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Lecture8.pdf-Homework Reading: Chap. 26 and Chap.
Lecture8.pdf-Homework Reading: Chap. 26 and Chap.
Lecture8.pdf-Homework Reading: Chap...
Lecture8.pdf-Homework Reading: Chap. 26 and Chap.
Page 12
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12
Kirchhoff’s Loop Law
Parallel Circuit
V
So,
23
24
Page 13
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13
Series Circuit
V
So,
Combinations of Capacitors
If capacitors
C
1
,
C
2
,
C
3
, … are in parallel, their equivalent
capacitance is
If capacitors
C
1
,
C
2
,
C
3
, … are in series, their equivalent
capacitance is
25
26
Page 14
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14
In-Class Activity 8.3
In-Class Activity 8.4
27
28
Page 15
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15
Example 8.2
In the arrangement shown
in the figure, a potential
difference
V
is applied,
and
C
1
is adjusted so that
the voltmeter between
points
b
and
d
reads zero.
This “balance” occurs
when
C
1
= 4.00
F. If
C
3
= 9.00
F and
C
4
= 12
.
0
F, calculate the value of
C
2
.
Energy Stored in a Capacitor
External work needs to be done in order to
charge a capacitor. Assume at one moment
during the charging of capacitor, the charge is
q
and the potential difference is
U
, if amount
of
dq
charge is moving from one plate to
another, the work done by external force (or
battery) is
The total work done (equal to the total
potential energy gained) by the external
force is
29
30
Page 16
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16
Energy Stored in a Capacitor
Since
Then
So a capacitor separated by electric potential
V
has a
potential energy
Energy Density
We can express
U
in terms of
E
(field) instead of
q
and
V
,
The energy density has units J/m
3
.
31
32
Page 17
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17
Energy Density
Interpretation:
placing a charge
Q
on a capacitor
requires external work, or building up a electric field
E
over a volume
Ad
requires external work
This works even when there are no
charge present, i.e., the electric
field
E
carries electric energy!
In-Class Activity 8.5
33
34
Page 18
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18
Dielectrics
Gauss’ Law in Dielectrics
: permittivity of the dielectric material
+
q
35
36
Page 19
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19
Figure shows a parallel plate capacitor of plate area
A
and
plate separation
d
. A potential difference
V
0
is applied
between the plates. The battery is then disconnected, and a
dielectric slab of thickness
b
and dielectric constant
is
placed between the plates as shown. Assume
A
= 115 cm
2
d
= 1.24 cm
V
0
= 85.5 V
b
= 0.78 cm
= 2.61
(a)What is the capacitance
C
0
before the dielectric slab is
inserted?
(b) What free charge appears on
the plates?
Example 8.3
(c) What is the electric field
E
0
in the gaps between the plates and
the dielectric slab?
(d) What is the electric field
E
1
in the dielectric slab?
(e) What is the potential difference
V
between the plates after the
slab has been introduced?
(f) What is the capacitance with the slab in place?
37
38
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