EXAM REVIEW ON
MONDAY
6:15 − 8:15 PM
McCarty A Room G186
By JJ Stankowicz
Also, formula sheet has been posted.
PHY2049: Chapter 25
1
Capacitance calculation review
+q
–q
Why do we
always consider
only +q and –q
pairs? Why not
just any q and Q?
PHY2049: Chapter 25
2
Why do we always consider only +q and –q
pairs? Why not just any q and Q?
Î Charging
capacitor
Battery just moves electrons
from one side to the other
until potential difference across
capacitor reaches battery’s emf
(aka “voltage”) V. Does not
add or remove charge.
Î In
general
In electronic circuits, capacitors are used in such ways
that +q and –q occur as a pair.
PHY2049: Chapter 25
3
Capacitors in parallel (derivation of formula)
Î
Three capacitors in parallel are charged by battery or
power supply
For given V applied by
battery/supply, the combo
stores more charge,
q1 + q2 + q3, than a single
capacitor.
Larger capacitance:
Ceq = C1 + C2 + C3
Î
Generalize to more than three capacitors
Ceq = C1 + C2 + C3 + ……
PHY2049: Chapter 25
4
Capacitors in series (derivation of formula)
Î
Two capacitors in series are charged by battery or power
supply
no charge
+q
no charge
still no net
charge
–q
no charge
before
Î
after
Induced charges appear immediately
+q
– q attracted to +q
+q attracted to – q
–q
after
PHY2049: Chapter 25
5
(continued)
Î
Potential difference
C1
+q
– q V1
+q
C2
V2
–q
Electric field is absent in the connecting
wire, a conductor.
→ No potential difference between two
ends of the wire. Potential differences are
only across the capacitors.
V=V1+ V2=q/C1 + q/C2
V=q/Ceq
Definition of Ceq
1/Ceq =1/C1 + 1/C2
Î
Generalize to more than two capacitors in series
1/Ceq =1/C1 + 1/C2 + ……
PHY2049: Chapter 25
6
Parallel and series capacitors—summary
Î
Capacitors in parallel
C eq = C1 + C 2 + ⋅ ⋅ ⋅
Î
Capacitors in series
1 1 1
= + +⋅⋅⋅
Ceq C1 C2
It is foolish to connect capacitors in series.
Example: 100 μF in series with 10 μF is 9 μF (check yourself).
The 100 μF capacitor will be totally wasted.
PHY2049: Chapter 25
7
Examples
Î Four
1 μF in parallel. Find Ceq. 4 μF
Î Four
1 μF in series. Find Ceq. 0.25 μF
Î 1.3
μF and 2.0 μF in series. Ceq is:
‹(a) 0.79 μF
‹(b) 1.65 μF
‹(c) 2.6 μF
‹(d) 3.3 μF
PHY2049: Chapter 25
8
(continued)
Î Capacitors
in series
1
1
1
1
=
+
+
+L
Ceq C1 C 2 C3
1
1
>
Ceq C1
Ceq < C1
smallest
‹ Ceq
is smaller than the smallest of all
PHY2049: Chapter 25
9
Example: parallel-series combo
Î Equivalent
capacitance?
1.0 μF
2.0 μF
6.0 μF
‹1
and 2 in parallel
1.0 + 2.0 = 3.0 μF
‹ Together,
in series with 3
1
1
1
3
1
=
+
=
=
Ceq 3.0 6.0 6.0 2.0
PHY2049: Chapter 25
2.0 μF
10
(continued)
Î
Charge on C1? (See
Sample Problem 25-2 1.0 μF
for an alternative
10 V
solution)
+q1
Va
+q2
2.0 μF Va
6.0 μF
q1= C1Va
Vb
– (q1+q2)
‹ Must find potential difference (aka voltage) Va across C1 (also
across C2.) Note:
Va + Vb = V (applied “voltage”)
‹
Need one more equation to relate Va to Vb. Must be through the
fact that charge stored in capacitor 3 is q1+q2.
q1 + q2 = C1Va + C2Va, q1 + q2 = C3Vb
C1Va + C2Va = C3Vb , i.e., Va:Vb=C3:(C1+C2)
Va = V C3/[C3+(C1+C2)]
q1 = VC1C3/[C3+(C1+C2)]
PHY2049: Chapter 25
6.7 μC
11
Energy stored in capacitor
Î In
‹
Î In
‹
terms of charge
Derived by considering work dW’ done by a fictitious process
which moves infinitesimally small amount of charge +dq’ from
conductor 1 to conductor 2 of capacitor, leaving behind –dq’ on
conductor 1:
terms of potential
Since q=CV (definition of C)
Î Similarities
with
1
K = mv 2
2
q2
U=
2C
1
U = CV 2
2
(kinetic energy) U =
PHY2049: Chapter 25
1 2
kx (spring)
2
12
Energy stored in electric field
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 with
electromagnetic waves)
Define energy density
‹ Show for parallel-plate capacitor
‹
Î
u=
U
volume
u=
1
ε0 E 2
2
This equation holds for any E field produced at any point
in space by any source
‹
Derivation requires vector calculus
PHY2049: Chapter 25
13
Equivalence of two views (by example)
Cylindrical capacitor
ÎView
1
‹ Energy
is stored in capacitor’s charge
q2
configuration
U=
2C
‹ Capacitance:
2
U=
C = 2πε0 L / ln(b / a )
ÎView 2
‹ Energy
is stored in E field
‹ In the gap E =
U = ∫ udv = ∫
gap
b
a
1 q
2πε0 Lr
q ln(b / a )
4πε0 L
1
u = ε0 E 2
2
Elsewhere
2
E=0
Agrees!
b dr
1  1 q 
q
q
 (2πrLdr ) =
=
ln(b / a )
ε0 
∫
a
2  4πε0 Lr 
4πε0 L r 4πε0 L
2
PHY2049: Chapter 25
2
14
Dielectrics
ÎDielectric
polarized.
is insulator. In E field, it becomes partly
‹ For
microscopic view, read Section 25-7.
‹ If dielectric fills the gap of charged capacitor, E0 due to charges +q
and –q partly polarizes it, inducing charges –q’ and +q’ near
surfaces.
‹ These in turn produce field that partly cancels E0.
‹ Net field E proportional to, and less than, E0.
ÎWhat’s
the point?
‹ E0
→ E = E0/κ
less than E0
‹ V0
→ V = V0/κ
from definition of V
‹ C0
→ C = κ C0
since C=q/V
Larger than C0, which means capacitor
stores more charge for given potential
difference applied by battery. Beneficial
to fill gap with dielectric.
PHY2049: Chapter 25
15
(continued)
ÎΚis
called dielectric constant. Larger than 1.
ÎInduced
charge q’.
‹ So
far, general to any capacitor. Now restrict
ourselves to parallel-plate capacitor
E0 A =
q
ε0
q + (− q′)
EA =
ε0
E q − q′
=
E0
q
κ
Gauss’ law
 1
q ′ = q 1 −  < q
 κ
Induced charge q’ is always less than q.
κ=1 (vacuum, no dielectric) → q’=0
No induced charge.
κ large (strong dielectric) → q’→q
PHY2049: Chapter 25
16
Concept Question
Î All
capacitors are identical. Across each
combo, the same voltage (potential
difference) is applied. Which combo stores
the highest energy?
(a)
(b)
(c)
PHY2049: Chapter 25
(d)
17
REMINDER
ÎExam
on Wednesday in Class (Chapters
21–25) Study sample exams posted
‹ Must
bring Gator1 ID card (Will take away points if you
forget ;< )
‹ Calculator (No formulae allowed on calculator)
‹ Pencil, eraser, and sharpner, as usual
PHY2049: Chapter 25
18