Two Conductors separated by a insulators form
CAPACITOR
If the two objects are charged, there will be an
electric field and a potential difference between them.
r
dl
∆V = ∫
b
a
r r
E ⋅ dl
1) a and b are conductor so they are
equipotential surface and it doesn’t
matter where I start from.
2) The electric field is conservative
so it doesn’t matter what path I take.
There is a single voltage difference between two charged conductors.
If I double the charge Q, I will double the voltage difference.
Define C = Q/V as the “capacitance
Capacitance depends only on the shape, size, and position of conductors
C = Q/V
Capacitance has units of “farads”
1 farad = 1 Coulomb/Volt
Capacitance between two parallel plates”
Va − Vb = ∫
b
a
r r
E ⋅ dl = Ed
σ
Q
=
E=
ε0 ε0 A
1 Qd
Vab = Ed =
ε0 A
Integral is easy because field is uniform
Recall result from infinite flat plate
C=
Q
A
= ε0
Vab
d
Spherical Capacitor
Example 24.3
From Gauss’s Law we know field between a and b
E (r ) =
Q
4πε 0 r 2
∆V = ∫
V=
Q
b
a
4πε 0 r
Q
b
4πε 0
∫
a
2
dr
1
dr
2
r
Q ⎛1 1⎞
⎜⎜ − ⎟⎟
V=
4πε 0 ⎝ rb ra ⎠
V
1 ⎛1 1⎞
⎜⎜ − ⎟⎟
=
Q 4πε 0 ⎝ rb ra ⎠
−1
⎛1 1⎞
⎛ ra rb ⎞
Q
⎜
⎟
⎟⎟
= 4πε 0 ⎜ − ⎟ = 4πε 0 ⎜⎜
V
⎝ rb ra ⎠
⎝ ra − rb ⎠
⎛ ra rb ⎞
⎟⎟
C = 4πε 0 ⎜⎜
⎝ ra − rb ⎠
IMPORTANT REMINDER
All points on a conductor are at the same potential
A particularly important case of this is two conductors
connected by a conducting wire
Capacitors in Parallel
“Equivalent Capacitors”
Q1 = C1V
Q2 = C2V
V is the same for both capacitors
Q = Q1 + Q2 = (C1 + C2 )V
Q
= C1 + C2
V
Cequiv = C1 + C2
Cequiv = C1 + C2 + C3 + ....
Capacitors in Series
Q
Vac = V1 =
C1
Vcb = V2 =
Q
C2
Q is the same for both capacitors
⎛ 1
1 ⎞
⎟⎟
Vac = V = V1 + V2 = Q⎜⎜ +
⎝ C1 C2 ⎠
V
1
1
=
+
Q C1 C2
1
Cequiv
1
Cequiv
=
1
1
+
C1 C2
=
1
1
1
+
+
+ ....
C1 C2 C3
Example 24.6 – What is equivalent capacitance for network “a”
C’
C’’
Ceq
1
1
1
=
+
C' 12 µF 6 µF
1
1
=
C' 4 µF
C' = 4 µF
C' ' = 3 µF + 11µF + 4 µF
C' ' = 18 µF
1
1
1
=
+
Ceq 18 µF 9 µF
1
1
=
Ceq 6 µF
Energy Storage in a Capacitor
+Q
Q
V=
C
-Q
To add more charge, I must do Work
To add a small amount of charge dq I must do work dW -
dW = Vdq =
q
dq
C
To add a finite charge Q, I must integrate Q final
Q final
Q final
Q final
q
1
1 ⎛1 2
1 2 ⎞
dq =
qdq = ⎜ Q final = Qinitial ⎟
W = ∫ dW = ∫ Vdq = ∫
∫
C
C Qinitial
C⎝2
2
⎠
Qinitial
Qinitial
Qinitial
If I start from NO net charge I get -
1 Q2 1
1
W=
= QV = CV 2
2 C 2
2
Stored Energy in a Charged Capacitor
End of Chapter 24
You are responsible for the material covered in T&F Sections 24.1-24.3
You are expected to:
•
Understand the following terms:
Capacitor, Capacitance, Equivalent Capacitors, Series, Parallel, Stored Energy
•
Be able to calculate capacitance in simple geometries (e.g. Parallel planes, spheres,…)
•
Be able to determine the equivalent capacitance of (relatively) simple networks of
capacitance using series and parallel rules (see example 24.6 for a typical network)
•
Calculate the energy stored in a capacitor using integration.
Recommended F&Y Exercises chapter 24:
1,3,4,14,15,20,22,23