V2 must decrease.

64. Example.
You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant.
What effect does adding the dielectric have on the energy stored in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide

65. Example.
You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant.
What effect does adding the dielectric have on the amount of charge on each of the capacitor plates?
A. The amount of charge increases.
B. The amount of charge remains the same.
C. The amount of charge decreases.
D. not enough information given to decide

66. Example.
You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference between the plates remains constant.
What effect does adding the dielectric have on the energy stored in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide

67. Norah Ali Al moneef

68. Example.
A parallel-plate capacitor is attached to a battery that maintains a constant potential difference V between the plates. While the battery is still connected, a glass slab is inserted so as to just fill the space between the plates. The stored energy
1. increases.
2. decreases.
3. remains the same.
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69. k Example. Increases Decreases Stays the Same
Week 04, Day 2
Example.
A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of material with dielectric constant k in inserted between the plates. The charge stored in the capacitor
Increases
Decreases
Stays the Same k
Answer: 3. Charge stays the same
Since the capacitor is disconnected from a battery there is no way for the amount of charge on it to change.
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70. k Example. Increases Decreases Stays the Same
Week 04, Day 2
Example.
A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of material with dielectric constant k in inserted between the plates. The energy stored in the capacitor k
Increases
Decreases
Stays the Same
Answer: 2. Energy stored decreases
The dielectric reduces the electric field and hence reduces the amount of energy stored in the field.
The easiest way to think about this is that the capacitance is increased while the charge remains the same so U = Q2/2C
Also from energy density:
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71. Calculating Capacitance
Units
1 F = 1 C2/N m (Note [e0]=C2/N m2)
1 mF = 10-6 F, 1 pF = F
e0 = 8.85 x F/m
Example : Size of a 1-F capacitor ,calculate its area

72. Example
Calculate the Capacitance ,the charge on the plates and the electric field

73. Example.
Find the capacitance of a 4.0 cm diameter sensor immersed in oil if the plates are separated by 0.25 mm.
The plate area is
The distance between the plates is
178 pF

75. Capacitors in Series and Parallel

76. example
In this problem you try to measure dielectric constant of a material. First
a parallel-plate capacitor with only air between the plates is charged by
connecting it to a battery. The capacitor is then disconnected from the
battery without any of the charge leaving the plates.
Express the capacitance C0 in terms of the potential difference V0
between the plates and the charge Q if air is between the plates.
(b) Express the dielectric constant k in terms of the capacitance C0 (air gap)
and the capacitance C with material of the dielectric constant k).
(c) Using the results of (a) and (b), express the ratio of the potential
difference V/V0 if Q is the same, where V is the potential difference
between the plates and a dielectric material dielectric constant is k
fills the space between them.
(d) A voltmeter reads 45.0 V when placed across the capacitor. When
dielectric material is inserted completely filling the space, the voltmeter
reads 11.5 V. Find the dielectric constant of this material.

77. (a)
(b)
(c)
(d) From (c)

78. Example.
If a 22 mF capacitor is connected to a 10 V source, the charge is
220 mC

79. Example.
If a mF capacitor is connected in series with an 800 pF capacitor, the total capacitance is
444 pF

80. Example.
If a mF capacitor is connected in parallel with an 800 pF capacitor, the total capacitance is
1800 pF

81. Example.
If a mF capacitor is in series with a 6800 pF capacitor, the total capacitance is
a pF
b pF
c pF
d mF

82. Example.
Two capacitors that are initially uncharged are connected in series with a dc source. Compared to the larger capacitor, the smaller capacitor will have
a. the same charge
b. more charge
c. less voltage
d. the same voltage

83. Example. Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C.
(a) Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm and are separated by a 1.0-mm air gap. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, given the same air gap d.
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x C.
c. E = V/d = 1.2 x 104 V/m.
d. A = C d /ε0 = 108 m2.

84. Example.
Determine the capacitance of a single capacitor that will have the same effect as the combination shown.
Solution: First, find the equivalent capacitance of the two capacitors in parallel (2C);
then the equivalent of that capacitor in series with the third (2/3 C).

85. Example
A camera flash unit stores energy in a 150-μF capacitor at 200 V. (a) How much electric energy can be stored?
Solution:
a. U = ½ CV2 = 3.0 J.

86. Example.
The plates of a parallel-plate capacitor have area A, separation x, and are connected to a battery with voltage V. While connected to the battery, the plates are pulled apart until they are separated by 3x. (a) What are the initial and final energies stored in the capacitor? (b) How much work is required to pull the plates apart (assume constant speed)? (c) How much energy is exchanged with the battery?
a. U = ½ CV2, and increasing the plate separation decreases the capacitance, so the initial and final energies can be calculated from this.
b. The work is equal to the negative of change in energy.
c. The energy of the battery is increased by the work done minus the change in potential energy of the capacitor (which is negative).

87. Example
A parallel-plate capacitor, filled with a dielectric with K = 3.4, is connected to a 100-V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A = 4.0 m2 and are separated by d = 4.0 mm
(a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor.
. (b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor.
Solution: a. C = Kε0A/d = 3.0 x 10-8 F. Q = CV = 3.0 x 10-6 C E = V/d = 25 kV/m. U = ½ CV2 = 1.5 x 10-4 J.
B. Now C = 8.8 x 10-9 F, Q = 3.0 x 10-6 C (no change), V = 340 V, E = 85 kV/m, U = 5.1 x 10-4 J. The increase in energy comes from the work it takes to remove the dielectric.

88. Example
Capacitor C1 is connected across a battery of 5 V. An identical capacitor C2 is connected across a battery of 10 V. Which one has more charge?
1) C1
2) C2
3) both have the same charge
4) it depends on other factors
Since Q = CV and the two capacitors are identical, the one that is connected to the greater voltage has more charge, which is C2 in this case.

89. Example
1) increase the area of the plates
2) decrease separation between the plates
3) decrease the area of the plates
4) either (1) or (2)
5) either (2) or (3)
What must be done to a capacitor in order to increase the amount of charge it can hold (for a constant voltage)? +Q –Q
Since Q = CV, in order to increase the charge that a capacitor can hold at constant voltage, one has to increase its capacitance. Since the capacitance is given by , that can be done by either increasing A or decreasing d.

90. Example
1) the voltage decreases
2) the voltage increases
3) the charge decreases
4) the charge increases
5) both voltage and charge change
A parallel-plate capacitor initially has a voltage of 400 V and stays connected to the battery. If the plate spacing is now doubled, what happens?
Since the battery stays connected, the voltage must remain constant! Since , when the spacing d is doubled, the capacitance C is halved. And since Q = CV, that means the charge must decrease. +Q –Q

91. Example
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) V
A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage? +Q –Q
Once the battery is disconnected, Q has to remain constant, since no charge can flow either to or from the battery. Since , when the spacing d is doubled, the capacitance C is halved. And since Q = CV, that means the voltage must double.

92. Example
1) Ceq = 3/2C
2) Ceq = 2/3C
3) Ceq = 3C
4) Ceq = 1/3C
5) Ceq = 1/2C
What is the equivalent capacitance, Ceq , of the combination below? o C
Ceq
The 2 equal capacitors in series add up as inverses, giving 1/2C. These are parallel to the first one, which add up directly. Thus, the total equivalent capacitance is 3/2C.

93. Follow-up: What is the current in this circuit?
Example
1) V1 = V2
2) V1 > V2
3) V1 < V2
4) all voltages are zero
How does the voltage V1 across the first capacitor (C1) compare to the voltage V2 across the second capacitor (C2)?
C1 = 1.0 mF
C3 = 1.0 mF
C2 = 1.0 mF
10 V
The voltage across C1 is 10 V. The combined capacitors C2 + C3 are parallel to C1. The voltage across C2 + C3 is also 10 V. Since C2 and C3 are in series, their voltages add. Thus the voltage across C2 and C3 each has to be 5 V, which is less than V1.
Follow-up: What is the current in this circuit?

94. Example
1) Q1 = Q2
2) Q1 > Q2
3) Q1 < Q2
4) all charges are zero
How does the charge Q1 on the first capacitor (C1) compare to the charge Q2 on the second capacitor (C2)?
We already know that the voltage across C1 is 10 V and the voltage across both C2 and C3 is 5 V each. Since Q = CV and C is the same for all the capacitors, we have V1 > V2 and therefore Q1 > Q2.
C1 = 1.0 mF
C3 = 1.0 mF
C2 = 1.0 mF
10 V

96. How much energy is stored in a 2
How much energy is stored in a 2.0 mF capacitor that has been charged to 5000 V? What is the average power dissipation if the capacitor is discharged in 10 ms?

97. The spacing between the plates of a 1.0 mF capacitor is 1 mm.
(a)What is the surface area A of the plates?
(b)How much charge is on the plates if this capacitor is attached to a 1.5 V battery?

98. Two flat parallel plates are d = 0. 40 cm apart
Two flat parallel plates are d = 0.40 cm apart. The potential difference between the plates is 360 V. The electric field at the point P at the center is approximately
90 kN/C.
180 N/C.
0.9 kN/C.
Zero.
3.6 ´ 105 N/C

99. Two large metallic plates are parallel to each other and charged
Two large metallic plates are parallel to each other and charged. The distance between the plates is d. The potential difference between the plates is V. The magnitude of the electric field E in the region between the plates and away from the edges is given by
d/V.
V2/d.
d ∙V.
V/d2.
V/d .

100. A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to 2Q,
the capacitance becomes (1/2)V.
the capacitance becomes 2C.
the potential changes to (1/2)V.
the potential changes to 2V.
the potential does not change.

101. If a capacitor of capacitance 2. 0 µF is given a charge of 1
If a capacitor of capacitance 2.0 µF is given a charge of 1.0 mC, the potential difference across the capacitor is
0.50 kV.
2.0 V.
2.0 µV.
0.50 V.
None of these is correct.

102. If the area of the plates of a parallel-plate capacitor is doubled, the capacitance is
not changed.
doubled.
halved.
increased by a factor of 4.
decreased by a factor of 1/4.

103. An 80-nF capacitor is charged to a potential of 500 V
An 80-nF capacitor is charged to a potential of 500 V. How much charge accumulates on each plate of the capacitor?
4.0 ´ 10–4 C
4.0 ´ 10–5 C
4.0 ´ 10–10 C
1.6 ´ 10–10 C
1.6 ´ 10–7 C

104. increases. decreases. does not change.
As the voltage in the circuit is increased (but not to the breakdown voltage), the capacitance
increases.
decreases.
does not change.

105. Doubling the potential difference across a capacitor
doubles its capacitance.
halves its capacitance.
quadruples the charge stored on the capacitor.
halves the charge stored on the capacitor.
does not change the capacitance of the capacitor.

106. If the area of the plates of a parallel plate capacitor is halved and the separation between the plates tripled, then by what factor does the capacitance change?
It increases by a factor of 6.
It decreases by a factor of 2/3.
It decreases by a factor of 1/6.
It increases by a factor of 3/2.
It decreases by a factor of ½.

107. Which of the following statements is false?
In the process of charging a capacitor, an electric field is produced between its plates.
The work required to charge a capacitor can be thought of as the work required to create the electric field between its plates.
The energy density in the space between the plates of a capacitor is directly proportional to the first power of the electric field.
The potential difference between the plates of a capacitor is directly proportional to the electric field.
None of these is false.

108. Which of the following statements about a parallel plate capacitor is false?
The two plates have equal charges of the same sign.
The capacitor stores charges on the plates.
The capacitance is proportional to the area of the plates.
The capacitance is inversely proportional to the separation between the plates.
A charged capacitor stores energy.

109. If you increase the charge on a parallel-plate capacitor from 3 µC to 9 µC and increase the plate separation from 1 mm to 3 mm, but keep all other properties the same, the energy stored in the capacitor changes by a factor of
27. 9. 3. 8.
1/3.

110. The energy stored in a capacitor is directly proportional to
the voltage across the capacitor.
the charge on the capacitor.
the reciprocal of the charge on the capacitor.
the square of the voltage across the capacitor.
None of these is correct.

111. A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a distance d = 2 mm and a voltage applied between the plates. The electric field strength within the plates is E = 4000 V/m. The energy stored in the capacitor is
0.71 nJ.
1.42 nJ.
2.83 nJ.
3.67 nJ.
Zero.

112. It increases. It decreases. It stays the same
A circuit consists of a capacitor, a battery, and a switch, all connected in series. Initially, the switch is open and the capacitor is uncharged. The switch is then closed and the capacitor charges. While the capacitor is charging, how does the net charge within the battery change?
It increases.
It decreases.
It stays the same

113. Several different capacitors are hooked across a DC battery in parallel. The charge on each capacitor is
directly proportional to its capacitance.
inversely proportional to its capacitance.
independent of its capacitance.

114. Several different capacitors are hooked across a DC battery in parallel. The voltage across each capacitor is
directly proportional to its capacitance.
inversely proportional to its capacitance.
independent of its capacitance.

115. Several different capacitors are hooked across a DC battery in series
Several different capacitors are hooked across a DC battery in series. The charge on each capacitor is
directly proportional to its capacitance.
inversely proportional to its capacitance.
independent of its capacitance.

116. Several different capacitors are hooked across a DC battery in series
Several different capacitors are hooked across a DC battery in series. The voltage across each capacitor is
directly proportional to its capacitance.
inversely proportional to its capacitance.
independent of its capacitance.

117. If C1 < C2 < C3 < C4 for the combination of capacitors shown, the equivalent capacitance is
less than C1.
more than C4.
between C1 and C4.

118. If C1 < C2 < C3 < C4 for the combination of capacitors shown, the equivalent capacitance is
less than C1.
more than C4.
between C1 and C4.

119. The equivalent capacitance of two capacitors in series is
the sum of their capacitances.
the sum of the reciprocals of their capacitances.
always greater than the larger of their capacitances.
always less than the smaller of the capacitances.
described by none of the above.

120. The equivalent capacitance of three capacitors in series is
the sum of their capacitances.
the sum of the reciprocals of their capacitances.
always greater than the larger of their capacitances.
always less than the smaller of the capacitances.
described by none of the above.

121. The equivalent capacitance of two capacitors in parallel is
the sum of the reciprocals of their capacitances.
the reciprocal of the sum of the reciprocals of their capacitances.
always greater than the larger of their capacitances.
always less than the smaller of the two capacitances.
described by none of the above.

122. The capacitance of a parallel-plate capacitor
is defined as the amount of work required to move a charge from one plate to the other.
decreases if a dielectric is placed between its plates.
is independent of the distance between the plates.
has units of J/C.
is independent of the charge on the capacitor.

123. A capacitor is connected to a battery as shown
A capacitor is connected to a battery as shown. When a dielectric is inserted between the plates of the capacitor,
only the capacitance changes.
only the voltage across the capacitor changes.
only the charge on the capacitor changes.
both the capacitance and the voltage change.
both the capacitance and the charge change.

124. Two identical capacitors A and B are connected across a battery, as shown. If mica (k = 5.4) is inserted in B,
both capacitors will retain the same charge.
B will have the larger charge.
A will have the larger charge.
the potential difference across B will increase.
the potential difference across A will increase.

125. If a dielectric is inserted between the plates of a parallel-plate capacitor that is connected to a 100-V battery, the
voltage across the capacitor decreases.
electric field between the plates decreases.
electric field between the plates increases.
charge on the capacitor plates decreases.
charge on the capacitor plates increases.

126. A. Ed > E0; Vd > V0. D. Ed < E0; Vd > V0.
A charged capacitor has an initial electric field E0 and potential difference V0 across its plates. Without connecting any source of emf, you insert a dielectric (k > 1) slab between the plates to produce an electric field Ed and a potential difference Vd across the capacitor. The pair of statements that best represents the relationships between the magnitude of the electric fields and potential differences is
A. Ed > E0; Vd > V0. D. Ed < E0; Vd > V0.
B. Ed = E0; Vd > V0. E. Ed < E0; Vd < V0.
C. Ed > E0; Vd = V0.

127. Does the capacitance always increase when a dielectric is inserted into the gap of a capacitor?
Yes, it always increases.
No, it always decreases.
No, it may increase or decrease depending on the dielectric constant of the material.

128. An external electric field, E, is applied to a region which contains a dielectric. Which of the following statements is true?
The electric field within the dielectric is less than E.
The dielectric produces an electric field in the opposite direction to E.
The molecules in the dielectric become polarized.
The electric field will produce a torque on molecules in the dielectric that have permanent dipoles.
All the above statements are true.

129. Increased area, decreased separations, “stronger” insulators
In real life we want to store more charge at lower voltage, hence large capacitances are needed
Increased area, decreased separations, “stronger” insulators
Electronic circuits – like a shock absorber in the car, capacitor smoothes power fluctuations
Response on a particular frequency – radio and TV broadcast and receiving
Undesirable properties – they limit high-frequency operation

130. Summary
Capacitance says how much charge is on an arrangement of conductors for a given potential.
Capacitance depends only on geometry
Parallel Plate Capacitor
Cylindrical Capacitor
Spherical Capacitor
Isolated Sphere
Units, F (farad) = C2/Nm or C/V (note e0 = 8.85 pF/m)
Energy and energy density stored by capacitor .

131. Capacitors: series equivalent
Phys Chapter 30

132. Capacitors: parallel equivalent

133. Capacitors
series
parallel V
V1+V2= Veq
V1=V2 Q
Q1=Q2
Q1+Q2= Qeq

134. Dielectrics change the potential difference
The potential between to parallel plates of a capacitor changes when the material between the plates changes.
k is a unitless number

135. Field lines as dielectrics change
Moving from part (a) to part (b) of Figure shows the change induced by the dielectric.
In vacuum, energy density is
In dielectric

136. With battery attached, V=const, so more charge flows to the capacitor
With battery disconnected, q=const, so voltage (for given q) drops.
Norah Ali Al moneef

137. , capacitors with and without dielectrics
If capacitor is disconnected from circuit, inserting a dielectric changes decreases electric field, potential and increases capacitance, but the amount of charge on the capacitor is unchanged.
If the capacitor is hooked up to a power supply with constant voltage, the voltage must remain the same, but capacitance and charge increase
Norah Ali Al moneef

138. Applications of Capacitors – Camera Flash
The flash attachment on a camera uses a capacitor
A battery is used to charge the capacitor
The energy stored in the capacitor is released when the button is pushed to take a picture
The charge is delivered very quickly, illuminating the subject when more light is needed
Defibrillators
When fibrillation occurs, the heart produces a rapid, irregular pattern of beats
A fast discharge of electrical energy through the heart can return the organ to its normal beat pattern
In general, capacitors act as energy reservoirs that can slowly charged and then discharged quickly to provide large amounts of energy in a short pulse