Ch 17 student_201516

Topic --- Capacitor & Dielectrics
17.1 Capacitance & Capacitors In Series &
Parallel (1 Hour)
17.2 Charging & Discharging Of Capacitors
(1 Hour)
17.3 Capacitors With Dielectrics (1 Hour)

Capacitor&Dielectrics
Capacitance
Capacitors
In series
In parallel
Charging & discharging
of capacitor
Dielectric

17.1 CAPACITANCE & CAPACITORS
IN SERIES & PARALLEL
(a) Define and use capacitance,
(b) Derive and determine the effective
capacitance of capacitors in series and
parallel
(c) Derive and use energy stored in a capacitor,
V
Q
C 
C
Q
QVCVU
2
2
2
1
2
1
2
1


Capacitor
(Condenser)
A device that is
capable of
storing electric
charges or
electric
potential
energy
Consist of two
conducting
plates separated
by a small air gap
or a thin
insulator
(dielectric) such
as mica, ceramics
or even oil
Uses: Photo-
flash, on-off
switches,
smoothen
direct current
(d.c) voltages
OR 
17.1 CAPACITANCE & CAPACITORS IN SERIES & PARALLEL

CVQ 
V
Q
C 
VQ 

V
+Q1
QIC1,V1
+Q2
Q2
C2,V
2
+Q3
Q3C3,V3
+Q
Q
Ceff,
V
V
321
111
CCCQ
V

eff
1
CQ
V

capacitors
in series
Where and
11
1
1
C
Q
C
Q
V 
22
2
2
C
Q
C
Q
V 
The magnitude of the charge Q on each plate is
the same
The potential difference across each capacitor
C1, C2 and C3 are V1,V2 and V3 respectively
The total potential difference V is given by
321 QQQQ 
33
3
3
C
Q
C
Q
V 
321 VVVV 
321 C
Q
C
Q
C
Q

nC
...
CCCC
11111
321eff


V +Q1
QI
C1,V1
+Q2
Q2
C2,V2
+Q3
Q3
C3,V3
+Q
Q
Ceff,
V
V
The potential difference across each capacitor
is the same as the supply voltage (V)
321 VVVV 
The charges stored by each capacitor C1,C2 and C3 are Q1,Q2 and Q3 respectively
The total charge Q on the effective capacitor is given by
where
;1111 VCVCQ  ;2222 VCVCQ  VCVCQ 3333 
321 QQQQ  VCVCVC 321 
and321 CCC
V
Q
 effC
V
Q
 nC...CCCC  321eff
capacitors
in parallel

Calculate the total capacitance in (a), (b) and (c).

Determine the potential difference across and
charges on capacitors X, Y and Z.

Determine the
effective capacitance
of the configuration
shown in Figure 17.1.
All the capacitors are
identical and each has
a capacitance of 2 F.
Figure 17.1
μF
11
30
eff C OR μF73.2

In Figure 17.2, C1= 100 F, C2
= 200 F and C3 = 300 F. The
applied potential difference
between points A and B is
VAB = 8.0 V. Calculate
(a) the charge on each
capacitor
(b) the potential difference
across each capacitor
(c) the potential difference
between points A and D.
A
B
D
1C
2C
3C
Figure 17.2
μC1200123  QQQ
V0.4321  VVV
V0.412AD VV

• When a switch in circuit
closed, charges begin to
accumulate on the plates
• A small amount
of work (dW) is done in
bringing a small amount of
charge (dQ) from the
battery to the capacitor
• This is given by
• The total work W required to
increase the accumulated
charge from zero to Q is given
by
dQ
VandVdQdW 
C
Q
V 
dQ
C
Q
dW 
 
QW
dQ
C
Q
dW
00
C
Q
WU
2
2
1

2
2
1
CVU 
QVU
2
1

 
  

Figure 17.3 shows a combination of three capacitors where C1=
100 F, C2 = 22 F and C3 = 47 F. A 20 V
supply is connected to the combination.
Determine
(a) the effective capacitance in the circuit,
(b) the charge stored in the capacitor C1,
(c) the potential difference across the
capacitor C2,
(d) the energy stored in the capacitor C3,
(e) the area of the each plate in capacitor C1 if the distance
between two plates is 0.02 m and the region between
plates is vacuum.
(Permittivity of free space, 0 = 8.85  1012 C2 N1 m2)
1C
2C 3C
V02
Figure 17.3

14
Consider the circuit shown in Figure 17.4, where C1= 50 F, C2 =
25 F and V = 25.0 V.
Capacitor C1 is first charged by
closing a switch S1. Switch S1 is
then opened, and then the
Charged capacitor is connected
to the uncharged capacitor C2 by
closing a switch S2.
Calculate the initial charge acquired by C1 and the final charge
on each capacitor.
1C 2C
V
1S 2S
Figure 17.4

Given 0 = 8.85  1012 C2 N1 m2
1. Four capacitors are connected as shown in Figure 17.5.
Figure 17.5
Calculate
(a) the equivalent capacitance
between points a and b,
(b) the charge on each capacitor if
Vab=15.0 V.
(Physics for scientists and engineers,6th
edition,Serway&Jewett, Q21, p.823)
ANS: 5.96 F; 89.5 C on 20 F, 63.2
C on 6 F, 26.3 C on 15 F and on 3
F

2. Determine the equivalent capacitance between
points a and b for the group of capacitors
connected as shown in Figure
17.6.
Figure 17.6
Take C1 = 5.00 F, C2 = 10.0
F and C3 = 2.00 F.
(Physics for scientists and engineers,6th
edition,Serway&Jewett, Q27,
p.824)
ANS: 6.04 F

17
3. An electronic flash unit for a camera contains a capacitor of
capacitance 850 F. When the unit is fully charged and ready for
operation, the potential difference across the plates is 330 V.
(a) What is the magnitude of the charge on each plate of the fully
charged capacitor?
(b) Calculate the energy stored in the “charged-up” flash unit.
(Physics,3rd edition, J.S.Walker, Q59, p.692)
ANS: 0.28 C; 46 J
4. A parallel-plate capacitor has plates with an area of 405 cm2 and an
air-filled gap between the plates that is 2.25 mm thick. The capacitor
is charged by a battery to 575 V and then is disconnected from the
battery.
(a) How much energy is stored in the capacitor?
(b) The separation between the plates is now increased to 4.50 mm.
How much energy is stored in the capacitor now?
(c) How much work is required to increase the separation of the
plates from 2.25 mm to 4.50 mm?
(Physics,3rd edition, J.S.Walker, Q60, p.692)
ANS: 2.63  105 J; 5.27  105 J; 2.63  105 J

17.2 CHARGING & DISCHARGING
OF CAPACITORS
18
(a) Define and use time constant,  = RC
(b) Sketch and explain the characteristics of Q-t and I-t
graph for charging and discharging of a capacitor
(c) Use
for discharging and
for charging
RC
t
eQQ

 0










RC
t
eQQ 10

Charging Discharging
17.2 CHARGING & DISCHARGING OF CAPACITORS

20
Time constant, 
• Scalar
quantity
• Unit: s
• A
measurement
of how
quickly the
capacitor
charges or
discharges
RC

• The voltage V across the
capacitor, increase from
zero at t = 0 to maximum
values V0 after a very long
time
•  is defined as the time
required for the capacitor to
reach (1 e1) = 0.63 or 63% of
its maximum voltage
Charging
0 t,time
V
0V
063.0 V
RCτ 










RC
t
eVV 10
0
0
V
C
Q











RC
t
e
C
Q
C
Q
V 10
and

• The charge Q across the
capacitor, increase from
zero at t = 0 to maximum
values Q0 after a very long
time
required for the capacitor to
reach (1 e1) = 0.63 or 63% of
its maximum charge
Charging
0 t,time
Q
0Q
063.0 Q
RCτ 










RC
t
eQQ 10
00 CVQ and

• the current drops
exponentially in time constant

required for the current drops
to 1/e = 0.37 or 37% of its
initial value(I0)
Q0: maximum charge
V0: maximum (supply) voltage
I0: maximum current
R: resistance of the resistor
C: capacitance of the capacitor
23
Charging
23
0I
0 t,time
I
037.0 I
RCτ 
and
RC
t
eII

 0
R
V
I 0
0 

24
0V
0 t,time
V
037.0 V
RCτ 
RC
t
eVV

 0
• the charge Q, the voltage V
and the current I is seen to
decrease exponentially in
time constant 
required for the charge on
the capacitor (or voltage
across it or current in the
resistor) decreases to 1/e =
0.37 or 37% of its initial value
Discharging: V-t, Q-t & I-t graph
Discharging

25
Charge on capacitor Current through resistor
0I
0 t,time
I
RCτ 
037.0 I
The charge on the capacitor
decreases exponentially with time
The current through the resistor
decreases exponentially with time
0Q
0 t,time
Q
037.0 Q
RCτ 
RC
t
eQQ

 0
RC
t
eII

 0
Discharging

In the RC circuit shown in Figure 17.15, the battery has
fully charged the capacitor. At time t = 0 s, a switch S is
thrown from position a to b. The battery voltage V0 is
12.0 V and the capacitance
C = 3.00 F. The current I is
observed to decrease to 0.45
of its initial value in 60 s.
Determine
(a) the value of R.
(b) the time constant, 
(c) the value of Q, the charge on the capacitor at t = 0.
(d) the value of Q at t = 100 s
C
R
0V
S
b
a
Figure 17.7

17.3 CAPACITORS WITH
DIELECTRICS
(a) Calculate capacitance of air-filled parallel plate
capacitor,
(b) Define and use dielectric constant
(c) Describe the effect of dielectric on a parallel plate
capacitor.
(d) Use capacitance with dielectric, 0CC r
0
r
ε
ε
ε 
d
Aε
C 0


• Consider a two parallel metallic plates capacitor of equal area
A are separated by a distance d and the space between plates
is vacuum or air as shown in Figure 17.8.
• When the capacitor is
charged, its plates have
charges of equal
magnitudes but opposite
signs: + Q and Q then
the potential difference V
across the plates is
produced.
d
Q
Q
V
A
positive
terminal
negative
terminal
E

Figure 17.8
17.3 CAPACITORS WITH DIELECTRICS

• Since d << A so that the electric field strength E is uniform
between the plates.
• The magnitude of the electric field strength within the plates
is given by
where, : surface charge density on either side
(Unit: C m-2 and a scalar quantity)
• Since Q=CV and V=Ed (Uniform E)
then equation (17.1) can be written as
0ε
σ
E 
A
Q
σ and
0Aε
Q
E  (17.1)
0Aε
CV
E 
0Aε
CEd
E 

0 = permittivity of free space = 8.85 x 10-12 C2 N-1 m-2
 = permittivity of dielectric material
d = distance between the two plates
A = area of each plate
• From equations (17.2) and (17.3), The capacitance, C of a
parallel-plate capacitor is proportional to the area, A of its plates
and inversely proportional to the plate separation, d.
(17.2)
d
Aε
C 0

and
(17.3)
d
εA
C 
Parallel-plate capacitor
separated by a vacuum
Parallel-plate capacitor
separated by a dielectric
material

A parallel-plate capacitor has plates of area 280 cm2 are separated by
a distance 0.550 mm. The plates are in vacuum. If a potential
difference of 20.1 V is supplied to the capacitor, determine
a. the capacitance of the capacitor.
b. the amount of charge on each plate.
c. the electric field strength was produced.
d. the surface charge density on each plate.

A circular parallel-plate capacitor with radius of 1.2 cm is connected to
a 6.0 V battery. After the capacitor is fully charged, the battery is
disconnected without loss of any of the charge on the plates. If the
separation between plates is 2.5 mm and the medium between plates
is air.
a. Calculate the amount of charge on each plate.
If their separation is increase to 8.0 mm after the battery is
disconnected, determine
b. the amount of charge on each plate.
c. the potential difference between the plates.
d. the capacitance of the capacitor.
(Given permittivity of free space, 0 = 8.85  1012 C2 N1 m2)

1. a. A parallel-plate, air-filled capacitor has circular
plates separated by 1.80 mm. The charge per unit area
on each plate has magnitude of 5.60 pC m2. Calculate
the potential difference between the plates of the
capacitor.
(University physics,11th edition, Young&Freedman, Q24.4, p.934)
b. An electric field of 2.80  105 V m1 is desired
between two parallel plates each of area 21.0
cm2 and separated by 0.250 cm of air. Determine the
charge on each plate. (Physics for scientist & engineers ,3rd
edition, Giancoli, Q14, p.628) ANS: 1.14 mV; 5.20  109 C
2. When the potential difference between the plates of a
capacitor is increased by 3.25 V, the magnitude of the
charge on each plate increases by 13.5 C. What is
the capacitance of this capacitor?
(Physics,3rd edition, J.S.Walker, Q86, p.694) ANS: 4.15 F

3. A 10.0 F parallel-plate, air-filled capacitor with circular
plates is connected to a 12.0 V battery. Calculate
a. the charge on each plate.
b. the charge on each plate if their separation were twice
while the capacitor remained connected to the battery.
c. the charge on each plate if the capacitor were
connected to the 12.0 V battery after the radius of each
plate was twice without changing their separation.
(University physics,11th edition, Young&Freedman, Q24.5, p.934)
ANS: 120 C; 60 C; 480 C
4. A capacitor stores 100 pC of charge when it is connected
across a potential difference of 20 V. Calculate
a. the capacitance of the capacitor,
b. the amount of charge to be removed from the
capacitor to reduce its potential difference to 15 V.
ANS: 5.0 pF; 25 pC

• Consider a parallel-plate capacitor as shown in Figure 17.19
Figure 17.9
Q Qd
0E


• Initially the plates are separated by a
vacuum and connected to a battery,
giving the charge on the plates +Q and –Q
• The battery is now removed and the
charge on the plates remains constant
• The electric field between the plates is
uniform and has a magnitude of E0
• Meanwhile the separation between plates
is d
• When a dielectric is placed in the electric
field between the plates, the molecules of
the dielectric tend to become oriented
with their positive ends pointing toward
the negatively charged plate and their
negative ends pointing toward the
positively charged plated as shown in
Figure 17.10
Q Q
Figure 17.10
0E

E


• The result is a buildup of positive charge on one surface of
the dielectric and of negative charge on the other as shown in
Figure 17.11 Q Q
Figure 17.11
E


38
• From Figure 17.11, the number of field lines within the
dielectric is reduced thus the applied electric field E0 is
partially canceled.
• Because the new electric field strength (E < E0) is less then
the potential difference, V across the plates is less as well.
• Since V is smaller while Q remains the same the capacitance,
is increased by the dielectric.
EdV 
V
Q
C 

• is defined as a ratio between the capacitance of a given
capacitor with space between plates filled with dielectric, C
and the capacitance of the same capacitor with plates in a
vacuum, C0
Mathematically,
No unit for dielectric constant.
• For parallel-plates capacitor:
0
r
C
C
ε  (17.10)
d
εA
C 
d
Aε
C 0
0 
and













d
Aε
d
εA
ε
0
r

then the equation (17.10) can be written as
: permittivity of dielectric material
• From the definition of the capacitance,
hence the equation (17.10) can be written as
V: potential difference across capacitor with dielectric
V0: potential difference across capacitor in vacuum
0
r
ε
ε
ε  0rεεε OR (17.11)
V
Q
C 
0
0
V
Q
C and where Q is constant
V
V
ε 0
r  (17.12)

• From the relationship between E and V for uniform electric
field,
thus the equation (17.12) can be written as
E0: electric field strength of the capacitor in vacuum
E0: electric field strength of the capacitor with dielectric
• The dielectric constant depends on the insulating material
used.
• Table 17.1 shows the value of dielectric constant and the
dielectric strength for several materials.
EdV  dEV 00 and
E
E
ε 0
r  (17.13)
Ed
dE
V
V
ε 00
r 

• The dielectric strength is defined as the electric field
strength at which dielectric breakdown occurs and the
material becomes a conductor.
Material Dielectric constant, r
Dielectric Strength
(106 V m1)
Air 1.00059 3
Mylar 3.2 7
Paper 3.7 16
Silicone oil 2.5 15
Water 80 -
Teflon 2.1 60
Table 17.1
Note:
E
E
V
V
ε
ε
C
C
ε 00
00
r 

A vacuum parallel-plate capacitor has plates of area A = 150 cm2 and
separation d = 2 mm. The capacitor is charged to a potential difference
V0 = 2000 V. Then the battery is disconnected and a dielectric sheet of
the same area A is placed between the plates as shown in Figure
17.12. In the presence of the dielectric, the potential difference across
the plates is reduced to 500 V. Determine
(a) the initial capacitance of the capacitor,
(b) the charge on each plate before the dielectric is inserted,
(c) the capacitance after the dielectric is in place,
(d) the relative permittivity,
(e) the permittivity of dielectric material,
(f) the initial electric field,
(g) the electric field after the dielectric is inserted.
(0 = 8.85  1012 C2 N1 m2)
Figure 17.12
dielectric
d

A parallel-plate capacitor has the space between the plates filled with two
slabs of dielectric constants 1 and 2 as shown in Figure 17.13.
Each slab has thickness d/2, where d is the plate separation. Show that
the capacitance of the capacitor is
Figure 17.13








21
2102


d
A
C
d/2
d/22
1

Given 0 = 8.85  1012 C2 N1 m2
1. What are the maximum and minimum equivalent capacitances that
can be obtained by combinations of three capacitors of 1.5 F, 2.0
F and 3.0 F?
(College Physics,6th edition, Wilson, Buffa & Lou, Q97, p.566)
ANS: 6.5 F; 0.67 F
2. The dielectric of a parallel-plate capacitor is to be constructed from
teflon that completely fills the volume between the plates. The area
of each plate is 0.50 m2.
a. What is the thickness of the teflon if the capacitance is to
be 0.10 F?
b. Calculate the charge on the capacitor if it is connected to a
12 V battery.
(Dielectric constant for teflon is 2.1)
ANS: 92.9 m; 1.2 C

3. Explain clearly why the electric field between two parallel plates of
a capacitor decreases when a dielectric is inserted if the capacitor
is not connected to a power supply, but remains the same when it
is connected to a power supply.
(College Physics,6th edition, Wilson, Buffa & Lou, Q79, p.566)
4. An air-filled parallel-plate capacitor has rectangular plates with
dimensions of 6.0 cm  8.0 cm. It is connected to a 12 V battery.
While the battery remains connected, a sheet of 1.5 mm thick
paper is inserted and completely fills the space between the
plates.
a. Explain briefly what is happen to the charge on the plates
of the capacitor while the dielectric was being inserted.
b. Determine the change in the charge storage of the
capacitor because of the dielectric insertion.
(Dielectric constant for paper is 3.7)
ANS: 0.92 nC

47
Next Chapter…
CHAPTER 18 :
Electric current
&
direct-current circuits

Ch 17 student_201516

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