8.022 Lecture Notes Class 13 - 09/28/2006
Find work needed
to bind together
W =
=
= 12 (q ·
=
=
qV
1 4
�i )
2 Σi=1 Qi · V (r
q
−q
−q
q
−q
1
1
√
√
4π�0 ( a 2 + a + a ) · 2 + 2 · −q · 4π�0 ( a + a 2
q 2√
q2
q 2√
q2
1
a ( 4π�0 2 − 2 · 4π�0 + 4π�0 2 − 2 · 4π�0 )
√
2
q2
√2 − 4) = q ( 2 − 4)
(
4π�0 a
4π�0 a
2
Do sides , then diagonals
U=
=
=
=
=
�0
2(4π�0 )2
2
( qr2 )2 r2 sin θdθdφdr
�∞ 1
q2
8π�0 0 r2 dr
�
q 2 −1 ∞
8π�0 r |0
2
q
1
1
8π�0 (− ∞ − (− 0 ))
∞
+ aq ))
2
Charge Up a Capacitor
dW = ( Cq )dq
W =
work gets harder as there is more charge already on it
�Q q
0 C dq
q2 Q
2C �
0
=
Q2
2C
=
C=
Q
V
1
2
2 CV
=
⎧
⎨�2 V = − ρ
Poisson’s
�0
⎩�2 V = 0 Laplace
In one dimension:
d2 V
dx2
= 0 , so V = mx + b (line).
First Uniqueness Theorem
Given V on boundary S of volume V , Laplace’s equation gives a
unique solution for V in V . ( V does not need to be finite )
.
�2 V1 =
V3 =
0 = �2 V2
V1 − V2
So �2 V3 = 0, and V3 = 0 on boundary
Also no local min/max , so V3 = 0 in V.
Thus V1 = V2
3
Second Uniqueness Theorem
In a volume surrounded by conductors, the total charge on each
conductor determines the E-field uniquely.
Will charges spread out ?
Yes!
Thus case below
no charges - (total on each) stable
solution
Still no charge on each conductor,
so by Second Uniqueness Theorem,
same solution as above