Lecture 10: Chapter 30, October 4 2005 - UNC Charlotte Pages · Lecture 10: Chapter 30, October 4...
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Lecture 10: Chapter 30, October 4 2005 • Calculating capacitance using steps discussed previously: a) Parallel-Plate Capacitor b) A Cylindrical Capacitor c) A Spherical Capacitor • Combination of capacitors • The Energy Stored in a Capacitor • Review problem for Quiz 5 C = Q/∆V c Which capacitance is larger? Density η and separation d are the same, areas are different. + + + + + - - - - - + + + + + - - - - - - - d Density η Area A d Area A’>A
Transcript of Lecture 10: Chapter 30, October 4 2005 - UNC Charlotte Pages · Lecture 10: Chapter 30, October 4...
Lecture 10: Chapter 30, October 4 2005• Calculating capacitance using steps discussed previously:a) Parallel-Plate Capacitorb) A Cylindrical Capacitorc) A Spherical Capacitor
• Combination of capacitors• The Energy Stored in a Capacitor• Review problem for Quiz 5
C = Q/∆Vc
Which capacitance is larger?Density η and separation d are the same, areas are different.
+ +++ +
- --- -
++
+++
- - ---- -d
Density ηArea A
d
Area A’>A
It will be shown that the structure of the expression for C:
)_(0 FactorlGeometricaC ε=
dAC 0ε=
)/ln(2 0 ab
LC πε=
Different Geometries
Parallel-Plate Cylindrical Spherical
+ +++ +
- --- -
Density ηArea A
ababC−
= 04πε
L
2b
2a
2b2a
Parallel-Plate Capacitor• First Step: Assume Charge Q• Second Step: Calculating E from Q using Gauss’s Law
AzE
facestheonSince
EdAAdE
e
Facese
)(21cos180
:___
2
0
−=Φ−=⇒=
==Φ ∫∫
θθ
rr
z
Electric Field (Enet) inside a capacitor:
• Third Step: Finding ∆V using ∫−=∆f
iEdsV
• Last Step: Calculate capacitance C = Q/∆V
Cylindrical Capacitor
L
2b
2a
a
b
++++
+
+ +
-
-
-
-
-
-
-
-
--
-
Total charge +Q
Gaussiansurface
r
Total charge -Q
Path ofintegration
We did not solve the cylindrical geometry in the class, but I placed the solution below
Cross-sectional view
• First Step: Assume Charge Q
• Second Step: Calculating E from Q using Gauss’s Law.As a Gaussian surface, we choose a cylinder of length L and radius r, closed by end caps.
LrQE
rLEAdEwhere
q
enc
e
ence
0
0
2
)2(_
/
πε
π
ε
=
=
==Φ
=Φ
∫rr
• Third Step: Finding ∆V using
⎟⎠⎞
⎜⎝⎛=−
==−=
−=
∫
∫ ∫+
−
ab
LQ
rdr
LQ
EdssdEV
ThusEdssdE
a
b
f
i
ln22
:_.
00 πεπε
rr
rr
∫−=∆f
isdEV rr
We select the path from negative to the positive plate. For thispath, the vectors E and ds will have opposite direction, so:
Where we used the fact that here ds = -dr (we integrated radially inward).
• Last Step: Calculate capacitance C = Q/∆V
)/ln(2 0 ab
LC πε=
A Spherical Capacitor
• First Step: Assume Charge Q
The cross-sectional image is indistinguishable from the cylindrical case. However in 3-D case the spherical geometry is quite different from the cylindrical one.
a
b
++++
+
+ +
-
-
-
-
-
-
-
-
--
-
Total charge +Q
Gaussiansurface
r
Total charge -Q
Path ofintegration
2b2a
20
0
2
2
0
41
4)(
4)(
/
rQE
QrrE
rrEAdE
q
enc
e
ence
πε
επ
π
ε
=
=
=
==Φ
=Φ
∫rr
Second Step: Calculating E from Q using Gauss’s Law.As a Gaussian surface, we choose a sphere with radius r:a < r < b
• Third Step: Finding ∆V using ∫−=∆f
isdEV rr
ababQ
baQ
rdrQEdsV
a
b
−=⎟
⎠⎞
⎜⎝⎛ −
=−== ∫∫+
−
00
20
411
4
4
πεπε
πε
Where again we substituted –dr for ds
• Last Step: Calculate capacitance C = Q/∆V
ababC−
= 04πε
An Isolated Sphere
baaC
/14 0 −
= πε
Let us first divide numerator and denominator by b:
If we now let b → ∞, external sphere infinitely expands leaving just one central sphere. Substituting R for a:
RC 04πε=
This is the capacitance of isolated single sphere
Combinations of Capacitors
Parallel Capacitors Series Capacitors
Replacing two parallel capacitors with an equivalent one
Equivalent capacitance is a single element which has same property, charge and potential as a pair of capacitors
Ceq = Q/(∆Vc) = (Q1+Q2)/ ∆Vc = C1 + C2
Ceq = C1 + C2 + … (parallel capacitors)
Replacing two series capacitors with an equivalent one
1/Ceq = ∆Vc/Q = (∆ V1 + ∆ V2)/Q = 1/C1 + 1/C2
Ceq = (1/C1 + 1/C2 + …)-1 (series capacitors)
In this case the magnitude of charges (Q) on all capacitors including equivalent one are equal – “chain” charging process.
A Capacitor Circuit
Typical question: to find charges and potentials across each capacitor
Analyzing the Capacitor Circuit
The Energy Stored in a Capacitor
( )22
0 21
21
c
Q
c VCC
QqdqC
U ∆=== ∫
Each little charge dq added to the charge q(t) on the capacitor increases its potential energy: dU = dq ∆V = qdq/C
The energy stored in a charged capacitor:
Since ∆Vc = Q/C
End of Lecture 10Reading: Chapter 30Review for Quiz 5Home Work 5
