The first midterm!!!
The first common hour midterm exam will be held on
Thursday October 1, 9:50 to 11:10 PM (at night) on the
Busch campus. You should go to the room corresponding to
the first 3 letters of your last name. If you have a conflict
with the exam time, please contact Prof. Cizewski
Cizewski@physics.rutgers.edu with your entire schedule for
the week of September 28 at your earliest convenience but
not later than 5:00 pm on Wednesday, September 23.
Aaa – Hoz
Hua – Moz
Mua – Shz
Sia – Zzz
ARC 103
Hill 114
PHY LH
SEC 111
The first midterm!!!
Next time: Review
Email me questions by Wednesday noon!!
Lecture 8. Capacitors and Dielectrics
equipot. @ V=V
+
V=V
-
+
equipot. @ V=0
“ground”
Static conditions (no currents)
3
Two Types of Problems
+
+
V=V
-
V=V
C
-
C
Capacitor is connected to
a voltage source.
Capacitor is charged and
disconnected from the voltage
source.
The voltage difference is fixed,
the charge may vary if the
capacitance varies.
The charge is fixed, the voltage
difference may vary if the
capacitance varies.
𝑉 = 𝑐𝑜𝑛𝑠𝑡
𝑄 = 𝐶𝑉
𝑄 = 𝑐𝑜𝑛𝑠𝑡
𝑄
𝑉=
𝐶
4
Parallel Connection of Capacitors
equipot. @ V=V
+
V=V
-
C1
What quantity is the same for both capacitors?
+ Q1
-
C2
+ Q2
𝑄1 𝑄2
𝑉=
=
𝐶1 𝐶2
-
equipot. @ V=0
𝑄1 = 𝐶1 𝑉
𝑄2 = 𝐶2 𝑉
- the charges on capacitors
are redistributed to keep
𝑉1 = 𝑉2 = 𝑉
𝑄Σ = 𝑄1 + 𝑄2 = 𝐶1 + 𝐶2 𝑉
𝐶Σ
𝐶Σ = 𝐶1 + 𝐶2
9
Series Connection of Capacitors
equipot. @ V=V
+
V=V
C1
-
C2
What quantity is the same for both capacitors?
+
Q1=Q
- equipot.
@ V=V*
𝑄1 = 𝑄2 = 𝑄
+
- Q2=Q
𝑉1 = 𝑉 − 𝑉 ∗ =
equipot. @ V=0
+Q
-Q
+Q
-Q
- the “central”
conductor is
uncharged, it can only
be polarized.
𝑉2 =
𝑉∗
𝑄
𝐶1
𝑄
−0=
𝐶2
𝑉1 + 𝑉2 = 𝑉
𝑄
𝑄
𝑄
𝑉= + =
𝐶1 𝐶2 𝐶Σ
𝐶1 𝐶2
𝐶Σ =
𝐶1 + 𝐶2
1
1
𝐶Σ =
+
𝐶1 𝐶2
−1
12
More Complex Circuits
C4
C5
C1
C2
C3
+
C5
V=V
-
C1
C4
C2
C3
𝐶1 = 0.33 + 0.33 + 0.33 = 1𝜇𝐹
3∙3
𝐶2 =
= 1.5𝜇𝐹
3+3
1.5 ∙ 3
𝐶2 … 𝐶5 =
= 1𝜇𝐹
1.5 + 3
𝐶4 ∥ 𝐶2 … 𝐶5 = 1 + 1 = 2𝜇𝐹
𝐶3… 𝐶4 ∥ 𝐶2 … 𝐶5
𝐶Σ = 𝐶1 ∥ 𝐶3… 𝐶4 ∥ 𝐶2 … 𝐶5
= 𝐶4 ∥ 𝐶2 … 𝐶5
= 1 + 1 = 2 𝜇𝐹
2⋅2
=
= 1𝜇𝐹
2+2
17
Redistribution of Charges
1
2
+
+
V=V
-
𝑉∗
V=V
C1
-
C2
C1
C2
Initially, only C1 was connected to the voltage source, C2 was uncharged. We
disconnect C1 from the source and connect it to C2. Find V*, Q1 and Q2.
After C1 was disconnected from the voltage source, its charge was 𝑄1 𝑖𝑛𝑖𝑡 = 𝐶1 𝑉.
Connection to C2 results in redistribution of charges, but the net charge (𝑄Σ = 𝑄1 𝑖𝑛𝑖𝑡 )
is conserved:
𝑄Σ = 𝑄1 𝑖𝑛𝑖𝑡 = 𝐶1 𝑉 - switch in position 1
𝑉∗ =
𝐶1
𝑉
𝐶1 + 𝐶2
𝐶2 ≫ 𝐶1
𝑄1 =
𝑉 ≫ 𝑉∗
𝑄Σ = 𝐶1 𝑉 ∗ + 𝐶2 𝑉 ∗ - switch in position 2
𝐶1
𝐶𝑉
𝐶1 + 𝐶2 1
𝑄1 ≪ 𝑄Σ
𝑄2 =
𝐶1
𝐶 𝑉
𝐶1 + 𝐶2 2
𝑄2 ≈ 𝑄Σ
18
Linear Dielectrics in External Electric Fields
Uncompensated charges reside on
the surfaces, the total field is the
sum of the external field and the
field due to uncompensated charges.
𝐸0
Has the field outside been changed?
𝐸
𝐸0
𝐸𝑖𝑛
Dielectric constant:
𝜎𝑖
𝜀0
𝑥
𝐾≡
𝐸0
𝐸𝑖𝑛
𝜎𝑖
1
= 𝐸0 − 𝐸𝑖𝑛 = 𝐸0 1 −
𝜀0
𝐾
>1
What would
be K for
metal?
19
Comment
Student: “After we introduced the concept of dielectric constant, why bother to
distinguish among conductors, vacuum, and dielectrics? Can’t we just say “a
spherical region of 𝐾 = ∞” instead of “a metal sphere”?
Distinctions:
(a) Physical mechanisms responsible for polarization (emerging induced
charges) of metals and dielectrics are quite different (redistribution of
mobile electrons vs. polarization of bound charges). This becomes
especially important when the electric field is time-dependent: the free
and bound electrons respond differently to high-frequency electric fields.
(b) Boundary conditions are different for metals and dielectrics, this becomes
clear when we’ll consider steady currents (supported by metals, not
dielectrics).
20
Interaction Between Charges Surrounded by Dielectric
+++
+ -+
+++
𝜎𝑖 = 𝜀0 𝐸0
+
- - - +
+ - +
1
1
1−
= 𝜎0 1 −
𝐾
𝐾
Two uniformly charged spherical
( 𝜎 = 𝜎0 ) shells are placed in a
dielectric. The dielectric is polarized:
the polarization charge partially
“screens” the charges on the spheres.
- absolute value, the sign is opposite to 𝜎0
Net charge (initial charge + polarization “cloud”):
1
𝑄
𝑄Σ = 𝑄 − 𝑄 1 −
=
𝐾
𝐾
The electric field due to the screened +Q acting upon –Q:
𝐸=𝑘
The force between +Q and –Q charges surrounded by dielectric:
𝑄
𝐾𝑟 2
𝐹 = 𝑄𝐸 =
𝐹0
𝐾
(𝐹0 is the force between these charges in vacuum)
Polar (large K) solvents at work: strongly polar compounds (e.g., sugars) or ionic
compounds (e.g., table salt) dissolve only in very polar solvents like water.
21
Dielectric-filled Capacitors
+
+
V=V
-
-
+
+
V=V
-
-
(1)
The voltage source (V=const) is
connected to (1) a “vacuum”
capacitor and (2) the same
capacitor filled with dielectric.
(2)
charge on metallic electrodes
(that’s what we measure/control)
(1)
𝐸1 =
𝑉
=𝐸
𝑑
(2)
𝐸2 =
𝑉
=𝐸
𝑑
𝜎𝑚
−𝜎𝑖
𝜎1 = 𝜎𝑚 = 𝜀0 𝐸 = 𝜀0
𝑉
𝑑
𝐶1 =
𝑄 𝜎𝑚 𝐴
𝐴
=
= 𝜀0
𝑉
𝑉
𝑑
𝜎2 = 𝜎𝑚 − 𝜎𝑖 = 𝜀0 𝐸
𝜎𝑖 = 𝜀0 𝐸𝑖𝑛 𝐾 − 1 = 𝜀0 𝐸 𝐾 − 1
𝑄 𝜎𝑚 𝐴
𝐴
𝐶2 = =
= 𝜀0 𝐾 = 𝐾𝐶1
𝑉
𝑉
𝑑
𝐶 = 𝐾𝜀0
𝐴
𝐴
=𝜀
𝑑
𝑑
𝜎𝑚 = 𝜎2 + 𝜎𝑖 = 𝜀0 𝐸𝐾
K times greater
than that for a
vacuum capacitor
𝜀 ≡ 𝐾𝜀0 - permittivity of a dielectric
22
Energy Stored in a Dielectric-filled Capacitor
𝑄𝑚 = 𝑄
𝑄𝑚 = 0
Initial
state
Final
state
𝐸=0
𝛿𝑞𝑚
𝑞𝑚 - charge of mobile electrons in metal
𝑞𝑖 - polarization charge in dielectric
We want to calculate the total
reversible work done on the free
charges in the charging circuit (these
are the charges that we control); this
work can be retrieved from the
capacitor.
𝜎𝑚 − 𝜎𝑖
∙𝑑
𝜀0
𝜎𝑚 𝑞𝑚
𝜎𝑚 − 𝜎𝑖 =
=
𝐾
𝐾𝐴
𝛿𝑊 = 𝛿𝑞𝑚 ∙ 𝑉 𝑞𝑚 , 𝑞𝑖 = 𝛿𝑞𝑚 ∙
𝜎𝑖 = 𝜎𝑚
1
1−
𝐾
We took into account the reaction of bound (polarization) charges: V depends on 𝜎𝑖 .
𝑞𝑚 =𝑄
𝑊=
𝑞𝑚 =0
𝑞𝑚
𝑑
𝑄2 1
𝛿𝑞𝑚 ∙
∙𝑑 =
∙
= 𝐶𝑓𝑖𝑙𝑙𝑒𝑑 𝑉 2
𝜀0 𝐾𝐴
𝜀0 𝐾𝐴 2
2
𝐶𝑓𝑖𝑙𝑙𝑒𝑑 =
𝜀0 𝐾𝐴
𝑑
Thus, if we take two identical capacitors - one empty, another one filled with a dielectric
– and charge them from the same voltage source, the energy stored in the filled one
will be greater (extra reversible work on polarization of the dielectric).
23
Energy Stored in a Dielectric-filled Capacitor (cont’d)
Different experiment: we charge an
“empty” capacitor to charge Q,
disconnect it from the voltage
source, and insert dielectric.
𝑈𝑓𝑖𝑙𝑙𝑒𝑑 /𝑈𝑒𝑚𝑝𝑦 =?
Now Q is fixed (rather than V).
𝑈𝑒𝑚𝑝𝑡𝑦
𝑑 𝑄2
𝑄2
=
∙
=
𝜀0 𝐴 2
2𝐶𝑒𝑚𝑝𝑡𝑦
𝑈𝑓𝑖𝑙𝑙𝑒𝑑
𝑑
𝑄2
𝑄2
=
∙
=
𝜀0 𝐾𝐴 2
2𝐶𝑓𝑖𝑙𝑙𝑒𝑑
𝑈𝑓𝑖𝑙𝑙𝑒𝑑 1
=
𝑈𝑒𝑚𝑝𝑦 𝐾
Now the final energy is smaller than the initial one. The energy was wasted: we have to
do some “negative” work in order to counteract the force on dielectric (which is pulled in
the region of stronger field).
A non-uniform electric field exerts a force on
𝑑𝐸
dipoles (𝐹 = 𝑞𝐸1 − 𝑞𝐸2 = 𝑞𝑑 𝑑𝑥 ):
24
Dielectric-Filled Capacitors
+
V=V
-
𝐴1
𝐴2
𝑘1 = 1
𝑘2
+
V=V
𝑑
-
C1
C2
𝜀0
𝐶Σ = 𝐶1 + 𝐶2 =
𝐴 + 𝑘 2 𝐴2
𝑑 1
+
V=V
-
𝐴
𝑘1 = 1
𝑘2
𝑑1
C1
-
C2
V=V
𝑑2
1
1
𝐶Σ =
+
𝐶1 𝐶2
+
−1
𝑑1
𝑑2
=
+
𝜀0 𝐴 𝜀0 𝑘2 𝐴
−1
27
Conclusion
Parallel and serial connection of capacitors.
Dielectrics in electric field.
Capacitance of dielectric-filled capacitors
Energy stored in dielectric-filled capacitors.
Next time: Lecture 9. Review.
Please prepare/email your questions!
28