Chapter 23 Summary: Electric Potential
• Differentiate Electric Potential Energy and Electric
Potential:
• Electric Potential Energy:work done change in
potential energy:
(q0 in point charge q/
q0 in multiple charges/
PE between all charges)
• Electric Potential: U per unit charge
U
1 q
• Where is zero potential?
V= =
q0 4 πε 0€r
• Infinity?Surface?
V=
• From electric potential to
electric field:


∂V ˆ ∂ V ˆ ∂V
ˆ
E = −(i
+j
+k
) = −∇V
∂x
∂y
∂z
€
€
U
1
q
=
∑
q0 4 πε 0 i ri
1 qq0
4 πε 0 r
q
q
U= 0 ∑ i
4 πε 0 i ri
U=
U=
qi q j
1
∑
4 πε 0 i< j rij
 
U
1
dq
1
ρ ( r )dr
V= =
=
∫
∫ r
q0 4 πε 0 r
4 πε 0

b 
b
Va − Vb = ∫ a E • dl = ∫ a E cos φdl
Chapter 24:
•  energy in a capacitor
•  capacitors and capacitance
•  capacitors in series and capacitors in parallel
•  Dielectrics
•  Announcement: Test on Monday
(CH21-24), HMWK for CH24 will be
assigned today (reduced load)
Where is the electric potential energy stored?
•  Example of storing gravitational
potential energy: Killing robber of
your tomb robber with a heavy stone
at the door top.
–  Who did that energy come
from?
•  Example of storing elastic potential
energy: Killing that robber with an
Your camera flash use energy
arrow on a cross-bow!
–  Who provided that energy?
•  Example of storing electric
potential energy?
stored in a device called
Capacitor, charged by the battery
Capacitor and Capacitance
•  Capacitor: two conductors separated by an insulator
•  Charging conductors: electrons moving from one to the other:
–  Total charge is 0
–  +Q one one, -Q on the other
–  The capacitor stores charge Q
•  Definition of Capacitance:
–  C=Q/Vab
–  If Q->2Q, how will C and Vab change?
–  Measuring the ability of storing energy
•  Unit: Farad
–  1F=1 farad = 1C/V = 1 coulomb /volt (a big one!)
Capacitance of a Parallel-Plate Capacitor
Parallel-Plate Capacitor:
(Two large conducting plates
Separated by a small distance d)
σ Q/A
=
ε0
ε0
Qd / A
Vab = Ed =
ε0
Q
Q
⇒C =
=
= ε 0 A /d
Vab Qd / A
ε0
E=
€
Key points: capacitance is only
dependent on the geometry!
What if one of the plate is flexible, and it can be moved by sound wave?
The unit of capacitance, the farad, is very large
•  Is a 1-F capacitor possible/practical?
•  Suppose a parallel-plate capacitor has capacitance of 1.0F.
If d=1.0mm? What is the area of the plates?
•  How to obtain large capacitance, practically?
–  Increase A (smartly)
–  Decrease d
–  Add material in between
•  Typical units: µF, pF
•  How much energy is stored?
Q24.1
The two conductors a and b are
insulated from each other, forming a
capacitor. You increase the charge on
a to +2Q and increase the charge on b
to –2Q, while keeping the conductors
in the same positions.
As a result of this change, the
capacitance C of the two conductors
A. becomes 4 times great. B. becomes twice as great.
C. remains the same.
D. becomes 1/2 as great.
E. becomes 1/4 as great.
Q24.2
You reposition the two plates of a capacitor so that the
capacitance doubles. There is vacuum between the plates.
If the charges +Q and –Q on the two plates are kept constant in
this process, what happens to the potential difference Vab
between the two plates?
A. Vab becomes 4 times as great
B. Vab becomes twice as great
C. Vab remains the same
D. Vab becomes 1/2 as great
E. Vab becomes 1/4 as great
Energy Storage in Capacitors
A capacitor has charge Q. How much energy is stored in the capacitor?
Method 1: How much work is needed to move the top plate upwards by
 
a distance of d?
W = ∫ F • dl
+Q


F + QE = 0
E

σ
E =−
yˆ
2
ε
+Q
E

Qσ
Qσ
Q σs
⇒ W = ∫ −(−
yˆ ) • dl = ∫
dy =
-Q
2ε
2ε
2ε
−Q
−Q
0
0
0
0
2
ΔU = W =
Q σs
Q
=
2ε 0 2ε 0 A /d
Method 2: Let the plates already be separated by distance d. They are
Then charged to Q. The electric potential is changing during the process:
€
Q2 CV 2 1
U=
=
= QV
2C
2
2
W =
∫
W
0
Q
€
Q
dW = ∫ 0 vdq = ∫ 0
Q2
Q2
ΔU = W =
=
2C 2ε 0 A /d
It’s not QV
€
dW = vdq
v = q /C
Q2
q /Cdq =
2C
Electric-Field Energy
For parallel plate (capacitors), the electric potential energy is stored
in the electric filed between them!
The density is:
CV 2 /2
u=
Ad
C = ε 0 A /d
V = Ed
(ε 0 A /d)(Ed) 2 1
∴u =
= ε0E 2
2Ad
2
Implication: Vacuum is not really empty
Everyday example:
Microwave
€
Cylindrical Capacitors (your Entertainment center cables)
A long cylindrical capacitor (right).
How is the capacitance determined?
Start from definition:
C=Q/Vab, Q=λL
Vab=?
The outer cylinder has no effect for the
Electric potential inside,
Therefore Vab is totally determined by
The inner cylinder:
Coaxial cylinders capacitance is totally
determined by its dimension
€
Vb = 0
λ
r
ln b
2πε 0 r
λ
r
Vab = Va =
ln b
2πε 0 ra
λL
2πε 0
C /L =
/L =
λ
r
r
ln b
ln b
2πε 0 ra
ra
∴Vr (ra < r < rb ) =
Capacitors may be connected one or many at a time
•  Capacitors in series:
•  One at a time,
head to tail
Q1 = Q2 = Q
C1V1 = C2V2 = Q
Why? V = V1 + V2 ⇒
V = Q /C1 + Q /C2 ⇒
V /Q = 1/C1 +1/C2 ≡ 1/C
•  Capacitors in
parallel:
•  Head to head,
tail to tail
Why?
€
C = C1 + C2
Q24.3
A 12–µF capacitor and a 6–µF capacitor
are connected together as shown. What is
the equivalent capacitance of the two
capacitors as a unit?
a
12 µF
A. Ceq = 18 µF
6 µF
B. Ceq = 9 µF
C. Ceq = 6 µF
D. Ceq = 4 µF
E. Ceq = 2 µF
b
Calculations regarding capacitance and example
•  These calculations involves simple algebra, and are similar to
what we will learn later about resistance
•  Five capacitors are connected according to the diagram below,
between point a and b. What is the equivalent capacitance Cab?
Series :
1
1
∑C = C
i
eq
i
€
Parallel :
∑C
i
€
i
= Ceq
Q24.4
A 12–µF capacitor and a 6–µF capacitor
are connected together as shown. If the
charge on the 12–µF capacitor is 24
microcoulombs (24 µC), what is the
charge on the 6–µF capacitor?
a
12 µF
A. 48 µC
6 µF
B. 36 µC
C. 24 µC
D. 12 µC
E. 6 µC
b
Q24.5
A 12–µF capacitor and a 6–µF capacitor
are connected together as shown. What is
the equivalent capacitance of the two
capacitors as a unit?
a
12 µF
6 µF
A. Ceq = 18 µF
B. Ceq = 9 µF
C. Ceq = 6 µF
D. Ceq = 4 µF
E. Ceq = 2 µF
b
Q24.6
A 12–µF capacitor and a 6–µF capacitor
are connected together as shown. If the
charge on the 12–µF capacitor is 24
microcoulombs (24 µC), what is the
charge on the 6–µF capacitor?
a
12 µF
6 µF
A. 48 µC
B. 36 µC
C. 24 µC
D. 12 µC
E. 6 µC
b
A slightly challenging Example: A capacitor matrix
NxM identical capacitors, each with C0 capacitance, are connected
in the matrix show below. What is the equivalent capacitance of the
capacitor network?
A
B
N capacitors, connected in series, on each row.
There are M rows in total. Both M and N are integers.
Energy Storage in Capacitors
A capacitor has charge Q. How much energy is stored in the capacitor?
Method 1: How much work is needed to move the top plate upwards by
 
a distance of d?
W = ∫ F • dl
+Q


F + QE = 0
E

σ
E =−
yˆ
2
ε
+Q
E

Qσ
Qσ
Q σs
⇒ W = ∫ −(−
yˆ ) • dl = ∫
dy =
-Q
2ε
2ε
2ε
−Q
−Q
0
0
0
0
2
ΔU = W =
Q σs
Q
=
2ε 0 2ε 0 A /d
Method 2: Let the plates already be separated by distance d. They are
then charged to Q. The electric potential is changing during the process:
€
Q2 CV 2 1
U=
=
= QV
2C
2
2
W =
∫
W
0
Q
€
Q
dW = ∫ 0 vdq = ∫ 0
Q2
Q2
ΔU = W =
=
2C 2ε 0 A /d
It’s not QV
€
dW = vdq
v = q /C
Q2
q /Cdq =
2C
Electric-Field Energy
For parallel plate (capacitors), the electric potential energy is stored
in the electric filed between them!
The density is:
CV 2 /2
u=
Volume = Ad
C = ε 0 A /d
V = Ed
(ε 0 A /d)(Ed) 2 1
∴u =
= ε0 E 2
2Ad
2
Implication: Vacuum is not really empty
Everyday example:
Microwave
€
The Z Machine—capacitors storing large amounts of energy
•  This large array of capacitors in parallel can store huge
amounts of energy. When directed at a target, the
discharge of such a device can generate temperatures
on the order of 109K!
Dielectrics change the potential difference
•  What if it’s not vacuum
between the two
conducting plates of a
capacitor?
(You know you can’t put a
conductor in between)
•  A dielectric (nonconducting material) in
between reduces the
potential difference, hence
the field, and the
capacitance
•  Let measure the change,
then figure out how it
happens.
How does dielectrics do this magic of increasing capacitance
Dielectrics and Insulators
Always insulator
Might not be a dielectrics,
which should have
high polarizability
•  Electric field (E0) induces
charge and polarization:
E = E 0 /K
V = Ed = (E 0 /K)d = (E 0 d) /K = V0 /K
C = Q /V = K(Q /V0 ) = KC0
Induced charge reduces the field
K: Dielectric Constant, property of specific material
€
Table 24.1—Dielectric constants
How do you propose to measure the
dielectric constants of a material?
How does the energy density changes?
Kε0 (=ε)is usually the
permittivity of the dielectric
€
2
CV
) /(Ad) = (
2
1
V
u = (Kε 0 )( ) 2
2
d
1
u = (Kε 0 )E 2
2
1
u = εE 2
2
u=(
Kε 0
A 2
V
d
) /(Ad)
2
Dielectric breakdown: What does it take for these
insulators to become (partly) conductors?
•  A very strong electrical field can exceed the strength of the
dielectric to contain it:
•
Electrons become loose 
removing more electrons 
causing avalanche
creating conducting path
Q24.7
You reposition the two plates of a capacitor so that the
capacitance doubles. There is vacuum between the plates.
If the charges +Q and –Q on the two plates are kept constant in
this process, the energy stored in the capacitor
A. becomes 4 times greater.
B. becomes twice as great.
C. remains the same.
D. becomes 1/2 as great.
E. becomes 1/4 as great.
Q24.8
You slide a slab of dielectric between the plates of a parallel-plate
capacitor. As you do this, the charges on the plates remain
constant.
What effect does adding the dielectric have on the potential
difference between the capacitor plates?
A. The potential difference increases.
B. The potential difference remains the same.
C. The potential difference decreases.
D. not enough information given to decide
Q24.9
You slide a slab of dielectric between the plates of a parallel-plate
capacitor. As you do this, the charges on the plates remain
constant.
What effect does adding the dielectric have on the energy stored
in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide
Q24.10
You slide a slab of dielectric between the plates of a parallel-plate
capacitor. As you do this, the potential difference between the
plates remains constant.
What effect does adding the dielectric have on the amount of
charge on each of the capacitor plates?
A. The amount of charge increases.
B. The amount of charge remains the same.
C. The amount of charge decreases.
D. not enough information given to decide
Q24.11
You slide a slab of dielectric between the plates of a parallel-plate
capacitor. As you do this, the potential difference between the
plates remains constant.
What effect does adding the dielectric have on the energy stored
in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide
CH24 Summary: Capacitance and dielectrics
Definition of capacitance:
Capacitors in series:
in parallel:
C=
Q1 = Q2 = Q
V = V1 + V2 ⇒
1/C +1/C ≡€1/C
1
2
V1 = V2 = V ⇒
C = C1 + C2
Energy in a capacitor:
€
Dielectrics:
€
Q
= ε 0 A /d
Vab
What if you have both
kinds of connections?
Q2
CV 2
1
=
= QV
2C
2
2
1
u = ε0 E 2
2
U =
C =€
KC0 = Kε 0 A /d = εA /d
1
1
u = Kε 0 E 2 = εE 2
2
2
When dielectrics fills the capacitor,
What changes? What stays the same?