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General Physics (PHY 2140)
Lecture 6
 Electrostatics
Capacitance and capacitors
 parallel-plate capacitor
 capacitors in electric circuits
 energy stored in a capacitor
 capacitors with dielectrics
http://www.physics.wayne.edu/~apetrov/PHY2140/
Chapter 16
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1
Lightning Review
Last lecture:
1. Potential and potential energy
q
 Potential and potential energy of a system of point charges V  ke
r
 Superposition principle for potential (algebraic sum)
 Potentials and charged conductors (V is the same in a conductor)
 Equipotential surfaces (surfaces of constant potential)
2. Capacitance and capacitors
C
Q
V
Review Problem: A cylindrical piece of insulating material is
placed in an external electric field, as shown. The net electric
flux passing through the surface of the cylinder is
a. negative
b. positive
c. zero
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16.7 The parallel-plate capacitor
The capacitance of a device
depends on the geometric
arrangement of the conductors
A
A
C  e0
d
where A is the area of one of
the plates, d is the separation,
e0 is a constant (permittivity of
free space),
+Q
d
A
-Q
ke 
1
4e 0
e0= 8.8510-12 C2/N·m2
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Problem: parallel-plate capacitor
A parallel plate capacitor has plates 2.00 m2 in area, separated by a
distance of 5.00 mm. A potential difference of 10,000 V is applied
across the capacitor. Determine
the capacitance
the charge on each plate
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A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A
potential difference of 10,000 V is applied across the capacitor. Determine
the capacitance
the charge on each plate
Solution:
Given:
V=10,000 V
A = 2.00 m2
d = 5.00 mm
Find:
C=?
Q=?
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Since we are dealing with the parallel-plate capacitor,
the capacitance can be found as
2
A
2.00
m
C  e 0  8.85 1012 C 2 N  m 2
d
5.00 103 m
 3.54 109 F  3.54 nF


Once the capacitance is known, the charge can be
found from the definition of a capacitance via charge
and potential difference:


Q  C V  3.54 109 F 10000V   3.54  105 C
5
16.8 Combinations of capacitors
It is very often that more than one capacitor is used in an
electric circuit
We would have to learn how to compute the equivalent
capacitance of certain combinations of capacitors
C2
C1
C3
C2
C5
C3
C1
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C4
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a. Parallel combination
Connecting a battery to the parallel
combination of capacitors is equivalent to
introducing the same potential difference for
both capacitors,
a
C1
V=Vab
+Q1 C2
+Q2
Q1
Q2
V1  V2  V
A total charge transferred to the system
from the battery is the sum of charges of
the two capacitors,
b
Q1  Q2  Q
By definition,
Q1  C1V1
Thus, Ceq would be
Ceq 
Q2  C2V2
Q Q1  Q2 Q1 Q2 Q1 Q2





V
V
V
V
V1 V2
CCeqeq CC11CC22
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Q1  Q2  Q
V1  V2  V
7
Parallel combination: notes
Analogous formula is true for any number of capacitors,
Ceq  C1  C2  C3  ...
(parallel combination)
It follows that the equivalent capacitance of a parallel
combination of capacitors is greater than any of the
individual capacitors
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Problem: parallel combination of capacitors
A 3 mF capacitor and a 6 mF capacitor are connected in parallel
across an 18 V battery. Determine the equivalent capacitance
and total charge deposited.
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A 3 mF capacitor and a 6 mF capacitor are connected in parallel across an 18 V
battery. Determine the equivalent capacitance and total charge deposited.
a
Given:
C1
V=Vab
V = 18 V
C1= 3 mF
C2= 6 mF
Find:
Ceq=?
Q=?
+Q1 C2
+Q2
Q1
Q2
b
First determine equivalent capacitance of C1 and C2:
C12  C1  C2  9 m F
Next, determine the charge


Q  C V  9 106 F 18V   1.6 104 C
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b. Series combination
Connecting a battery to the serial
combination of capacitors is equivalent to
introducing the same charge for both
capacitors,
a
V=Vab
Q1
c
C2
b
Q1  Q2  Q
+Q1
C1
+Q2
Q2
A voltage induced in the system from the
battery is the sum of potential differences
across the individual capacitors,
V  V1  V2
By definition,
Q1  C1V1 Q2  C2V2
Thus, Ceq would be
1 V V1  V2 V1 V2 V1 V2
 
   
Ceq Q
Q
Q Q Q1 Q2
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11 11 1 1
 
CCeqeq CC
1 1 CC
2 2
Q1  Q2  Q
V1  V2  V
11
Series combination: notes
Analogous formula is true for any number of capacitors,
1
1
1
1
 
  ...
Ceq C1 C2 C3
(series combination)
It follows that the equivalent capacitance of a series
combination of capacitors is always less than any of the
individual capacitance in the combination
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Problem: series combination of capacitors
A 3 mF capacitor and a 6 mF capacitor are connected in series
across an 18 V battery. Determine the equivalent capacitance.
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A 3 mF capacitor and a 6 mF capacitor are connected in series across an 18 V
battery. Determine the equivalent capacitance and total charge deposited.
a
Given:
V=Vab
V = 18 V
C1= 3 mF
C2= 6 mF
Find:
Ceq=?
Q=?
+Q1
C1
Q1
c
C2
+Q2
Q2
b
First determine equivalent capacitance of C1 and C2:
Ceq 
C1C2
 2 mF
C1  C2
Next, determine the charge


Q  C V  2 106 F 18V   3.6 105 C
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16.9 Energy stored in a charged capacitor
Consider a battery connected to a capacitor
A battery must do work to move electrons
from one plate to the other. The work done
to move a small charge q across a voltage
V is W = V q.
As the charge increases, V increases so the
work to bring q increases. Using calculus
we find that the energy (U) stored on a
capacitor is given by:
V
V
1
Q2 1
U  QV 
 CV 2
2
2C 2
q
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Q
15
Example: electric field energy in parallel-plate
capacitor
Find electric field energy density (energy per unit volume) in a
parallel-plate capacitor





















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
1
U  CV 2
2
e A
C 0
volume  Ad V  Ed
d
Thus,
u  U / volume  energy density
1e A
= 0 ( Ed ) 2 /( Ad )
2 d
1
and so, the energy density is
u  e0E2
2
Recall
16
Example: stored energy
In the circuit shown V = 48V, C1 = 9mF, C2 = 4mF and C3 = 8mF.
(a) determine the equivalent capacitance of the circuit,
(b) determine the energy stored in the combination by calculating
the energy stored in the equivalent capacitance.
C1
V
C2
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C3
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In the circuit shown V = 48V, C1 = 9mF, C2 = 4mF and C3 = 8mF.
(a) determine the equivalent capacitance of the circuit,
(b) determine the energy stored in the combination by
calculating the energy stored in the equivalent capacitance,
Given:
V = 48 V
C1= 9 mF
C2= 4 mF
C3= 8 mF
Find:
Ceq=?
U=?
C1
V
C2
C3
First determine equivalent capacitance of C2 and C3:
C23  C2  C3  12 m F
Next, determine equivalent capacitance of the circuit
by noting that C1 and C23 are connected in series
Ceq  C123 
C1C23
 5.14 m F
C1  C23
The energy stored in the capacitor C123 is then


1
1
2
U  CV 2  5.14 106 F  48V   5.9 103 J
2
2
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16.10 Capacitors with dielectrics
A dielectrics is an insulating material (rubber, glass, etc.)
Consider an insolated, charged capacitor
Q
Q
V0
Q
Q
Insert a dielectric
V
Notice that the potential difference decreases (k = V0/V)
Since charge stayed the same (Q=Q0) → capacitance increases
Q0
Q0
 Q0
C


  C0
V V0 
V0

dielectric constant: k = C/C0
Dielectric constant is a material property
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Capacitors with dielectrics - notes
Capacitance is multiplied by a factor k when the
dielectric fills the region between the plates completely
E.g., for a parallel-plate capacitor
A
C  e 0
d
The capacitance is limited from above by the electric
discharge that can occur through the dielectric material
separating the plates
In other words, there exists a maximum of the electric
field, sometimes called dielectric strength, that can be
produced in the dielectric before it breaks down
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Dielectric constants and dielectric
strengths of various materials at room
temperature
Material
Vacuum
Air
Water
Fused quartz
Dielectric
constant, k
Dielectric
strength (V/m)
1.00
--
1.00059
3 106
80
--
3.78
9 106
For a more complete list, see Table 16.1
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Example
Take a parallel plate capacitor whose plates have an area of
2000 cm2 and are separated by a distance of 1cm. The capacitor
is charged to an initial voltage of 3 kV and then disconnected
from the charging source. An insulating material is placed
between the plates, completely filling the space, resulting in a
decrease in the capacitors voltage to 1 kV. Determine the
original and new capacitance, the charge on the capacitor, and
the dielectric constant of the material.
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Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a
distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then
disconnected from the charging source. An insulating material is placed between the
plates, completely filling the space, resulting in a decrease in the capacitors voltage to
1 kV. Determine the original and new capacitance, the charge on the capacitor, and the
dielectric constant of the material.
Given:
V1=3,000 V
V2=1,000 V
A = 2.00 m2
d = 0.01 m
Find:
C=?
C0=?
Q=?
k=?
5/22/2017
Since we are dealing with the parallel-plate capacitor,
the original capacitance can be found as
2
A
2.00
m
C  e 0  8.85 1012 C 2 N  m2
 18 nF
3
d
1.00 10 m


The dielectric constant and the new capacitance are
V1
C   C0 
C0  0.33 18nF  6nF
V2
The charge on the capacitor can be found to be


Q  C V  18 109 F  3000V   5.4  105 C
23
How does an insulating dielectric material reduce electric fields
by producing effective surface charge densities?
Reorientation of polar molecules

















Induced polarization of non-polar molecules
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Dielectric Breakdown: breaking of molecular bonds/ionization of
molecules.
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