PHY2049: Chapter 25 1
EXAM REVIEW ON
MONDAY
6:15 −8:15 PM
McCarty A Room G186
By JJ Stankowicz
Also, formula sheet has been posted.
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PHY2049: Chapter 25 - Capacitance and Energy Storage, Exams of Physics
A collection of review notes for chapter 25 of phy2049, focusing on capacitance calculation, charging and discharging capacitors in parallel and series, and energy storage in capacitors. It includes formulas, examples, and explanations.
Typology: Exams
Pre 2010
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PHY2049: Chapter 25
EXAM REVIEW ON
MONDAY
8:15 PM
McCarty A
Room G
By JJ Stankowicz
Also, formula sheet has been posted.
PHY2049: Chapter 25
Capacitance calculation review
+q
–q
Why do wealways consideronly +q and –
q
pairs? Why notjust any
q and Q?
PHY2049: Chapter 25
Î
Three capacitors in parallel are charged by battery orpower supply
Î
Generalize to more than three capacitors
Capacitors in parallel (derivation of formula)
For given V applied bybattery/supply, the combostores more charge,q
1
+ q
2
+ q
3
, than a single
capacitor.
Larger capacitance:
C
eq
= C
1
- C
2
- C
3
C
eq
= C
1
- C
2
- C
3
……
PHY2049: Chapter 25
Î
Two capacitors in series are charged by battery or powersupply
Î
Induced charges appear immediately
before
Capacitors in series (derivation of formula)
after
no charge no charge
no charge
+q
- q still no netcharge
after
+q
- q
- q attracted to +q +q attracted to – q
PHY2049: Chapter 25
Î
Capacitors in parallel
Î
Capacitors in series
It is foolish to connect capacitors in series.
eq
C
C
C
⋅ ⋅ ⋅ + + =
2
1
eq
C
C
C
Example: 100
μ
F in series with 10
μ
F is 9
μ
F (check yourself).
The 100
μ
F capacitor will be totally wasted.
Parallel and series capacitors—summary
PHY2049: Chapter 25
Î
Four 1
F in parallel.
Find C
eq.
Î
Four 1
F in series.
Find C
eq.
Î
F and 2.
F in series.
C
eq
is:
(a) 0.
F
(b) 1.
F
(c) 2.
F
(d) 3.
F
Examples
F
F
PHY2049: Chapter 25
10
Î
Equivalent capacitance?
1 and 2 in parallel
Together, in series with 3
Example: parallel-series combo
F
F
F
0
.
2
1
0
.
6
3
0
.
6
1
0
.
3
1
1
=
=
=
eq
C
F
F
PHY2049: Chapter 25
11
Î
Charge on C
1
? (See
Sample Problem 25-2for an alternativesolution)
Must find potential difference (aka voltage) Va across C
1
(also
across C
2
Note:
Need one more equation to relate Va to Vb. Must be through thefact that charge stored in capacitor 3 is q
1
+q
2
(continued)
μ
F
μ
F
μ
F
V
a
+ V
b
= V (applied “voltage”)
10 V
q
1
+q
2
q
1
- q
2
= C
1
V
a
+ C
2
V
a
q
1
- q
2
= C
3
V
b
μ
C
+q
2
+q
1
V
a
V
a
V
b
C
1
V
a
+ C
2
V
a
= C
3
V
b
, i.e., V
a
:V
b
=C
3
:(C
1
+C
2
V
a
= V C
3
/[C
3
+(C
1
+C
2
)]
q
1
= C
1
V
a
q
1
= VC
1
C
3
/[C
3
+(C
1
+C
2
)]
PHY2049: Chapter 25
2
0
E
u
Î
Two alternative views
Energy is stored in charge configuration in capacitor
Energy is stored in E field
Î
Second view (will be important later in dealing withelectromagnetic waves)
Define energy density
Show for parallel-plate capacitor
Î
This equation holds for any E field produced at any pointin space by any source
Derivation requires vector calculus
Energy stored in electric field
volume
U
u
PHY2049: Chapter 25
Equivalence of two views (by example)
Î
View 1
Energy is stored in capacitor’s charge configuration
Capacitance:
Î
View 2
Energy is stored in E field
In the gap
Elsewhere
Lr
q
E
0
Agrees!
Cylindrical capacitor
ln(
0
a
b
L
C
C
q
U
2
L
a
b
q
U
0
2
ln(
(
)
ln(
0 2
0 2
2
0
0
gap
a
b
L
q
r
dr
L
q
rLdr
Lr
q
udv
U
b a
b a
2
0
2 1
E
ε
u
=
0
=
E
PHY2049: Chapter 25
(continued)
Î
is called dielectric constant.
Larger than 1.
Î
Induced charge q’.
So far, general to any capacitor.
Now restrict
ourselves to parallel-plate capacitor
0
0
ε
q
A
E
0
ε
q
q
EA
q
q
q
E
E
′
−
=
0
κ
q
κ
q
q
<
−
=
′
1
1
Gauss’ law
Induced charge q’ is always less than q.
κ
=1 (vacuum, no dielectric)
q
No induced charge
κ
large (strong dielectric)
q
q
PHY2049: Chapter 25