Transcript Document

5) Coulomb’s Law
a) form
F  kq1q2 /r
2
F  kq1q2 /r
2
b) Units

Two possibilities:
- define k and derive q (esu)
- define q and derive k (SI) √
“Define” coulomb (C) as the quantity of charge that
produces a force of 9 x 109 N on objects 1 m apart.
F  kq
9 10
 10 N
k(1C)
1q2 /r/(1m)
9
2 2
 k  9  10 N
9
2
• For practical reasons, the coulomb is defined using
current and magnetism giving
k = 8.988 x 109 Nm2/C2
• Permittivity of free space
1
0 
 8.84 1012 C2 /Nm2
4k
Then

1 q1q2
F
2
4 0 r
c) Fundamental unit of charge
e = 1.602 x 10-19 C
Example: Force between two 1 µC charges
1 mm apart
F = kq1q2/r2 = 9•109(10-6)2/(10-3)2 N = 9000 N
• ~weight of 1000-kg object (1 tonne))
• same as force between two 1-C charges 1 km apart
Example: Coulomb force vs gravity for electrons
m, e
m, e
Fg
FC
FC = ke2/r2
FN = Gm2/r2
Ratio:
19 2
FC
ke
(9 10 )(1.6 10 ) N


2
FN Gm (6.710-11 )(9.11031 )2 N
2
9
 4 10
42
Example: Velocity of an electron in the Bohr Atom
Coulomb force: F = kq1q2/r2 (attractive)
Circular motion requires: F = mv2/r
So,
v2 = kq1q2/mr
For r = 5.29 x 10-11m, v = 2.18 x 106 m/s
d) Superposition of electric forces
Net force is the vector sum of forces from each
charge
F3
q1
q2
q
q3
F2
F1
Net force on q:
F = F1 + F2 + F3
F
6) Electric Field
- abstraction
- separates cause and effect in Coulomb’s law
a) Definition
r
r F
E
q0
Units: N/C
b) Field due to a point charge
q0
F
Q
r
Coulomb’s law:
Electric Field:
r r
E //
F
Qq0
Fk 2
r
Q
E  F /q0  k 2
r
direction is radial
r kQ
E  2 rˆ
r
c) Superposition of electric fields
Net field is the vector sum of fields from each charge
E3
q1
q2
P
q3
E2
E1
Net field at P: E = E1 + E2 + E3
E
Example
D=3m
d
16 µC
q1
Find d to give E = 0 at P
E P  E1  E 2  0
q411
q122
kq
kq
2 
2
d
(D  d)

P
4 µC
q2
E2
 E1  E 2
P
E1
2
2
2(D

d)

d
4(D
d
d  2D or
2
3
D  6m or 2m
7) Electric Field Lines (lines of force)
a) Direction of force on positive charge
radial for point charges
out for positive (begin)
in for negative (end)
b) Number of lines proportional to charge
Q
2Q
c) Begin and end only on charges; never cross
E?
d) Line density proportional to field strength
Line density at radius r:
N
1

2  2
4r
r
Number of lines
area of sphere


Lines of force model <==> inverse-square law
8) Applications of lines-of-force model
a) dipole
b) two positive charges
c) Unequal charges
d) Infinite plane of charge
+
+
+ +
+
+ +
+
+ +
+ q,
+
E
A
By comparison with the
field from a point charge,
we find:
Field is uniform and constant to ∞,
in both directions
Electric field is proportional to the line
density, and therefore to the charge
density, =q/A

E
2 0
e) Parallel plate capacitor (assume separation small compared to the size)
E+
E-
EL=0
+
+
+
+
E+
E-
E=2E+
-
+
-
+
-
• Strong uniform field between:
• Field zero outside
-
E+
E-
ER=0
E   / 0
• Fringing fields near the edges
f) Spherically symmetric charge distribution
• Symmetry ==> radial
• number of lines prop. to charge
+
+
+
+
+
+
+
+
Outside the sphere:
r kq
E  2 rˆ
r
as though all charge concentrated at the
centre (like gravity)
9) Electric Fields and Conductors
• Excess charge resides on surface
at equilibrium
• Field inside is zero at eq’m;
charges move until |E1| = |E2|
E1
E1
E2
• Closed conductor shields external fields
E
E=0
• Field lines
perpendicular at
surface
• Field outside
conducting shell
not shielded
• Field outside grounded
shell is shielded
• Field larger for smaller
radius E = kq/r2
(concentrated at sharp
tips)
Demonstration: Van de
Graaf generator
- purpose: to produce high field by
concentrating charge -- used to
accelerate particles for physics expts
- principle: charge on conductors
moves to the surface