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26.1
Chapter 26: Capacitors
When a spring mousetrap is set, the work done is stored as spring PE. In a similar fashion, a capacitor (two conducting plates separated by an insulator) is a device that stores EPE. To store EPE in this device, one transfers charge to those conductors such that one has a positive charge and the other a negative. Work must be done to move the charges onto the plates which results in a potential difference between the conductors. It is this work that is stored as EPE.
DEMO mousetrap vs. charging a capacitor with lightbulb
Capacitors have a tremendous number of applications (which I will only speaking about ) three categories: 1. Quick recovering of energy (use “small-valued” capacitances)
• Triggers the flash bulb in a camera • Keyboards • Pulsed laser such as the NOVA lasers (possible image of capacitor banks) • Stub-finder • Fuel Cell Power for Lift Trucks
Fermilab power supply. The blue and orange device is a capacitor, among the largest in the world. It stores electrical power for the accelerator at Fermilab. The Tevatron was the most powerful in the world in 2005, colliding particles at energies of 1.8TeV. Magnetic fields to guide the particle beam use a current of 4000A. Cooling the magnets uses 13 MW of power. The annual electricity bill is $12-$18 million.
Capacitive Touch Screens One of the more futuristic applications of capacitors is the capacitive touch screen. These are glass screens that have a very thin, transparent metallic coating. A built-in electrode pattern charges the screen so when touched, a current is drawn to the finger and creates a voltage drop. This exact location of the voltage drop is picked up by a controller and transmitted to a computer. These touch screens are commonly found in interactive building directories and more recently in Apple’s iPhone.
2. Slow release of energy (use “large-valued” capacitances)
Analogy: springs in the suspension of an automobile help smooth out the ride by absorbing the energy from sudden jots. The shocks then release this energy gradually. Capacitors are used in electronic circuits to protect sensitive components by smoothing out variations in voltage due to power surges.
3. Resonant Circuits (use “variable-valued” capacitances) Analogy: springs in mechanical systems have natural frequencies. When the system is driven at the natural frequency (ωdriving = ωnatural), the system is at resonance and large amplitudes occur. The presences of capacitors in a RLC circuit are also resonant circuits that lead to large currents at the resonant frequency. This is the basis for constructing a radio or TV receivers.
Definition: conventional current is the “flow” of positive charge Definition: any two conductors separated by an insulator is a CAPACITOR
26.2
Suppose each conductor starts initially has no charge. When connected to a power supply, positive charge is transferred to the upper plate where as positive charge leaves the bottom plate. As a consequence, the upper plate becomes positively charged and the lower plate is negatively charged. Key point: in-between the plates is a dielectric and therefore, there is NO charge transfer through the capacitor. I will speak about current “through” a capacitor later in the lecture notes. The amount of positive charge that is stored on the upper plate will have exactly the same negative charge on the lower plate, hence, the two conductors have an equal and opposite charge:
a anetb bnet charge 0q q q q q→ = == − = +
The upper plate has a higher potential than the lower plate. Symbolically, in circuit diagrams a capacitor is represented by horizontal lines representing the conductors and the vertical lines are the wires.
The Capacitor Equation The potential field in-between the parallel plates is proportional to the charge on each of the plates, q ∝ ∆V. The more/less charge is stored on the plates, the higher/lower the voltage ∆V rises. The ability or capacity to store charges on the plate must be related to the charge and voltage of the plates. In other words, this must be a characteristic of the capacitor plates themselves. A new quantity must be defined called the Capacitance of the parallel plates, which is measure of the ability of a capacitor to store charge/energy. This can be summarized as the ratio between the two quantities and written as
oq constant C capacitanceV
fability a capacitor tostore charge / energy
qV
= = ≡ = →
∆ ∆
where the Capacitor equation is q C V capacitor equation= ∆
The larger capacitance, the greater amount of EPE (or charge) can be store for a given voltage ∆V. Units: [C] = [q]/[V] = C/V ≡ 1 Farad = 1F
The Capacitance of Different Capacitors The mathematical relationships for the capacitance must be derived for the particular shape of the capacitor (parallel plates, spherical or cylindrical). As we will see, capacitance is the characteristic quantity that defines how much charge can be stored on the capacitor plates and is independent of both charge and voltage. However, one can use the capacitor equation to calculate capacitance because it is related to the ratio of charge to voltage. Capacitance has only two characteristics, geometry and the property material of the dielectric inset. To derive any capacitance, one must use the capacitor equation where the charge and voltage must be calculated. We start with the parallel plate capacitor.
Parallel Plate Capacitance Cparallel plate ≡ Cpp DEMO PASCO parallel plate capacitor
Suppose we have two infinite parallel plates (i.e., one side of the square plates is much larger than the plate separation). The E-field is almost completely localized in the region in-between the plates and given by
0 0 0 between plates 0 outside plates
/2 /2 /E
+
η η = η=
26.3
The charge q and voltage ∆V of the parallel plates must be determined, where taking the ratio of the two will give the capacitance Cpp. 1. The charge on the parallel plates that creates the E-field is related to the charge
density η:
00 0
solving for q
q/A
q/AE q AEη=
η→= = =
2. The voltage in-between the plates is solving for VV
sE V Ed∆
→ ∆∆
= − =
Combining these together, the capacitance is
pp0A EqC
V= =∆
E PP 0
0A A Cd dd
→= =
Note that the capacitance does not depend on charge or voltage, which implies that the ability to store charge on depends on its characteristics, material and geometry dependence. Interpretation of Capacitance To get an idea of what capacitance is, we will first need to answer the questions of (1) how current flow charges up a capacitor and then answer (2) how geometry and the dielectric inset effects capacitance.
Example 26.1 A parallel-plate capacitor has circular plates of 8.20 cm radius and 1.30 mm separation. (a) Calculate the capacitance. (b) What charge will appear on the plates if a potential difference of 120 V is applied? (c) How much electrical potential energy is stored?
Solution a. The capacitance of a parallel-plate capacitor is given by C = ε0A/d, where A is the
area of each plate and d is the plate separation. Since the plates are circular, the plate area is A = πR2, where R is the radius of a plate. Thus,
( ) ( )212 22
30 10
8.85 10 F m 8.2 10 mRC 1.44 10 F 144 pF Cd 1.3 10 m
− −−
−
× π ×π= = = × = =
×
b. The charge on the positive plate is given by q = CV, where V is the potential difference across the plates. Thus,
10 18q C V 1.44 10 F 120V 1.73 10 C 17.3 nC q− −= ∆ = × ⋅ = × = = c. The electrical potential in the capacitor is
9E EU q V 17.3 10 120V 2.08 C U−∆ = ∆ = × ⋅ = µ = ∆
Interpreting how a capacitor is charged
DEMO capacitor and light bulb When the charging of the capacitor occurs, there are three-time intervals to think about: (i) t = 0− (zero minus 0− implies the time just before the switch is closed) when the capacitor starts uncharged. (ii) t = 0+ (zero plus 0+ implies the time just after the switch is closed) current flows and the capacitor starts charging. The circuit is in a state of nonequilibrium during the charging state. Finally, “t = ∞” the capacitor is fully charged and no current flows in the circuit.
What does it mean for a current flow in a capacitive circuit?
Current can only flow when there is a potential difference across two elements. In this case, there is a potential difference (voltage) between the power supply and the capacitor (∆Vsource ≡ ∆VS ≠ ∆Vcap = 0). As the current places more and more charge onto
26.4
the plates, the potential of the capacitor increases but the potential difference between the power supply and capacitor decreases. At “t = ∞”, the potential difference between the power supply and the capacitor are equal (∆VS = ∆Vcap) and the current goes to zero. since there is no potential difference between the two, all current flow stops and we say that the capacitor is now charged.
S cap C
S cap C
V V i (0 ) 0 charging a capacitor
V V i ( ) 0 charged capacitor
+∆ ≠ ∆ → ≠ →
∆ = ∆ → ∞ = →
From the potential picture, how does a capacitor charge up? An uncharged capacitor has a zero potential height initially. When the switch is close, current flows everywhere simultaneously, as and charge starts to build-up on the capacitor. As charge builds up, the capacitor increases in electrical height and at “t = ∞”, the electrical height between the power supply and the capacitor are equal (∆VS = ∆Vcap) and the current goes to zero. The capacitor is now fully charged.
Things to write-up in the future: what does it mean for a current flow in a capacitive circuit?
If I suddenly increase the voltage of the power supply, once again, there is a potential difference and current will continue flowing again, and the capacitor stores more charge. The higher the voltage, the more charge can be placed on the capacitor plates.
The Capacitance characteristics: geometry and dielectric insert dependence
Geometry Dependence The capacitance depends on the area and plate separation:
more area to place charge
stronger field, more induced charge
PP
PP
PP
larger area higher capacitance
smaller separation higher capacitance
A
CAC 1d C
d
∝ → →∝ →
∝ → →
Example: Size of a 1-F capacitor DEMO Pasco parallel plate capacitor Suppose that a parallel plate capacitor where the plates are separated 1mm apart. What would the area have to be in order for the capacitance to be 1-Farad?
According the parallel plate capacitance equation, 2
38 2
12PP
PP0
0(10 km)1F 10 m A 10 m
8.85 10 F/mA C dC d
−
−→ ≈⋅= ≈×
⋅= =
This corresponds to a square area with side of 6 miles! The area of Aptos is 20 km2 while Santa Cruz is 40 km2. It used to be considered a good joke to send an undergraduate or new graduate student to the stockroom for a 1F-capacitor. That’s not
26.5
as funny anymore since farad capacitors are now common. In fact, physics department just purchased a 20F capacitor. The key to this is in the development of activated carbon granules for which a 1-gram sample can have areas of about 1000 m2 but be packaged in a few cm2 area.
Dielectric Dependence A dielectric that is different from the vacuum value, has a different permittivity constant dictated by its dielectric constant κ:
0 0 0 and 1→ →κ ≡ ≥ κ ≥
The capacitance of a parallel plate capacitor is directly proportional to the permittivity constant:
more the polarized field reduces the field inside the dielectric
PP larger dielectric constant higher capacitance C →∝ →
The units for the permittivity constant are written as [ϵ0] = [C][d]/[A] = F/m where ϵ0 = 8.85×10-12 F/m.
We will compare two different capacitive circuits where one capacitor has a vacuum dielectric ϵ0 insert while the has a dielectric insert given by ϵ = κϵ0. The capacitor with capacitance C0 (the one with a vacuum (ϵ0) between its plates) will charge up to charge q0 and voltage ∆V0 = ∆VS. Suppose now that one inserts a dielectric ϵ ≠ ϵ0 in-between the plates – how does the capacitor’s voltage and charge change in this new situation? The E-field between the plates Epp polarizes the dielectric insert and an opposing Epolarized reduces the net field Enet, which in turn, reduces the overall voltage of the capacitor such Vsource > ∆Vcap:
net PP polarized S C CV VE E E i 0∆= − → ≠ → ≠
Current begins flowing again from the power supply to the capacitor and a larger build-up of charge is now stored in this capacitor such that C(ϵ) > C(ϵ0). That is, a dielectric insert different from the vacuum value, has a higher capacity to store charge (C ∝ ϵ). How is the human body capacitance measured?
The external human body resistance is about 500kΩ while the internal resistance is 300Ω to 1000Ω. Only a thin layer of dry skin separates the internal resistance from an external object. The inside of your body can be considered a conductor, and thus if you
26.6
place your hand flat on a metal plate, you will form a capacitor with an area of perhaps 100 cm2, with a thin (maybe 1/8 mm) insulating layer of dry skin, which will form a capacitor. Unfortunately, capacitance is a frequency dependent quantity that we will revisit in the future to determine which hurts more, AC or DC voltages. Stay tuned.
Analogy: storing water in tanks What determines the capacity of cylinders to hold a volume of water? The volume of water that a cylinder can hold is equal to the surface area of the cross section of the cylinder (which is analogous to the surface area of the capacitor's plates) multiplied by the height of the cylinder (which is analogous to the voltage that the capacitor's dielectric (insulator) can withstand). Increasing the surface area and/or height of the cylinder (the maximum voltage rating) will increase the maximum volume (charge) the cylinder can hold.
Area Dependence
larger area capacitance
larger area capacitance
larger diameter pipe larger capacity to hold water
larger plate area larger capacity to hold charge
→
→
Height Dependence The water pump does work on the water and pushes it up the cylinder to some particular height. If the work done by the pump is less than the work done by gravity, then the water level is lower, storing less water and having an overall lower capacity to store water; if it is higher than the water level will rise, storing more water, and increasing its capacity.
work done by the pump PE acquired by the water height of the cylinderwork done by the battery PE acquired betwen the plates Voltage across the plates
= ∝= ∝
Spherical Capacitor Two concentric spherical shells are separated by dielectric where the inner shell (radius a) has charge +q and outer shell (radius b) has charge –q. To determine the capacitance C = q/∆V, the voltage for this charge distribution, must be determined. The only voltge that is not zero is in-between the spherical shells and therefore, we determine ∆Vab:
b baab b sphere 2a 0 0 0a
q dr q 1 1 q b aV V V E dr4 4 a b 4 abr
∆
−= − = ⋅ = = − =π π π∫ ∫
Substituting
this voltage into the capacitance equation,
sphere sphere0a
0
b
q q 4 abC CV b aq b a
4 ab
= = π =
=−−
π
Remarks 1. A spherical capacitor has the general mathematical structure as a parallel plate
capacitor:
26.7
sphere
materiag
0 0 0
leometry
area 4 distance
ab 4 abCb a b a
π = =
π=− −
The quantity 4πab is the geometric area average between the two spheres and the b−a is the plate separation between the two plates. Physically, the way to interpret this is if the distance between the spherical plates is very small in comparison to their radii (b-a ≪a and b, then they behave like a parallel plate.
2. If the outer radius is much larger than the inner radius (b ≫ a), then the outer shell is said to be at infinity (that’s where the E-field lines of flux terminate at) and what’s left is an isolated sphere. The capacitance of an isolated sphere of radius a = R is
0 0sphere isolatedb large b large sphere04 4 a 4 R Calim C lim 1 a / b→ →
π π
= ≡ π ==
−
The capacitance of objects can be determine just knowing its radius using
( )1210sphere
R 10 R(cm) F10
C −≈ =
The capacitance of a Van de Graaff (R = 12.5 cm) is 12 2
VdG 10 10 0.1 pFC −≈ ⋅ =
DEMO Van de Graaff
Cylindrical Capacitor DEMO Coaxial cable
A coaxial cable is used for transmission of high-frequency (broadband) electrical signals. Shown here is a stripped coaxial cable revealing its internal structure. The innermost conducting wire is insulated from the cylindrical conducting sheath by the non-conducting spacer (typically made of polyethylene). The whole cable is usually covered by an insulating jacket. Such a construction allows for confining of the electromagnetic signal inside the cable, between the inner and the outer conductors, and therefore makes the cable immune to RF electromagnetic interference (Faraday Cage). The most common household use of coaxial cables is to bring the cable service to the home TV set. However, they are also used extensively in industrial radio-frequency applications.
A long coaxial cable has an inner conducting cylinder with radius a and linear charge density +λ; the outer shell has radius b and charge density –λ. What is the capacitance of this geometry?
As we did earlier, we must start with the capacitance equation in order to derive the capacitance of a coaxial cable. So let do this:
ref surfaceb b
aab b cylindera a0 0 0
r Rln lnr r
drV V V E dr2 r 2 2
λ λ λ= − = ⋅ = = =π π π∫ ∫
For the coaxial cable, the reference location where we will measure the potential is at the outer shell radius b so
ab0
blna
V2
λ=π
Substituting into the capacitance equation,
cylinder cylinder0
ab
0
Lq Lblna
2C CV ln(b / a)
2
λ= = =
π=
λπ
Remarks 1. The capacitance per unit length is
26.8
cylinder0
materialgeometry
L
C 2 1ln(b / a) ln(b / a)
= ∝π
The capacitance of a coaxial cylinder is determined entirely by its dimensions. This is similar to a parallel plate and spherical capacitors.
2. Ordinary coaxial cables have an insulating material in-between the conductors. Since the coaxial cable is a hollow conductor, it shields the electrical signals from any possible external influences. A typical cable for TV antennas and VCR connections have C/L = 70 pF/m.
3. The cable carries electrical signals on the outer and inner shells which this geometry The Nature of Circuit Elements in Series or Parallel When a combination of circuit elements is added together, the two most basic combinations are series and parallel. Series or parallel combinations always have the same basic behavior, regardless of the circuit element.
Series combination For a series combination along a branch of a circuit, all circuit elements (capacitors, resistors, and inductors) that are in series will always divide the voltage across the branch (known as Kirchhoff’s Voltage Law – KVL) and currents through series elements are the same. These basic characteristic behaviors are summed up as
series 1 2 3
series 1 2 3
V V V V
and
i i i i
∆ = ∆ + ∆ + ∆
= = =
As current flows through the first element, current does not building-up at anyone place. This is analogous to water flowing through a water hose, no water ever builds up at anyone location, it just keeps flowing. On the other hand, the voltage must divide since the electric height change from the ground is different for each element.
Parallel combination For a parallel combination along a branch of a circuit, all circuit elements (capacitors, resistors, and inductors) that are in parallel will always divide the current through the element (known as Kirchhoff’s Current Law – KCL) and voltages across parallel elements are the same. These basic characteristic behaviors are summed up as:
parallel 1 2 3
parallel 1 2 3
V V V V
and
i i i i
∆ = ∆ = ∆ = ∆
= + +
As current flows into a parallel branch, the junction (or node) forces the current to divide or split into three different paths. On the other hand, the voltage does not divide since the electric height change from the ground is all the same for each element. We now apply these ideas to parallel and series capacitors while keeping an eye on the capacitance equation: C = ϵA/d. Series Capacitors
26.9
From our previous discussion, capacitors in series immediately imply that (i) voltages are divided among series capacitors and (ii) the current is the same through series capacitors. What does this mean for series capacitance? When capacitors are added in series, the effective plate separation increases as more capacitors are added in series. The whole combination of series capacitors can be replaced with the effective capacitance Cseries:
where
1 1series n
series 1 2
A 1 1 1 1C or C Cd d C C C
− −= ∝ → = + + = ∑
Therefore, adding more and more capacitors in series decreases the capacitance: series 1 2C C , C , ...<
For series capacitors, the current “through” each capacitor is the same. However, it is customary at this level to speak about the amount of charge qcap each capacitor receives at steady state (or equilibrium), which is directly proportional to current icap: q ∝ i. In other words, each capacitor stores the same amount of charge given by
series 1 2 3 series 1 2 3i i i i q q q q= = = → = = = In the same though, capacitor in series divide the voltage, so the latter relation can be written as
series 1 2 n series branch 1 1 2 2 n nq q q q C V C V C V C V= = = = → ∆ = ∆ = ∆ = = ∆ Note that this relation can be written to derive the Voltage Divider Rule (VDR) for capacitors:
seriesseries branch 1 1 n n n branch
n
VDR forCC V C V C V V V capacitorsC
∆ = ∆ = = ∆ → ∆ = ∆
The manner in which capacitors behave voltage wise is
n
seriesn branch
C
seriesn n branch
nCsmaller capacitance C larger voltage drop V
Clarger capacitance C smaller voltage drop V Vn
V
C
→ = ∆
→ ∆ = ∆
∆
Physically, because of the condition that series capacitors have the charges, a small capacitance forces the voltage drop across it to be large while a large capacitance has a large voltage drop. The capacitance equation reads
q constant C V V C VC= = ∆ = ∆ = ∆ Parallel Capacitors From our previous discussion, capacitors in parallel immediately imply (i) currents are divided among parallel capacitors and (ii) voltage is the same across parallel capacitors. What does this mean for capacitance? When capacitors are added in parallel, the effective plate area increases as more capacitors are added in parallel. That is, capacitors in parallel can be replaced by the effective capacitance Cparallel:
26.10
where
parallel 1 2 n parallel nAC A C C C C or C Cd
= ∝ → = + + = ∑ = ∑
Since adding more and more capacitors in parallel increases the effective area, the parallel capacitance increases such that
parallel 1 2C C , C , ...> For parallel capacitors, the voltage across each capacitor is the same,
parallel 1 2 nV V V V∆ = ∆ = ∆ = = ∆ but the currents are divided. It is customary at this level to speak about the amount of charge qcap each capacitor receives at steady state (or equilibrium), which is directly proportional to current icap: q ∝ i. In other words, each capacitor stores the same amount of charge given by
parallel 1 2 n parallel 1 2 ni i i i q q q q= + + + → = + + + How much charge is stored on parallel capacitors depend on its capacitance value. Using the fact that all parallel capacitors have the same voltage, one can write down the “charge divider rule” where
parallel 1 nparallel 1 2 n
parallel 1 n
nn parallel
parallel
q q qV V V VC C C
Charge Divider RuleC q q for capacitorsC
∆ = ∆ = ∆ = = ∆ → = = =
→ =
The manner in how capacitors divided the charge is n
n parallelparallel
parallelparallel
n
Csmaller capacitance smaller charge stored: q qC
larger capacitance larger charge stored: qCnCq
→ =
→ =
In the table, series and parallel capacitor characteristics are summarized:
Example: Series and Parallel Capacitors Let C1 = 6.0 µF, C2 = 3.0 µF, and Vab = 18V. Find the equivalent capacitance, and find the charge and potential difference for each capacitor when the two capacitors are connected in (a) series and (b) parallel.
Solution
26.11
a. Using the equivalent capacitance of a series combination, we find 1 2
eq serieseq 1 2 1 2
C C1 1 1 6 3C 2 F CC C C C C 6 3
⋅ ⋅= + → = = = µ =
+ +
The charge Q on each capacitor in series is the same as the charge on the equivalent capacitor:
( )( )series seriesQ C V 2.0 F 18 V 36 C Q= = µ = µ = The potential difference across each capacitor is inversely proportional to its capacitance:
1 1 2 21 2
Q 36 C Q 36 CV 6.0 V V and V 12.0 V VC 6.0 F C 3.0 F
µ µ= = = = = = = =
µ µ
b. To find the equivalent capacitance of the parallel combination, we use eq 1 2 parallelC C C 6.0 F 3.0 F 9.0 F C= + = µ + µ = µ =
The potential difference across each of the two capacitors in parallel is the same as that across the equivalent capacitor, 18 V. The charges Q1 and Q2 are directly proportional to the capacitance C1 and C2:
( )( )( )( )
1 1 1
2 2 2
Q C V 6.0 F 18 V 108 C Q
Q C V 3.0 F 18 V 54 C Q
= = µ = µ =
= = µ = µ =
Note that Cseries is less than either C1 or C2, while Cparallel is greater than C1 or C2. It’s instructive to compare the potential differences and charges in each part of the example. For two capacitors in series, the charge is the same on both capacitors and the larger potential difference appears across the capacitor with the smaller capacitance. Furthermore, the sum of the voltages in series add up to 18 V (= V1 + V2), as it must. By contrast, for two capacitors in parallel, each capacitor has the same potential difference and the larger charge appears on the capacitor with the larger capacitance.
Example 26.2 What are the (a) equivalent capacitance Ceq of the capacitors and the charge stored by Ceq? What are voltages and charges on the capacitors (b) CX, CY, and CZ?
Solution a. Using the equivalent capacitance of the upper left and upper
right combinations are
left leftleft right
1 1 1 4 12 1 1 1 6 12C 3 F and C 4 FC 4 12 4 12 C 6 12 6 12
⋅ ⋅= + → = = µ = + → = = µ
+ +
Now, Cleft and Cright are in parallel and add up to Cyz = Cleft + Cright = 7μF. Finally, the equivalent capacitance is Cyz and Cx in series:
x yzeq eq
eq yz x x yz
C C1 1 1 7 7C 3.5 F CC C C C C 7 7
⋅ ⋅= + → = = = µ =
+ +
The total charge on the equivalent capacitance is determined using the capacitance equation:
( ) ( )eq eq S eqQ C V 3.5 F 10 V 35 C Q= ∆ = µ = µ =
b. To Determine the voltages and charges of CX, CY, and CZ, one uses the VDR for capacitors. find the equivalent capacitance of the parallel combination, we use
eq 1 2 parallelC C C 6.0 F 3.0 F 9.0 F C= + = µ + µ = µ =
Capacitance Cx
26.12
The capacitors Cyz and Cx are in series and therefore, they (i) divided the 10V according to their capacitor values via the VDR and (ii) they have the same charge. The voltage and charge of capacitor Cx is
seriesx x x x x x x
x
C 3.5V V 10V 5 V V Q C V 7 F 10V 70 C QC 7
∆ = ∆ = ⋅ = = ∆ → = ∆ = µ ⋅ = µ =
Since the capacitors have the same capacitance values, they divided the voltage in half: ∆Vyz = ∆Vx = 5V.
Capacitance Cy The capacitor Cy is embedded in Cyz and we will have to work our ways “backwards” in the circuit. We have already determined the capacitance of this right branch: Cright = 4μF. Applying VDR to determine the voltage and charge of capacitor Cy is
seriesx y y
x
y y y x
53
53
C 4V V 5V V VC 12
Q C V 12 F V 20 C Q
∆ = ∆ = ⋅ = = ∆
→ = ∆ = µ ⋅ = µ =
Capacitance Cz The capacitor Cz is embedded in Cyz and we apply the same kind of thinking to get the voltage and charge of Cz:
seriesz z z
z
z z z z
154
154
C 3V V 5V V VC 4
Q C V 4 F V 15 C Q
∆ = ∆ = ⋅ = = ∆
→ = ∆ = µ ⋅ = µ =
Example 26.3 The circuit displays a battery and three uncharged capacitors. The switch is thrown to terminal-a until capacitor C1 is fully charged. Then the switch is thrown to terminal-b. What is the final charge and voltage for each capacitor when the capacitor system reaches equilibrium? Solution In this circuit there are three-time intervals: (i) switch is at terminal-a and capacitor C1 is in equilibrium with the power supply. (ii) Switch is now thrown to terminal-b, and the capacitors are in a nonequilibrium state. (iii) After sometime (t = ∞), the system of capacitors reach equilibrium. We are only concerned with the equilibrium states of (i) and (iii).
At terminal-a (t = 0−) Capacitor C1 fully charges up to 12V and acquires an initial charge q10:
10 10 1 10 10V 12V and q C V 4 12 48 F q∆ = = ∆ = ⋅ = µ =
At terminal-b (t = ∞) The switch has been at terminal-b for a suffice time to reach equilibrium, capacitor C1 discharges until the voltage (∆V1f) across C1 is equal to the voltage (∆V23) of the combination of capacitors C2 & C3 in series:
1f 23V V∆ = ∆ Since capacitor C1 discharged from its initial value q10, the total amount of charge cannot disappear and therefore, must remain constant. That is, the initial charge q10 discharges and leaves capacitor C1 with less charge while the two series capacitors gain charge q23. This charge conservation, can then be rewritten using the capacitor equation:
10 1f 23 10 10 1 1f 23 23q q q C V C V C V= + → ∆ = ∆ + ∆
26.13
However, the final voltages must be the same: ∆V1f = ∆V23. Substituting this into the above equation, and solving for ∆V1f, we get
1 101 10 1 1f 23 1f 1f
1 23
C VC V C V C V VC C
∆∆ = ∆ + ∆ →∆ =
+
Substituting in the numbers, the voltage and charge of C1 is
1f 1f
1f 1 1f 1f
4 12V 8V V4 6 3/(6 3)
q C V 4 8 32 C q
⋅∆ = = = ∆
+ ⋅ +
= ∆ = ⋅ = µ =
The voltage across capacitors C2 and C3 is ∆V1f = ∆V23 = 8V, and they are in series C23 = 2μF. The fact that C2 and C3 are in series implies the (i) sum of the voltages ∆V2 + ∆V3 = 8V must be true and (ii) the charges for series capacitors must also be the same, q2 = q3. Calculations must show this! Using the VDR for capacitors, the voltages and charges of C2 and C3 are
232 1f 2 2 2 2 2
2
233 1f 2 3 2 2 3
3
C 2V V 8V 2.67V V , q C V 6 F 2.67V 16 C qC 6C 2V V 8V 5.33V V , q C V 3 F 5.33V 16 C qC 3
∆ = ∆ = ⋅ = = ∆ = ∆ = µ ⋅ = µ =
∆ = ∆ = ⋅ = = ∆ = ∆ = µ ⋅ = µ =
Example 26.4 A parallel plate capacitor (0.12 m2, 1.2 cm) is charged by a battery of 120V and is then disconnected. A glass slab of thickness 4.0 mm and dielectric constant 4.8 is then placed symmetrically between the plates. What are the values (i) before and (ii) after the slab is inserted for (a) charge, (b) electric field, and (c) voltage across the plates, and (d) capacitance? (e) How much external work is involved in inserting the slab?
Solution a. Initially, the capacitance is
12 290
0 02A 8.85 10 0.12C 1.06 10 F 1.06 nF C
d 1.2 10
−−
−
× ⋅= = = × = =
×
and therefore, before the insertion the initial charge q0 is 9 9
0 0 0 0 fq C V 1.06 10 F 120V 127 10 F 127 nF q q− −= ∆ = × ⋅ = × = = = Since the battery is disconnected, the charge will remain the same after the insertion of the slab (where can it go?).
b. The initial E-field is related to the voltage across the plates: 30
0 02V 120VE 10 10 V/m 10 kV/m Ed 1.2 10 m−
∆= = = × = =
×
The E-field is reduced to Enet = E0 – Eploarized where the E-field of the dielectric slab is 3
30polarized polarized
E 10 10 VE 2.08 10 V/m 2.08 kV/m E4.8×
= = = × = =κ
c. The voltage before the insertion is of course ∆V0 = 120V. After the insertion, part of the region between the plates is filled with air (call this plate separation x) while the remaining space is filled with the glass slab. Using the definition of the voltage (magnitude only),
0f
E 10kV/m E 2.1kV/mf 0 f 0 fd 12mm x 4mmx 4mm
f
V E ds E ds E (d x) E (x 0)
V 88.4V
= == ==
∆ = ⋅ + ⋅ = − + −
∆ =
∫ ∫
d. The final capacitance is derived from the usual suspect:
26.14
f0 f f
f
q 127nFC 1.06 nF C 1.44 nF CV 88.4 V
= → = = = =∆
e. The work done is 2 9 2
7ext ext12 12
0
q 1 1 (11 10 ) 1 1W U 1.7 10 J W2 C C 2 89 10 120 10
−−
− −
× = ∆ = − = − = − × = × ×
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