Download C 1 - The Pimsat Student

CHAPTER 17
CAPACITOR
&
DIELECTRICS
(PST :3 hours)
(PDT : 7 hours)
17.1
17.2
17.3
17.4
Capacitors
Capacitors in series and parallel
Charging and discharging of capacitors
Capacitors with dielectrics
17.1 CAPACITORS
LEARNING OUTCOMES :
At the end of this lesson, the students should be able to :
a) Define capacitance.
b) Use formulae, C 
Q
V
b) Calculate the capacitance of parallel plate capacitor.
17.1 Capacitors
• A capacitor , sometimes called a condenser, is a
device that can store electric charge.
• It is consists of two conducting plates separated
by a small air gap or a thin insulator (called a
dielectric such as mica, ceramics, paper or even oil).
• The electrical symbol for a capacitor is
or
Capacitance, C
• The ability of a capacitor to store charge is
measured by its capacitance.
• Capacitance is defined as the ratio of the charge
on either plate to the potential difference
between them.
Q
C
V
Q : charge on one of the plates
V : potential difference across the two plates
•
The unit of capacitance is the farad (F).
1 farad is the capacitance of a capacitor if the charge on
either of the plates is 1C when the potential difference
across the capacitor is 1V.
i.e.
Q
1 coulomb
1 farad=
•
1 volt
C
V
By rearranging the equation from the definition of
capacitance, we get
Q  CV
where the capacitance of a capacitor, C is constant then
Q V
(The charges stored, Q is directly proportional to the
potential difference, V across the conducting plate.)
• One farad (1F) is a very large unit.
• Therefore in many applications the most
convenient units of capacitance are microfarad
and the picofarad where the unit conversion can
be shown below :
6
1 μF  10 F
12
1 pF  10 F
6
 10 μF
 1 μμF
Parallel-plate Capacitors
• A parallel–plate capacitor consists of a pair of parallel plates
of area A separated by a small distance d.
• If a voltage is applied to a capacitor (connected to a
battery), it quickly becomes charged.
• One plate acquires a negative charge, the other an equal
amount of positive charge and the full battery voltage
appears across the plates of the capacitor (12 V).
• The capacitance of a parallel-plate capacitor, C is
proportional to the area of its plates and inversely
proportional to the plate separation.
0 A
C
Parallel-plate capacitor separated by a vacuum
d
or
A
C
d
Parallel-plate capacitor separated by a dielectric material
ε0 : permittivity of free space
0 : 8.85 x 10-12 C2 N-1 m-2
A : area of the plate
d : distance between the two plates
Example 17.1
a) Calculate the capacitance of a capacitor whose plates are 20 cm x
3.0 cm and are separated by a 1.0-mm air gap.
b) What is the charge on each plate if the capacitor is connected to a
12-V battery?
c) What is the electric field between the plates?
Answer :
Example 17.2
An electric field of 2.80 x 105 V m-1 is desired between two parallel
plates each of area 21.0 cm2 and separated by 250 cm of air. Find the
charge on each plate.
(Given permittivity of free space, 0 = 8.85 x 10-12 C2 N-1 m-2)
Answer :
Exercise 17.1
The plates of a parallel-plate capacitor are 8.0 mm apart and
each has an area of 4.0 cm2. The plates are in vacuum. If the
potential difference across the plates is 2.0 kV, determine
a) the capacitance of the capacitor.
b) the amount of charge on each plate.
c) the electric field strength was produced.
C  4.42 x 10 F @ 0.44 pF
-13
Q  8.84 x 10-10C
E  2.50 x 10 N C @V m
5
-1
-1
17.2 Capacitors in series and parallel
LEARNING OUTCOMES :
At the end of this lesson, the students should be
able to :
a) Deduce and use the effective capacitance of
capacitors in series and parallel.
b) Derive and use equation of energy stored in
a capacitor.
17.2 (i) Capacitors connected in series
+Q
V1
V2
V3
Q1
Q2
Q3
-Q
Ceq,V
equivalent to
V
• Figure above shows 3 capacitors connected in series to a
battery of voltage, V.
• When the circuit is completed, the electron from the battery (-Q)
flows to one plate of C3 and this plate become negatively
charge.
• This negative charge induces a charge +Q on the other plate of C3
because electrons on one plate of C3 are repelled to the plate of C2.
Hence this plate is charged –Q, which induces a charge +Q on the other
plate of C2.
•
This in turn produces a charge –Q on one plate of C1 and a charge of +Q
on the other plate of capacitor C1.
•
Hence the charges on all the three capacitors are the same, Q.
• The potential difference across capacitor C1,C2 and C3 are
Q1 Q
V1 
 ;
C1 C1
Q2 Q
V2 

;
C2 C2
Q3 Q
V3 

C3 C3
•
The total potential difference V is given by
V  V1  V2  V3
Q Q Q
V


C1 C2 C3
V
1
1
1



Q C1 C2 C3
• If Ceq is the equivalent capacitance, then
• Therefore the equivalent (effective) capacitance Ceq for n
capacitors connected in series is given by
1
1 1 1
1
    ...
Ceq C1 C 2 C3
Cn
capacitors
connected in
series
17.2 (ii) Capacitors connected in parallel
+Q
-Q
Ceq,V
equivalent to
V
• Figure above shows 3 capacitors connected in parallel to a
battery of voltage V.
• When three capacitors are connected in parallel to a battery,
the capacitors are all charged until the potential differences
across the capacitors are the same.
• If not, the charge will flow from the capacitor of higher
potential difference to the other capacitors until they all
have the same potential difference, V.
• The potential difference across each capacitor is the
same as the supply voltage V.
• Thus the total potential difference (V) on the equivalent
capacitor is
V  V1  V2  V3
• The charge on each capacitor is
Q1  C1V1  C1V
Q2  C 2V2  C 2V
Q3  C3V3  C3V
• The total charge is
Q  Q1  Q2  Q3
Q  C1V  C2V  C3V
Q
Q
 Ceq
 C1  C2  C3 and
V
V
• Therefore the equivalent (effective) capacitance Ceq for n
capacitors connected in parallel is given by
C eq  C1  C 2  C 3  ...C n
capacitors connected
in parallel
Example 17.3
50 V
C1 = 1µF
C2 = 2µF
In the circuit shown above, calculate the
a) charge on each capacitor
b) equivalent capacitance
Example 17.4
In the circuit shown below, calculate the
a) equivalent capacitance
C1 = 1µF C2 = 2µF
V1
V2
50 V
b) charge on each capacitor
c) the pd across each capacitor
Example 17.5
In the circuit shown below, calculate the
a) equivalent capacitance
b) charge on each capacitor
c) the pd across each capacitor
C1 = 6.0µF
V1
C3 = 8.0µF
V2 =V3
a)
12 V
C1 = 6.0µF C23 = 12.0µF
V1
V2
b)
12 V
c)
Example 17.6
Find the equivalent capacitance between points a
and b for the group of capacitors connected as
shown in figure below.
Take
C1 = 5.00 F,
C2 = 10.0 F
C3 = 2.00 F.
Solution 17.6
C1 = 5.00 F, C2 = 10.0 F and C3 = 2.00 F.
Series a and Series b
Series a
Series b
C12
C12
parallel
C22
Parallel
Solution 17.6
C1 = 5.00 F, C2 = 10.0 F and C3 = 2.00 F.
•a
Parallel
Parallel
C3
C12
Ca
C22
•b
C12
Solution 17.6
•a
series
Ca
Ceq
C22
•b
Series
Example 17.7
Determine the equivalent capacitance of the
configuration shown in figure below. All the
capacitors are identical and each has capacitance
of 1 F.
1 F
1 F
1 F
1 F
1 F
1 F
Solution 17.7
series
1 F
series
1 F
1 F
1 F
1 F
1 F
Ca
1 F
series
1 F
series
Ca 1 F
Cb
Solution 17.7
parallel
Ceq
Cb
1 F
parallel
Exercise 17.2
a
C1
C2
b
d
C3
1. In the circuit shown in figure above, C1= 2.00 F, C2 = 4.00
F and C3 = 9.00 F. The applied potential difference between
points a and b is Vab = 61.5 V. Calculate
a) the charge on each capacitor.
b) the potential difference across each capacitor.
c) the potential difference between points a and d.
V3  24.6 V
Q  221 μC
Q  73.8 μC
1
Q2  147 μC
3
V1  V2  36.9 V
Vad  36.9 V
2. Four capacitors are connected as shown in figure below.
Calculate
a) the equivalent capacitance between points a and b.
b) the charge on each capacitor if Vab=15.0 V.
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.
3. A 3.00-µF and a 4.00-µF capacitor are connected in series
and this combination is connected in parallel with a 2.00-µF
capacitor.
a) What is the net capacitance?
b) If 26.0 V is applied across the whole network, calculate the
voltage across each capacitor.
3.71-µF, 26.0 V,
14.9 V, 11.1 V
Energy stored in a capacitor, U
• A charged capacitor stores electrical energy.
• The energy stored in a capacitor will be equal to the work done
to charge it.
• A capacitor does not become charged instantly. It takes time.
• Initially, when the capacitor is uncharged , it requires no work
to move the first bit of charge over.
• When some charge is on each plate, it requires work to add
more charge of the same sign because of the electric repulsion.
• The work needed to add a small amount of charge dq, when a
potential difference V is across the plates is, dW  Vdq
• Since V=q/C at any moment ,
where C is the capacitance,
the work needed to store a
total charge Q is
• Thus the energy stored in a
capacitor is
1 Q2
W U 
2 C
Q
W   Vdq
0
1 Q
W   qdq
C 0
1 Q2
W 
2 C
1
U  CV 2
2
Q
1 q 
W   
C  2 0
2
or
or
U
1
QV
2
Example 17.8
A camera flash unit stores energy in a 150 µF capacitor
at 200 V. How much energy can be stored?
Example 17.9
A 2 µF capacitor is charged to 200V using a battery.
Calculate the
a) charge delivered by the battery
b) energy supplied by the battery.
c) energy stored in the capacitor.
Solution 17.9
Exercise 17.3
Two capacitors, C1= 3.00 F and C2 = 6.00 F are connected
in series and charged with a 4.00 V battery as shown in figure
below.
4.00 V
C1
C2
Calculate
a) the total capacitance for the circuit above. 2.00 µF
b) the charge on each capacitor. 8.00 µC
c) the potential difference across each capacitor. V1 = 2.67 V, V2 = 1.33 V
d) the energy stored in each capacitor. U1 = 1.07 x 10 -5 J, U2 = 5.31 x 10-6 J
e) the area of the each plate in capacitor C1 if the distance
between two plates is 0.01 mm and the region between
plates is vacuum. 3.39 m 2
17.3 Charging and discharging of
capacitors (1 hour)
LEARNING OUTCOMES :
At the end of this lesson, the students should be able to :
a) Define and use time constant, τ = RC.
b) Sketch and explain the characteristics of Q-t and I-t graph
for charging and discharging of a capacitor.
 t / RC
Q

Q
e
b) Use formula
for discharging and
o
Q  Qo (1 e  t / RC ) for charging.
Charging a capacitor through a
resistor
 Figure below shows a simple circuit for charging a
capacitor.
 When the switch S is closed, current Io immediately
begins to flow through the circuit.
R
V0
switch , S
e
Electrons will flow out from the
negative terminal of the
battery, through the resistor R
and accumulate on the plate B
A   
C of the capacitor.
B  
 Then electrons will flow into
the positive terminal of the

e
battery, leaving a positive
charge on the plate A.

• As charge accumulates on the capacitor, the
potential difference across it increases and the
current is reduced until eventually the maximum
voltage across the capacitor equals the voltage
supplied by the battery, Vo.
• At this time, no further current flows (I = 0)
through the resistor R and the charge Q on the
capacitor thus increases gradually and reaches a
maximum value Qo.
Charge, Q (C)
Current , I ( A )
Q0
I0
0.63Q0
0.37 I 0
0
τ  RC
time, t ( s)
The charge on the capacitor
increases exponentially with time
Charge on charging
capacitor :

Q  Q0  1  e

t

RC
0
τ  RC
The current through the resistor
decreases exponentially with time
where




time, t ( s)
Q0 : maximum charge
I o : maximum current
R : resistance of the resistor
C : capacitance of the capacitor
Current in resistor :
I  I0e
t

RC
Discharging a capacitor through a resistor
 Figure below shows a simple circuit for discharging a
capacitor.
 When a capacitor is already
charged to a voltage Vo and it is
allowed to discharge through
R

e
the resistor R as shown in
figure below.
 When the switch S is closed,
V0






A
C electrons from plate B begin to
B  
flow through the resistor R and
C
neutralizes positive charges at
plate A.
switch , S
e
• Initially, the potential difference (voltage)
across the capacitor is maximum, V0 and then
a maximum current I0 flows through the
resistor R.
• When part of the positive charges on plate A is
neutralized by the electrons, the voltage
across the capacitor is reduced.
• The process continues until the current
through the resistor is zero.
• At this moment, all the charges at plate A is
fully neutralized and the voltage across the
capacitor becomes zero.
Charge, Q (C)
Q0
Current , I ( A )
Charge on discharging
capacitor :
t
Q  Q0 e

RC
0.37 Q0
0
τ  RC
0
The charge on the capacitor
decreases exponentially with time.
time, t ( s )
0.37 I 0
Current in resistor :
I0
time, t ( s)
τ  RC
I   I0e

t
RC
The current through the resistor
decreases exponentially with time.
The negative sign indicates that as the capacitor discharges, the current
direction opposite its direction when the capacitor was being charged.
For calculation of current in discharging process, ignore the negative
sign in the formula.
Time constant, 
•
•
•
It is a measure of how quickly the capacitor charges or discharges.
Its formula,   RC
Its unit is second (s).
Charging process
• The time constant is defined as the time required for the
capacitor to reach 0.63 or 63% of its maximum charge (Qo).
• The time constant is defined as the time required for the
current to drop to 0.37 or 37% of its initial value(I0).

Q  Q0  1  e

t

RC
when t=RC




RC



RC
Q  Q0  1 e



Q  Qo 1 0.37 
I  I0e
Q  0.63Qo
I  0.37 I o
I  I0e

t
RC
RC

RC
when t=RC
Discharging Process
• The time constant is defined as the time required for
the charge on the capacitor/current in the resistor
decrease to 0.37 or 37% of its initial value.
Q  Q0 e

when t=RC
Q  Q0 e
t
RC
RC

RC
Q  0.37Qo
I  I0e
when t=RC
I  I0e

t
RC

RC
RC
I  0.37 I o
Example 17.10
Consider the circuit shown in figure below, where C1= 6.00 F,
C2 = 3.00 F and V = 20.0 V.
Capacitor C1 is first charged by the 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
capacitor.
Solution 17.10
After the switch S1 is closed. The capacitor C1 is fully charged and the
charge has been placed on it is given by
+ +
V- -
+ ++
-- -
C1
S1
After the switch S2 is closed and S1 is opened. The capacitors C1 and
C2 (uncharged) are connected in parallel and the equivalent capacitance
is
The total charge Q on the circuit is given by
++
C1 - -
+
S2
C2
Solution 17.10
The charge from capacitor C1 flows to the capacitor C2 until the
potential difference V’ across each capacitor is the same (parallel) and
given by
++
+
C1 - -
- C2
S2
Therefore the final charge accumulates
- on capacitor C1 :
- on capacitor C2 :
Example 17.11
In the RC circuit shown in figure below, the battery has fully
charged the capacitor.
a
S
V0
b
R
C
Then at t = 0 s the switch S is thrown from position a to b. The
battery voltage is 20.0 V and the capacitance C = 1.02 F. The
current I is observed to decrease to 0.50 of its initial value in 40
s. Determine
a. the value of R.
b. the time constant, 
b. the value of Q, the charge on the capacitor at t = 0.
c. the value of Q at t = 60 s
Solution 17.11
V0
a S
b
R
C
17. 4 Capacitors With Dielectrics
LEARNING OUTCOMES :
At the end of this lesson, the students
should be able to :
a) Define dielectric constant.
b) Describe the effect of dielectric on a
parallel plate capacitor.
c) Use formula
C   r Co
17. 4 Capacitors with Dielectrics
• A dielectric is an insulating material. Hence no
free electrons are available in it.
• When a dielectric (such as rubber, plastics,
ceramics, glass or waxed paper) is inserted
between the plates of a capacitor, the capacitance
increases.
• The capacitance increases by a factor  or r
which is called the dielectric constant (relative
permittivity) of the material.
• Two types of dielectric :
i) non-polar dielectric
For an atom of non-polar dielectric, the center of the negative charge of
the electrons ‘coincides’ with the center of the positive charge of the
nucleus.
* It does not become a permanent dipole.
+
+
-
+
ii) polar dielectric
- Consider the molecule of waters.
- Its two positively charge hydrogen ions are ‘attracted’ to a negatively
charged oxygen ion.
- Such an arrangement of ions causes the center of the negative charge to
be permanently separated slightly away from the center of the positive
charge, thus forming a permanent dipole.
• Dielectric constant,  (r) is defined as the ratio between the
capacitance of given capacitor with space between plates
filled with dielectric, C with the capacitance of same
capacitor with plates in a vacuum, C0.
 : permittivity of dielectric material
ε
r 
or ε   r ε0
ε0
C
r 
C0
 εA 
 d 
r   
 ε0 A 
 d 


εA
C
d
0 A
C0 
d
•
From the definition of the capacitance,
Q
Q
C
Q is constant
and C0  V
V
0
where
V0
r 
V
•
r 
C
C0
V : potential difference across capacitor with dielectric
V0 : potential difference across capacitor in vacuum
From the relationship between E and V for uniform electric
field,
V  Ed and V0  E0 d
E0 d
r 
Ed
E0
r 
E
where
E0 : electric field strength of the capacitor in vacuum
E : electric field strength of the capacitor with dielectric
Material
•
Dielectric constant, εr
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
The dielectric strength is the maximum electric field before
dielectric breakdown (charge flow) occurs and the
material becomes a conductor.
C
 V0 E0
r 
 

C0 0 V
E
Example 17.12
A parallel-plate capacitor has plates of area A = 2x10-10
m2 and separation d = 1 cm. The capacitor is charged to
a potential difference V0 = 3000 V. Then the battery is
disconnected and a dielectric sheet of the same area A
is placed between the plates as shown in figure below.
dielectric
d
Example 17.12
In the presence of the dielectric, the potential difference across
the plates is reduced to 1000 V. Determine
a) the initial capacitance of the air-filled 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 sheet.
f) the initial electric field.
g) the electric field after the dielectric is inserted.
(Given permittivity of free space, 0 = 8.85 x 10-12 F m-1)
Solution 17.12
Dielectric effect on the parallel-plate capacitor
In part a, the region between the
charged plates is empty. The field
lines point from the positive toward
the negative plate
In part b, a dielectric has been inserted
between the plates. Because of the electric
field between the plates, the molecules of
the dielectric (whether polar or non-polar)
will tend to become oriented as shown in
the figure, the negative ends are attracted to
the positive plate and the positive ends are
attracted to the negative plate. Because of
the end-to-end orientation, the left surface
of the dielectric become negatively charged,
and the right surface become positively
charged.
• Because of the surface charges on the
dielectric, not all the electric field lines
generated by the charges on the plates
pass through the dielectric.
• As figure c shows, some of the field
lines end on the negative surface
charges and begin again on the
positive surface charges.
r 
E0
E
•
Thus, the electric field inside the dielectric is
less strong than the electric field inside the
empty capacitor, assuming the charge on the
plates remains constant.
• This reduction in the electric field is described
by the dielectric constant εr which is the ratio of
the field magnitude Eo without the dielectric to
the field magnitude E inside the dielectric:
Quantity
Capacitor
without
dielectric
Capacitor with
dielectric
Electric field
Eo
Potential
difference
Vo
E
V
Charge
Qo
Q
Capacitance
Co
C
C
 V0 E0
r 
 

C0 0 V
E
Relationship
E < Eo
V < Vo
Q = Qo
C > Co