24.4 Dielectrics

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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: 𝐢0 =
With dielectric:
𝐢=
𝑄
𝑉0
𝑄
𝑉
Κ: =
𝑄 = π‘π‘œπ‘›π‘ π‘‘
𝐢
𝑉0
=
𝐢0 𝑉
because V<V0
C>C0
K>1: relative dielectric constant
What happens with the E-field in the presence of dielectric material
𝑸 = 𝒄𝒐𝒏𝒔𝒕
𝐸0
𝑉0 𝐸0
We know V<V0
E<E0 specifically
𝐸=
Κ=
=
Κ
𝑉
𝐸
d
Recall: 𝐸 =
πœŽπ‘›π‘’π‘‘
πœ–0
πœŽπ‘›π‘’π‘‘ reduced with dielectric
material
The surface charge (density) σ on
conducting plates does not change but
induced charge σi of opposite sign
𝜎
𝐸0 =
πœ–0
and 𝐸 =
𝜎 − πœŽπ‘–
πœ–0
𝜎
πΎπœ–0
1
πœŽπ‘– = 𝜎(1 − )
𝐾
𝐸=
πœ– = πΎπœ–0
Definition of the
permittivity
𝐴
1
𝐢 = πœ– 𝑑 and 𝑒 = 2 πœ–πΈ 2
DIELECTRICS
+Q
Example:
E0
E1
E2
K1
K2
d/2
d/2
V
-Q
𝐸0
𝐸1
𝐸2
𝜎
𝑄
=
=
πœ–0 πœ–0 𝐴
𝐸0
𝑄
=
=
𝐾1
πœ–0 𝐴𝐾1
𝐸0
𝑄
=
=
𝐾2
πœ–0 𝐴𝐾2
𝜎1 = 𝜎(1 −
1
)
𝐾1
𝑄2 𝑑
π‘ˆ=
= 4πœ– 𝐴
0
πœ–0
1
= 4 (𝐸2 𝐾2 )2 𝐾 +
1
1
𝑄𝑉
2
𝑑
𝑑
+ 𝐸2
2
2
𝑄
𝑄
𝑑
𝑄𝑑 1
1
=
+
=
( + )
πœ–0 𝐴𝐾1 πœ–0 𝐴𝐾2 2 2πœ–0 𝐴 𝐾1 𝐾2
V = 𝐸1
𝑄
𝑄
2πœ–0 𝐴𝐾1 𝐾2
𝐢= =
=
𝑄𝑑 1
1
𝑉
( + ) 𝑑(𝐾1 + 𝐾2 )
2πœ–0 𝐴 𝐾1 𝐾2
1
𝜎2 = 𝜎(1 − )
𝐾2
1
𝐾1
1
𝐾2
1
π‘ˆ
1
𝑄
+ 𝐾 π‘Žπ‘›π‘‘ 𝑒 = 𝐴𝑑 = 4πœ– (𝐴 )2
2
0
1
𝐾1
1
πœ–
+ 𝐾 = 40 (𝐸1 𝐾1 )2
2
1
𝐾1
1
+𝐾
2
24.4 DIELECTRICS
Dielectric breakdown or Dielectric strength
High Voltage
Ground
Ground
Air
Cr2O3
GAUSS’S LAW IN DIELECTRICS
Recall:
𝑄𝑒𝑛𝑐𝑙
πœ–0
=
𝐸 βˆ™ 𝑑𝐴
𝐸≠0
𝐸=0
Dielectrics
Conductor
𝑄𝑒𝑛𝑐𝑙 = 𝜎 − πœŽπ‘– 𝐴
𝐸 βˆ™ 𝑑𝐴 = 𝐸𝐴
𝐸𝐴 =
𝜎 −πœŽπ‘–
𝜎 − πœŽπ‘– 𝐴
πœ–0
1
πœŽπ‘– = 𝜎(1 − )
𝐾
𝐸𝐴 =
𝜎𝐴
πΎπœ–0
𝑄𝑒𝑛𝑐𝑙−π‘“π‘Ÿπ‘’π‘’
=
πœ–0
A
A
A
𝐾𝐸 βˆ™ 𝑑𝐴
GAUSS’S LAW IN DIELECTRICS
𝑄𝑒𝑛𝑐𝑙−π‘“π‘Ÿπ‘’π‘’
=
πœ–0
Example:
K
E1
ra
r
rb
E2
𝑉=
π‘Ÿπ‘
π‘Ÿπ‘Ž
𝑉=
π‘Ÿπ‘
π‘Ÿπ‘Ž
𝑄1 (π‘Ÿπ‘ − π‘Ÿπ‘Ž )
𝐸1 π‘‘π‘Ÿ =
2πœ–0 πΎπœ‹π‘Ÿπ‘Ž π‘Ÿπ‘
𝐸2 π‘‘π‘Ÿ =
𝑄2 (π‘Ÿπ‘ − π‘Ÿπ‘Ž )
2πœ–0 πœ‹π‘Ÿπ‘Ž π‘Ÿπ‘
𝐾𝐸 βˆ™ 𝑑𝐴
𝑄
=
πœ–0
𝐾𝐸 βˆ™ 𝑑𝐴 = 𝐾𝐸1 2πœ‹π‘Ÿ 2 + 𝐸2 2πœ‹π‘Ÿ 2
𝑄2
𝑄1
πœ–0
πœ–0
𝑄1
𝐸1 =
2πœ–0 πΎπœ‹π‘Ÿ 2
𝑄2
𝐸2 =
2πœ–0 πœ‹π‘Ÿ 2
2πœ–0 πΎπœ‹π‘Ÿπ‘Ž π‘Ÿπ‘ 𝑉
𝑄1 =
(π‘Ÿπ‘ − π‘Ÿπ‘Ž )
𝑄2 =
2πœ–0 πœ‹π‘Ÿπ‘Ž π‘Ÿπ‘ 𝑉
(π‘Ÿπ‘ − π‘Ÿπ‘Ž )
2πœ–0 πœ‹π‘Ÿπ‘Ž π‘Ÿπ‘ 𝑉
𝑄 = 𝑄1 + 𝑄2 =
(𝐾 + 1)
π‘Ÿπ‘ − π‘Ÿπ‘Ž
𝐢=
𝑄 2πœ–0 πœ‹π‘Ÿπ‘Ž π‘Ÿπ‘ (𝐾 + 1)
=
𝑉
(π‘Ÿπ‘ − π‘Ÿπ‘Ž )
MOLECULAR MODEL OF INDUCED CHARGE
CLICKER QUESTION
A 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 = ο‚₯
C. A value depends on the material of the conductor
E
E0
ο€½
0 i
0
ο€½
1
K
8
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