Dielectrics
Experiment: Place dielectrics between plates of capacitor at Q=const condition
Observation: potential difference decreases to smaller value with dielectric material relative to air
Without dielectric:
With dielectric: 𝐶=𝑄𝑉
𝑄=𝑐𝑜𝑛𝑠𝑡 because V<V0
C>C0
Κ := 𝐶𝐶0
=𝑉 0
𝑉 K>1: relative dielectric constant
d
What happens with the E-field in the presence of dielectric material𝑸=𝒄𝒐𝒏𝒔𝒕We know V<V0
E<E0 specifically Κ=𝑉 0
𝑉 =𝐸0
𝐸 𝐸=𝐸0
Κ
Recall:
𝐸0=𝜎𝜖0
𝐸=𝜎−𝜎 𝑖
𝜖0and
𝜎 𝑖=𝜎 (1− 1𝐾 ) 𝐸=𝜎𝐾 𝜖0
𝜖=𝐾 𝜖0 Definition of the permittivity
and
The surface charge (density) σ on conducting plates does not change butinduced charge σi of opposite sign
𝜎 𝑛𝑒𝑡reduced with dielectric material
DIELECTRICSExample: K1
K2
d/2d/2
+Q
-Q
E0E1E2
‖𝐸0‖=𝜎𝜖0
=𝑄𝜖0 𝐴
‖𝐸1‖=‖𝐸0‖𝐾1
= 𝑄𝜖0𝐴 𝐾1
‖𝐸2‖=‖𝐸0‖𝐾2
= 𝑄𝜖0 𝐴𝐾 2
V
𝐶=𝑄𝑉 = 𝑄
𝑄𝑑2𝜖0 𝐴
( 1𝐾 1+ 1𝐾 2
)=2𝜖0 𝐴𝐾 1𝐾 2
𝑑 (𝐾 1+𝐾2)
𝜎 1=𝜎 (1− 1𝐾1
) 𝜎 2=𝜎 (1− 1𝐾 2
)
==
24.4 DIELECTRICSDielectric breakdown or Dielectric strength
Cr2 O3
Ground GroundHigh Voltage
Air
GAUSS’S LAW IN DIELECTRICSRecall:
Conductor Dielectrics�⃗�=0 �⃗�≠0
𝜎−𝜎 𝑖
𝑄𝑒𝑛𝑐𝑙=(𝜎−𝜎 𝑖 ) 𝐴
∮𝐸 ∙𝑑𝐴=𝐸𝐴
AA
A
𝐸𝐴=(𝜎−𝜎 𝑖 ) 𝐴
𝜖0
𝜎 𝑖=𝜎 (1− 1𝐾 )
𝐸𝐴=𝜎 𝐴𝐾 𝜖0
𝑄𝑒𝑛𝑐𝑙− 𝑓𝑟𝑒𝑒
𝜖0=∮𝐾 𝐸 ∙ �⃗�𝐴
GAUSS’S LAW IN DIELECTRICSExample:Capacitance of half filled spherical capacitor
Kra
rbr
𝑄𝑒𝑛𝑐𝑙− 𝑓𝑟𝑒𝑒
𝜖0=∮𝐾 𝐸 ∙ �⃗�𝐴
𝑄𝜖0
=∮ 𝐾 �⃗� ∙ �⃗�𝐴=𝐾 𝐸12𝜋𝑟2+𝐸22𝜋𝑟 2E1
E2
𝑄1
𝜖0𝑄2
𝜖0𝐸1=
𝑄1
2𝜖0𝐾 𝜋𝑟 2
𝐸2=𝑄2
2𝜖0 𝜋𝑟2
𝑉=∫𝑟 𝑎
𝑟 𝑏
𝐸1𝑑𝑟=𝑄1(𝑟𝑏−𝑟 𝑎)2𝜖0𝐾 𝜋 𝑟𝑎𝑟 𝑏
❑⇒𝑄1=¿
2𝜖0𝐾 𝜋𝑟 𝑎𝑟 𝑏𝑉(𝑟 𝑏−𝑟𝑎)
¿
𝑉=∫𝑟 𝑎
𝑟 𝑏
𝐸2𝑑𝑟=𝑄2(𝑟 𝑏−𝑟𝑎)2𝜖0 𝜋𝑟𝑎 𝑟𝑏
❑⇒𝑄2=
2𝜖0 𝜋𝑟𝑎𝑟𝑏𝑉(𝑟 𝑏− 𝑟𝑎)
𝑄=𝑄1+𝑄2=2𝜖0𝜋 𝑟𝑎𝑟 𝑏𝑉
(𝑟𝑏−𝑟 𝑎)(𝐾 +1)
𝐶=𝑄𝑉 =
2𝜖0𝜋𝑟 𝑎𝑟 𝑏(𝐾 +1)(𝑟 𝑏−𝑟𝑎)
Check: K->1 needs to reproduce empty =
MOLECULAR MODEL OF INDUCED CHARGE
EE
8
CLICKER QUESTIONA conductor is an extreme case of a dielectric, since if an electric field is applied to a conductor, charges are free to move within the conductor to set up “induced charges”. What is the dielectric constant of a perfect conductor?
A. K = 0
B. K =
C. A value depends on the material of the conductor
0
0 0
1iEE K
