CAPACITORS
• C - Capacitance
V= Va-Vb - voltage difference
Q
C=
V
• Capacitance, depends on the geometry of the
two conductors (size, shape, separation) and capacitance is always
a positive quantity by its definition (voltage difference and charge
of + conductor)
• UNITs: Coulomb/Volts or Farads, after Michael Faraday
Capacitor Summary
•
•
A Capacitor is an object with two spatially separated conducting
surfaces.
The definition of the capacitance of such an object is:
C ≡
Q
V
• The capacitance depends on the geometry :
-Q
A
++++
d
----Parallel Plates
C=
Aε 0
d
r
+Q
-Q
+Q
a
a
b
b
L
Cylindrical
Spherical
2 πε 0 L
⎛ b ⎞
ln ⎜
⎟
⎝ a ⎠
4 πε 0 ab
C =
b−a
C =
CAPACITORs in parallel
Want to find the equivalent capacitance Ceq:
• Voltage is same across each capacitor
Q2
Q1 Q2
Q1
=
; C1 =
; C2 =
Vab =
C1 C2
Vab
Vab
• Total charge and voltage ratios for
parallel capacitor,
Q1 + Q2
= C parallel = C1 + C2
Vab
For more parallel
capacitors:
C parallel = ∑ Ci
i
Note; 2 parallel capacitors C, doubles capacitance.
Example 1:
Two identical parallel plate
capacitors are shown in an endview in A) of the figure. Each
has a capacitance of C.
4) If the two are joined together as shown in B), forming a
single capacitor, what will be the final capacitance?
a) C/2
b) C
c) 2C
Example 1:
Two identical parallel plate
capacitors are shown in an endview in A) of the figure. Each
has a capacitance of C.
4) If the two are joined together as shown in B), forming a
single capacitor, what will be the final capacitance?
a) C/2
b) C
c) 2C
CAPACITORs in series
Want to find the equivalent capacitance Ceq:
• If a voltage is applied across a and b, then
a +Q appears on upper plate and –Q on lower
plate.
• A –Q charge is induced on lower plate of C1
and a +Q charge is induced on upper plate of
C2. The total charge in circuit c is neutral.
Q
Q
V1 =
; V2 =
C1
C2
Q Q
V = V1 + V2 =
+
C1 C2
V
1
1
1
=
= +
Q
Cseries C1 C2
Note; 2 series capacitors C, halves capacitance.
Examples:
Combinations of Capacitors
a
≡
C3
b
C1
a
C2
C
• How do we start??
• Recognize C3 is in series with the parallel
combination on C1 and C2. i.e.,
1 1
1
= +
C C3 C1 + C2
⇒
b
C=
C 3 (C1 + C 2 )
C1 + C 2 + C 3
Example 2:
C
C
C
C
C
Configuration A
Configuration B
Configuration C
Three configurations are constructed using identical capacitors
Î Which of these configurations has the lowest overall capacitance?
a) Configuration A
b) Configuration B
c) Configuration C
Example 2:
C
C
C
C
C
Configuration A
Configuration B
Configuration C
Three configurations are constructed using identical capacitors
Î Which of these configurations has the lowest overall capacitance?
a) Configuration A
b) Configuration B
c) Configuration C
Example 4
• What is the equivalent capacitance, Ceq, of the
combination shown?
o
Ceq
C
C
C
o
(a) Ceq = (3/2)C
(b) Ceq = (2/3)C
(c) Ceq = 3C
Example 4
• What is the equivalent capacitance, Ceq, of the
combination shown?
o
Ceq
C
C
C
o
(b) Ceq = (2/3)C
(a) Ceq = (3/2)C
C
C
1 1 1
= +
C1 C C
C
≡
C
C1 =
2
(c) Ceq = 3C
C1
C
C 3
Ceq = C + = C
2 2
Energy storage in CAPACITORs
Charge capacitor by transferring bits of
charge dq at a time from bottom to top plate.
Can use a battery to do this. Battery does
work which increase potential energy of
capacitor.
+q
-q
V
dq
q is magnitude of charge on plates
V= q/C
V across plates
dU = V dq
increase in potential energy
(
q
Q
CV )
1
=
= CV 2
U = ∫ dU = ∫ dq =
C
2C
2C
2
0
0
U
Q
2
2
two ways to write
Y&F, eqn. 24.9
Question!
A
++++
d
-----
•
Suppose the capacitor shown here is
charged to Q and then the battery is
disconnected.
•
Now suppose I pull the plates further apart so that the final
separation is d1.
•
How do the quantities Q, C, E, V, U change?
•
•
•
•
•
Q:
C:
E:
V:
U:
•
How much do these quantities change?.. exercise for student!!
remains the same.. no way for charge to leave.
decreases.. since capacitance depends on geometry
remains the same... depends only on charge density
increases.. since C ↓, but Q remains same (or d ↑ but E the same)
increases.. add energy to system by separating
Answers:
C1 =
d
C
d1
V1 =
d1
V
d
U1 =
d1
U
d
Related Question
•
Suppose the battery (V) is kept
attached to the capacitor.
A
++++
d
-----
V
•
Again pull the plates apart from d
to d1.
•
Now what changes?
•
•
•
•
•
C:
V:
Q:
E:
U:
•
How much do these quantities change?
decreases (capacitance depends only on geometry)
must stay the same - the battery forces it to be V
must decrease, Q=CV charge flows off the plate
must decrease ( E =
V
σ
, E= )
E0
D
must decrease (U = 12 CV 2 )
Answers:
d
C1 =
C
d1
E1 =
d
E
d1
U
1
=
d
U
d1
Example 5:
Two identical parallel plate capacitors are connected to a battery, as
shown in the figure. C1 is then disconnected from the battery, and the
separation between the plates of both capacitors is doubled.
Î What is the relation between the charges on the two capacitors ?
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
Example 5:
Two identical parallel plate capacitors are connected to a battery, as
shown in the figure. C1 is then disconnected from the battery, and the
separation between the plates of both capacitors is doubled.
Î What is the relation between the charges on the two capacitors ?
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
Example 6:
Two identical parallel plate capacitors are connected to a battery, as
shown in the figure. C1 is then disconnected from the battery, and the
separation between the plates of both capacitors is doubled.
Î How does the electric field between the plates of C2 change as
separation between the plates is increased ? The electric field:
a) increases
b) decreases
c) doesn’t change
Example 6:
Two identical parallel plate capacitors are connected to a battery, as
shown in the figure. C1 is then disconnected from the battery, and the
separation between the plates of both capacitors is doubled.
Î How does the electric field between the plates of C2 change as
separation between the plates is increased ? The electric field:
a) increases
b) decreases
c) doesn’t change
Example 7:
Two identical parallel plate capacitors are connected to a battery,
as shown in the figure. C1 is then disconnected from the battery,
and the separation between the plates of both caps is doubled.
Î What is the relation between the voltages on the two capacitors?
a) V1 > V2
b) V1 = V2
c) V1 < V2
Example 7:
Two identical parallel plate capacitors are connected to a battery,
as shown in the figure. C1 is then disconnected from the battery,
and the separation between the plates of both caps is doubled.
Î What is the relation between the voltages on the two capacitors?
a) V1 > V2
b) V1 = V2
c) V1 < V2
Where is the Energy Stored?
• Claim: energy is stored in the electric field itself.
Think of the energy needed to charge the capacitor
as being the energy needed to create the field.
•
To calculate the energy density in the field, first consider the
constant field generated by a parallel plate capacitor, where
-Q
-------- ------
++++++++ +++++++
+Q
•
1 Q2 1 Q2
U=
=
2 C 2 ( Aε 0 / d )
The electric field is given by:
σ
Q
E= =
ε0 ε0 A
•
⇒
This is the energy
density, u, of the
electric field….
1
U = ε0 E2 Ad
2
The energy density u in the field is given
by:
U
U 1 2
= = ε0E
u=
volume Ad 2
Units:
J
m3
Energy Density
Claim: the expression for the energy density of the
electrostatic field
1
u = ε0E 2
2
is general and is not restricted to the special case of the
constant field in a parallel plate capacitor.
• Example (and another exercise for the student!)
– Consider E- field between surfaces of cylindrical
capacitor:
– Calculate the energy in the field of the capacitor by
integrating the above energy density over the volume of
the space between cylinders.
U=
1
1
ε 0 ∫ E 2 dV = ε 0 ∫ ∫ E 2π r dr dl = etc.
2
2
– Compare this value with what you expect from the
general expression:
1
W = CV 2
2
For next time
• HW #3 Æ turn in if haven’t
• HW #4 Æ available
• Office Hours immediately after this class (9:30
– 10:00) in WAT214 [M 1:30-2; WF 1-1:30]
• Don’t fall behind – 2nd Quiz Friday