capacitance and dielectricsastro1.panet.utoledo.edu/~vkarpov/l05s.ch25.pdf · capacitance and...
TRANSCRIPT
Capacitance and Dielectrics
Capacitance
General Definition:
VqC /====Special case for parallel
plates:
d
AC 0εεεε====
Potential Energy
• I must do work to charge up a capacitor.
• This energy is stored in the form of electric
potential energy.
• We showed that this isC
QU
2
2
====
2
02
1Eu εεεε====
• Then we saw that this energy is stored in the
electric field, with a volume energy density
Potential difference and
Electric fieldSince potential difference is work per unit charge,
∫∫∫∫====∆∆∆∆b
aEdxV
For the parallel-plate capacitor E is uniform, so
EdV ====
Also for parallel-plate case Gauss’s Law gives
d
A
V
QCsoVd
A
QE 0
0
0/εεεε
εεεεεεεεσσσσ ====================
Spherical exampleA spherical capacitor has inner radius a = 3mm,
outer radius b = 6mm. The charge on the inner
sphere is q = 2 C. What is the potential difference?
From Gauss’s Law or the Shell
Theorem, the field inside is 2r
kqE ====
From definition of
potential difference
V33
33
99103
6
1
3
11018
106
1
103
1102109 ××××====
−−−−××××====
××××
−−−−××××
××××××××××××==== −−−−−−−−−−−−
∫∫∫∫
−−−−========b
a bakqdr
r
kqV
112
What is the capacitance?
FVCVQC4
107.6)3000/()2(/−−−−××××============
Q.25-1
(1) 3 µF
(2) 6 µF
(3) 12 µF
(4) 24 µF
A capacitor has capacitance C = 6 µF
and charge Q = 2 nC. If the charge is
increased to 4 nC what will be the
new capacitance?
Q. 25 - 1
A capacitor has capacitance C = 6 µF and a
charge Q = 2 nC. If the charge is increased to 4
nC what will be the new capacitance?
(1) 3 µF (2) 6 µF (3) 12 µF (4) 24 µF
Solution:
Capacitance depends on the structure of
the capacitor, not on its charge.
Q.25-2
1) 10,000 V/m
2) 900 V/m
3) 90 nV/m
A parallel-plate capacitor has plates of area
0.1 m2 and carries a charge of 9 nC.
What is the electric field between the plates?
Q. 25 - 2
Solution:Field at surface
of conductor:
mVmVmV /109)3(/900)2(/101)1(84 −−−−××××××××
?
109
1.0
9
2
====
××××====
====−−−−
E
CQ
mA
mVE
AQ
E
/101109
109
109101
109/
/
4
12
8
8
1
9
0
××××====××××××××
====
××××====××××××××
========
====
−−−−
−−−−
−−−−−−−−
−−−−
σσσσ
εεεεσσσσ
Capacitors in Parallel
If several capacitors are
connected in parallel as
shown, we can consider
the arrangement as a
single capacitor.
L++= 21 qqqtot
By charge conservation:
L============ 21 VVVtot
The wires are good
conductors, so:
tottottot
tot
CVqV
qqq
V
q
V
q
V
qCC
========++++++++++++
====
++++++++++++====++++++++
/321
3
3
2
2
1
121
L
LL
Therefore:
Capacitors in Series
What if several capacitors are
connected in series as shown?
Now charge conservation
gives:
L============ 21 qqqtot
And now it is the voltages that add:
L++++++++==== 21 VVVtot
So now the voltage sum,
when written in terms of the
capacitances, gives the result
for the series case:
L
L
L
++++++++====
++++++++====
++++++++====
21
2211
21
/1/1/1
///
CCC
CqCqCq
VVV
tot
tottot
tot
Results for Capacitor Combinations
• Capacitors in parallel:
L++++++++==== 21 CCCtot
• Capacitors in series:
L++++++++==== 21 /1/1/1 CCCtot
Example: Problem 25-8
What is effective capacitance of the device?
C12 FC
FC
FC
µµµµ
µµµµ
µµµµ
0.4
0.5
10
3
2
1
====
====
====
FC
CCC
µµµµ3.33/10
10/35/110/1/1/1/1
12
2112
========
====++++====++++====
FCCCtot µµµµ3.743.3312 ====++++====++++====
Dielectric Materials
• Applied electric field polarizes material
• This produces an induced surface charge
• This reduces field within material
• This reduces potential difference
• This increases capacitance
Effect of placing a dielectric between the
plates of a charged capacitor:
E=0
electronnucleus
Polarization of atoms and molecules
Non-polar molecules:
induced dipoles
Polar molecules:
preferential orientation
Dielectric materials
Dielectric materials polarize under external electric field E0
Polarization creates the opposing electric field: –E’
Total field E=E0-E’=E0/κ < E0
κ is called dielectric constant
It shows how the field in a dielectric is weaker than in vacuum
Dielectric Constant
• So the effect of a dielectric is to
decrease the field and increase the
capacitance.
• The factor by which E is decreased
and C is increased is defined to be the
dielectric constant κ.
0CC κκκκ====κκκκ/0EE ====
Example: Problem 25-48
Given a parallel-plate capacitor, dielectric-
filled, with area A = 100 cm2, charge Q = 890
nC, and electric field E = 1.4 kV/mm.
(a) Find dielectric constant of the material.
(b) Find the induced charge.
(a)
(((( )))) 00 // εεεεAqE ====
If there were no dielectric the field
between the plates would be given
by Gauss’s Law as
mVE /1001.11085.8
10/109.8 7
12
27
0 ××××====××××
××××==== −−−−
−−−−−−−−
2.7104.1/1001.1/67
0 ====××××××××======== EESoκκκκ
κ/0EE =But by definition:
0
/
εεεεAq
E
Gauss
tot====
⇒⇒⇒⇒
nCnCnCqqqso
qqqBut
tot
tot
766124890'
'
====−−−−====−−−−====
−−−−====
nCC
EAqSo tot
124104.1210104.11085.882612
0
====××××====××××××××××××××××====
====−−−−−−−−−−−−
εεεε
(b)